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All questions of Number System for Interview Preparation Exam
After the division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively. What will be the remainder if 84 divides the same number?- a)80
- b)75
- c)41
- d)53
Correct answer is option 'D'. Can you explain this answer?
After the division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively. What will be the remainder if 84 divides the same number?
a)
80
b)
75
c)
41
d)
53
|
Krishna Iyer answered |
Since after division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively, the number is of form ((((4*4)+1)*3)+2)k = 53K.
Let k = 1; the number becomes 53
If it is divided by 84, the remainder is 53.
Hence Option D is correct
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If the difference of (1025−7) and (1024+x) is divisible by 3 then x is equal to :- a)3
- b)2
- c)6
- d)1
Correct answer is option 'B'. Can you explain this answer?
If the difference of (1025−7) and (1024+x) is divisible by 3 then x is equal to :
a)
3
b)
2
c)
6
d)
1
|
Pallavi Datta answered |
We cannot provide an answer to this question as it is incomplete. Please provide the full problem.
Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product?
- a)1050
- b) 540
- c)1440
- d)1590
Correct answer is option 'D'. Can you explain this answer?
Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product?
a)
1050
b)
540
c)
1440
d)
1590
|
Meera Rana answered |
Let us assume that the number with which Anita has to perform the multiplication is 'x'.
Instead of finding 35x, she calculated 53x.
The difference = 53x - 35x = 18x = 540
Therefore, x = 540/18 = 30
So, the new product = 30 x 53 = 1590.
Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all the elements of S. With how many consecutive zeroes will the product end?- a)1
- b)4
- c)5
- d)10
Correct answer is option 'A'. Can you explain this answer?
Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all the elements of S. With how many consecutive zeroes will the product end?
a)
1
b)
4
c)
5
d)
10
|
|
Kavya Saxena answered |
For number of zeroes we must count number of 2 and 5 in prime numbers below 100.
We have just 1 such pair of 2 and 5.
Hence we have only 1 zero.
We have just 1 such pair of 2 and 5.
Hence we have only 1 zero.
Largest 3 digit composite number is- a)990
- b)998
- c)999
- d)995
Correct answer is option 'C'. Can you explain this answer?
Largest 3 digit composite number is
a)
990
b)
998
c)
999
d)
995
|
Aravind Mehta answered |
► 999 is the Largest composite number, it is divisible by 1, 3, 9, 111, 333, 999.
Find the highest power of 24 in 150!
- a)46
- b)47
- c)48
- d)49
Correct answer is option 'C'. Can you explain this answer?
Find the highest power of 24 in 150!
a)
46
b)
47
c)
48
d)
49
|
|
Anaya Patel answered |
24 = 8 × 3
Therefore, we need to find the highest power of 8 and 3 in 150!
8 = 23
Highest power of 8 in 150! is:
Therefore, we need to find the highest power of 8 and 3 in 150!
8 = 23
Highest power of 8 in 150! is:
= [(150 / 2) + (150 / 4) + (150 / 8) + (150 / 16) + (150 / 32) + (150 / 64) +(150 / 128)] / 3
= 48
= 48
Highest power of 3 in 150! is:
= [150 / 3] + [150 / 9] + [150 / 27] + [150 / 81]
= 72
= 72
As the powers of 8 are less, powers of 24 in 150! = 48
Tatto bought a notebook containing 96 pages leaves and numbered them which came to 192 pages. Tappo tore out the latter 25 leaves of the notebook and added the 50 numbers she found on those pages. Which of the following is not true?
- a)She could have found the sum of pages as 1990
- b)She could have found sum of pages as 1275
- c)She could have got sum of pages as 1375
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Tatto bought a notebook containing 96 pages leaves and numbered them which came to 192 pages. Tappo tore out the latter 25 leaves of the notebook and added the 50 numbers she found on those pages. Which of the following is not true?
a)
She could have found the sum of pages as 1990
b)
She could have found sum of pages as 1275
c)
She could have got sum of pages as 1375
d)
None of these
|
|
Tanishq Dey answered |
Information Given:
- The notebook has 96 leaves, which means it has 192 pages (since each leaf has two pages).
- The pages are numbered from 1 to 192.
- Tappo tore out the last 25 leaves of the notebook. Since each leaf has 2 pages, she tore out 50 pages.
Step 1: Determine the page numbers torn out
The last 25 leaves correspond to the last 50 pages in the notebook. Since the total number of pages is 192, the page numbers torn out would be from 143 to 192.
Step 2: Calculate the sum of the torn-out pages
The sum of an arithmetic series (in this case, the page numbers) is given by:
For the torn-out pages from 143 to 192:
- First term (aaa) = 143
- Last term (lll) = 192
- Number of terms (nnn) = 50
So, the sum is:
Step 3: Analyze each option
-
Option 1: She could have found the sum of pages as 1990.
- To find if this is possible, subtract 1990 from the total sum of all pages (1 to 192):
-
- Since the remaining sum does not match with any realistic remaining pages, this option is not possible.
-
Option 2: She could have found the sum of pages as 1275.
- Subtracting 1275 from the total sum: Remaining sum=18528−1275=17253
- The sum is possible and reasonable, so this option is possible.
-
Option 3: She could have found the sum of pages as 1375.
- Subtracting 1375 from the total sum: Remaining sum=18528−1375=17153
- The sum is possible and reasonable, so this option is possible.
Conclusion:
Option 1: She could have found the sum of pages as 1990 is not true because the sum 1990 cannot realistically be the sum of the pages torn out in this context.
Answer: Option 1
In some code, letters a, b, c, d and e represent numbers 2, 4, 5, 6 and 10. We just do not know which letter represents which number. Consider the following relationships:I. a + c = e,
II. b – d = d and
III. e + a = bWhich of the following options are true?- a)b = 4, d = 2
- b)a = 4, e = 6
- c)b = 6, e = 2
- d)a = 4, c = 6
Correct answer is option 'B'. Can you explain this answer?
In some code, letters a, b, c, d and e represent numbers 2, 4, 5, 6 and 10. We just do not know which letter represents which number. Consider the following relationships:
I. a + c = e,
II. b – d = d and
III. e + a = b
II. b – d = d and
III. e + a = b
Which of the following options are true?
a)
b = 4, d = 2
b)
a = 4, e = 6
c)
b = 6, e = 2
d)
a = 4, c = 6
|
|
Aditya Kumar answered |
We have a + c = e so possible summation 6+4=10 or 4+2 = 6.
Also b = 2d so possible values 4 = 2 * 2 or 10 = 5 * 2.
So considering both we have b = 10 , d = 5, a= 4 ,c = 2, e = 6.
Hence the correct option is B .
Also b = 2d so possible values 4 = 2 * 2 or 10 = 5 * 2.
So considering both we have b = 10 , d = 5, a= 4 ,c = 2, e = 6.
Hence the correct option is B .
How many factors of 1080 are perfect squares?- a)6
- b)4
- c)8
- d)12
Correct answer is option 'B'. Can you explain this answer?
How many factors of 1080 are perfect squares?
a)
6
b)
4
c)
8
d)
12
|
Bank Exams India answered |
The factors of 1080 which are perfect square:
1080 → 23 × 33 × 5
For, a number to be a perfect square, all the powers of numbers should be even number.
Power of 2 → 0 or 2
Power of 3 → 0 or 2
Power of 5 → 0
Power of 3 → 0 or 2
Power of 5 → 0
So, the factors which are perfect square are 1, 4, 9, 36.
Hence, Option B is correct.
Hence, Option B is correct.
(16a + 6) will fall under which of the following categories, a is an Integer- a)Odd
- b)Even
- c)Can be Even or Odd
- d)None of the Above
Correct answer is option 'B'. Can you explain this answer?
(16a + 6) will fall under which of the following categories, a is an Integer
a)
Odd
b)
Even
c)
Can be Even or Odd
d)
None of the Above
|
|
Ashutosh Singh answered |
► 16 is even
► Irrespective of whether a is even or odd, Even multiplied by any number = Even
► 6 is even
► (16a + 6) = Even + Even = Even
► Irrespective of whether a is even or odd, Even multiplied by any number = Even
► 6 is even
► (16a + 6) = Even + Even = Even
The sum of the first 100 natural numbers, 1 to 100 is divisible by- a)2, 4 and 8
- b)2 and 4
- c)2
- d)100
Correct answer is option 'C'. Can you explain this answer?
The sum of the first 100 natural numbers, 1 to 100 is divisible by
a)
2, 4 and 8
b)
2 and 4
c)
2
d)
100
|
Machine Experts answered |
The sum of the first 100 natural numbers is:
= (n * (n + 1)) / 2
= (100 * 101) / 2
= 50 * 101
= (100 * 101) / 2
= 50 * 101
101 is an odd number and 50 is divisible by 2.
Hence, 50 * 101 will be divisible by 2.
Hence, 50 * 101 will be divisible by 2.
All the page numbers from a book are added, beginning at page 1.
However, one page number was added twice by mistake. The sum obtained was 1000. Which page number was added twice?
- a)44
- b)45
- c)10
- d)12
Correct answer is option 'C'. Can you explain this answer?
All the page numbers from a book are added, beginning at page 1.
However, one page number was added twice by mistake. The sum obtained was 1000. Which page number was added twice?
a)
44
b)
45
c)
10
d)
12
|
|
Kavya Saxena answered |
The Correct Answer is C: 10
If a three digit number ‘abc’ has 2 factors (where a, b, c are digits), how many factors does the 6-digit number ‘abcabc’ have?
- a)16
- b)24
- c)18
- d)30
Correct answer is option 'A'. Can you explain this answer?
If a three digit number ‘abc’ has 2 factors (where a, b, c are digits), how many factors does the 6-digit number ‘abcabc’ have?
a)
16
b)
24
c)
18
d)
30
|
|
Krishna Iyer answered |
The correct option is A
16
'abc' has 2 factors.
This means 'abc' is a prime number (Only a prime number can have exactly 2 factors).
Now, 'abcabc' = 'abc'×1001
'abcabc' = 'abc' × 7 × 11 × 13
Since 'abc' is prime we can write 'abcabc' as - p1×71×111×131
No. of factors = (1+1) (1+1) (1+1) (1+1) = 16 factors.
In a four-digit number, the sum of the first 2 digits is equal to that of the last 2 digits. The sum of the first and last digits is equal to the third digit. Finally, the sum of the second and fourth digits is twice the sum of the other 2 digits. What is the third digit of the number?
- a)5
- b)8
- c)1
- d)4
Correct answer is option 'A'. Can you explain this answer?
In a four-digit number, the sum of the first 2 digits is equal to that of the last 2 digits. The sum of the first and last digits is equal to the third digit. Finally, the sum of the second and fourth digits is twice the sum of the other 2 digits. What is the third digit of the number?
a)
5
b)
8
c)
1
d)
4
|
|
Arun Sharma answered |
Let the 4 digit no. be xyzw.
According to given conditions we have x + y = z + w, x + w = z, y + w = 2x + 2z.
With help of these equations, we deduce that y = 2w, z = 5x.
Now the minimum value x can take is 1 so z = 5 and the no. is 1854, which satisfies all the conditions. Hence option A.
According to given conditions we have x + y = z + w, x + w = z, y + w = 2x + 2z.
With help of these equations, we deduce that y = 2w, z = 5x.
Now the minimum value x can take is 1 so z = 5 and the no. is 1854, which satisfies all the conditions. Hence option A.
Rohan purchased some pens, pencils and erasers for his young brothers and sisters for the ensuing examinations. He had to buy atleast 11 pieces of each item in a manner that the number of pens purchased be more than the number of pencils, which is more than the number of erasers. He purchased a total of 38 pieces.If the number of pencils cannot be equally divided among his 4 brothers and sisters, how many pens did he purchase?- a)11
- b)12
- c)13
- d)14
Correct answer is option 'D'. Can you explain this answer?
Rohan purchased some pens, pencils and erasers for his young brothers and sisters for the ensuing examinations. He had to buy atleast 11 pieces of each item in a manner that the number of pens purchased be more than the number of pencils, which is more than the number of erasers. He purchased a total of 38 pieces.
If the number of pencils cannot be equally divided among his 4 brothers and sisters, how many pens did he purchase?
a)
11
b)
12
c)
13
d)
14
|
|
Bank Exams India answered |
- Different possibilities for the number of pencils = 12 or 13.
- Since it cannot be divided into his 4 brothers and sisters, it has to be 13.
- The number of erasers should be less than the number of pencils and greater than or equal to 11. So the number of erasers can be 11 or 12.
- If the number of erasers is 12, then the number of pens = 38 - 13 - 12 = 13, which is not possible as the number of pens should be more than the number of pencils.
- So the number of erasers = 11 and therefore the number of pens = 14
Two players A and B are playing a game of putting ‘+’ and '-'signs in between any two integers written from 1 to 100. A starts the game by putting a plus sign anywhere between any two integers. Once all the signs have been put, the result is calculated. If it is even then A wins and if it is odd then B wins, provided they are putting signs by taking turns one by one and either of them can put any sign anywhere between any two integers. Who will win at the end?
- a)A
- b)B
- c)Either A or B
- d)Cannot be determined
Correct answer is option 'A'. Can you explain this answer?
Two players A and B are playing a game of putting ‘+’ and '-'signs in between any two integers written from 1 to 100. A starts the game by putting a plus sign anywhere between any two integers. Once all the signs have been put, the result is calculated. If it is even then A wins and if it is odd then B wins, provided they are putting signs by taking turns one by one and either of them can put any sign anywhere between any two integers. Who will win at the end?
a)
A
b)
B
c)
Either A or B
d)
Cannot be determined
|
|
Meera Rana answered |
Whatever is the sign between two consecutive integers starting from 1 to 100, it will be odd. So, we are getting 50 sets of odd numbers. Now, whatever calculation we do among 50 odd numbers, result will always be even. So, A will win always.
Find the remainder when 4^96 is divided by 6.
a)0
b)2
c)3
d)4
Correct answer is option 'D'. Can you explain this answer?
|
Faizan Khan answered |
496/6, We can write it in this form
(6 - 2)96/6
Now, Remainder will depend only the powers of -2. So,
(-2)96/6, It is same as
([-2]4)24/6, it is same as
(16)24/6
Now,
(16 * 16 * 16 * 16..... 24 times)/6
On dividing individually 16 we always get a remainder 4.
So,
(4 * 4 * 4 * 4............ 24 times)/6.
Hence, Required Remainder = 4.
NOTE: When 4 has even number of powers, it will always give remainder 4 on dividing by 6.
(6 - 2)96/6
Now, Remainder will depend only the powers of -2. So,
(-2)96/6, It is same as
([-2]4)24/6, it is same as
(16)24/6
Now,
(16 * 16 * 16 * 16..... 24 times)/6
On dividing individually 16 we always get a remainder 4.
So,
(4 * 4 * 4 * 4............ 24 times)/6.
Hence, Required Remainder = 4.
NOTE: When 4 has even number of powers, it will always give remainder 4 on dividing by 6.
The last two digits in the multiplication of 122×123×125×127×129 is
- a)20
- b)50
- c)30
- d)40
Correct answer is option 'B'. Can you explain this answer?
The last two digits in the multiplication of 122×123×125×127×129 is
a)
20
b)
50
c)
30
d)
40
|
Malavika Sharma answered |
The last two digits in the multiplication of 122 depend on the number that 122 is multiplied by. Could you please provide more information?
The total number of 3 digit numbers which have two or more consecutive digits identical is:
- a)171
- b)170
- c)90
- d)180
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
The total number of 3 digit numbers which have two or more consecutive digits identical is:
a)
171
b)
170
c)
90
d)
180
e)
None of these
|
|
Faizan Khan answered |
In each set of 100 numbers, there are 10 numbers whose tens digit and unit digit are same. Again in the same set there are 10 numbers whose hundreds and tens digits are same. But one number in each set of 100 numbers whose Hundreds, Tens and Unit digit are same as 111, 222, 333, 444 etc
Hence, there are exactly (10 + 10 - 1) = 19 numbers in each set of 100 numbers. Further there are 9 such sets of numbers
Therefore such total numbers = 19 × 9 = 171
Alternatively,
9 × 10 × 10 - 9 × 9 × 9 = 900 - 729 = 171
Hence, there are exactly (10 + 10 - 1) = 19 numbers in each set of 100 numbers. Further there are 9 such sets of numbers
Therefore such total numbers = 19 × 9 = 171
Alternatively,
9 × 10 × 10 - 9 × 9 × 9 = 900 - 729 = 171
The integers 34041 and 32506,when divided by a three digit integer N, leave the same remainder. What can be the value of N?- a)289
- b)307
- c)317
- d)319
Correct answer is option 'B'. Can you explain this answer?
The integers 34041 and 32506,when divided by a three digit integer N, leave the same remainder. What can be the value of N?
a)
289
b)
307
c)
317
d)
319
|
Ishani Rane answered |
Let the common remainder be x. Then numbers (34041 – x) and (32506 – x) would be completely divisible by n. Hence the difference of the numbers (34041 – x) and (32506 – x) will also be divisible by n or (34041 – x – 32506 + x) = 1535 will also be divisible by n. Now, using options we find that 1535 is divisible by 307.
When a number is successively divided by 7,5 and 4, it leaves remainders of 4,2 and 3 respectively. What will be the respective remainders when the smallest such number is successively divided by 8,5 and 6 ?
- a)3,0,3
- b)2,2,4
- c)5,0,3
- d)2,4,2
Correct answer is option 'A'. Can you explain this answer?
When a number is successively divided by 7,5 and 4, it leaves remainders of 4,2 and 3 respectively. What will be the respective remainders when the smallest such number is successively divided by 8,5 and 6 ?
a)
3,0,3
b)
2,2,4
c)
5,0,3
d)
2,4,2
|
Preeti Khanna answered |
The number would be in the form of (7X+4) as when this number is divide by 7, will give remainder 4.
Now, we will try hit and trial method to obtained the number.
Put, X=17, then
7X+4=7×17+4=119+4=123
Now, when 123 divided by 7, gives quotient 17 , remainder =4
17 divided by 5, quotient =3, remainder =2
3 divide by 4 gives remainder 3.
So for first condition satisfied.
Now, we will try hit and trial method to obtained the number.
Put, X=17, then
7X+4=7×17+4=119+4=123
Now, when 123 divided by 7, gives quotient 17 , remainder =4
17 divided by 5, quotient =3, remainder =2
3 divide by 4 gives remainder 3.
So for first condition satisfied.
Now, 123 divided by 8, quotient =15, remainder =3
15 divided by 5, quotient =3, remainder =0
3 divided by 6, remainder =3.
15 divided by 5, quotient =3, remainder =0
3 divided by 6, remainder =3.
Three distinct prime numbers, less than 10 are taken and all the numbers that can be formed by arranging all the digits are taken. Now, difference between the largest and the smallest number formed is equal to 495. It is also given that sum of the digits is more than 13. What is the product of the numbers?
- a)30
- b)70
- c)105
- d)315
Correct answer is option 'B'. Can you explain this answer?
Three distinct prime numbers, less than 10 are taken and all the numbers that can be formed by arranging all the digits are taken. Now, difference between the largest and the smallest number formed is equal to 495. It is also given that sum of the digits is more than 13. What is the product of the numbers?
a)
30
b)
70
c)
105
d)
315
|
Pritam Saha answered |
Prime numbers less than 10 = 2, 3, 5, 7.
If the difference between the largest and the smallest number is ending in 5, the prime numbers in the end position have to be 7 and 2.
The smallest and largest numbers are of form 2_7 and 7_2
Since it is given that the sum of the digits is >13, x will be 5.
Verifying, 752-257 = 495. Answer is option (b).
Since it is given that the sum of the digits is >13, x will be 5.
Verifying, 752-257 = 495. Answer is option (b).
as 7*5*2 = 70
Convert 
in p/q form- a)2095/99
- b)2116/99
- c)2116/100
- d)2195/100
Correct answer is option 'A'. Can you explain this answer?
Convert in p/q form
in p/q forma)
2095/99
b)
2116/99
c)
2116/100
d)
2195/100
|
|
Aryan Kapoor answered |
► All the digits written once = 2116
► All the digits without Bar written once = 21
► No of digits with bar after decimal = 2
► No of digits without bar after decimal = 0
► Rational form =
► All the digits without Bar written once = 21
► No of digits with bar after decimal = 2
► No of digits without bar after decimal = 0
► Rational form =
A nursery has 363, 429 and 693 plants respectively of 3 distinct varieties. It is desired to place these plants in straight rows of plants of 1 variety only so that the number of rows required is the minimum. What is the size of each row and how many rows would be required? - a)33 and 45
- b)37 and 48
- c)41 and 56
- d)45 and 55
Correct answer is option 'A'. Can you explain this answer?
A nursery has 363, 429 and 693 plants respectively of 3 distinct varieties. It is desired to place these plants in straight rows of plants of 1 variety only so that the number of rows required is the minimum. What is the size of each row and how many rows would be required?
a)
33 and 45
b)
37 and 48
c)
41 and 56
d)
45 and 55
|
Debanshi Chakraborty answered |
Solution:
To find the size of each row and the number of rows required, we need to find the HCF (highest common factor) of the given numbers.
1. Find the prime factors of the given numbers:
- 363 = 3 x 11 x 11
- 429 = 3 x 11 x 13
- 693 = 3 x 3 x 7 x 11
2. Identify the common factors of the given numbers:
- The common factor is 3 x 11 = 33
3. Divide each number by the common factor:
- 363 ÷ 33 = 11
- 429 ÷ 33 = 13
- 693 ÷ 33 = 21
The size of each row is 33 plants and the number of rows required are 11, 13, and 21 for the three varieties respectively.
Therefore, the correct answer is option A: 33 and 45.
To find the size of each row and the number of rows required, we need to find the HCF (highest common factor) of the given numbers.
1. Find the prime factors of the given numbers:
- 363 = 3 x 11 x 11
- 429 = 3 x 11 x 13
- 693 = 3 x 3 x 7 x 11
2. Identify the common factors of the given numbers:
- The common factor is 3 x 11 = 33
3. Divide each number by the common factor:
- 363 ÷ 33 = 11
- 429 ÷ 33 = 13
- 693 ÷ 33 = 21
The size of each row is 33 plants and the number of rows required are 11, 13, and 21 for the three varieties respectively.
Therefore, the correct answer is option A: 33 and 45.
What is the least number of soldiers that can be drawn up in troops of 12, 15, 18 and 20 soldiers and also in form of a solid square?
a)900
b)400
c)1600
d)2500
Correct answer is option 'A'. Can you explain this answer?
|
|
Hridoy Mehra answered |
In this type of question, We need to find out the LCM of the given numbers.
LCM of 12, 15, 18 and 20;
12 = 2*2*3;
15 = 3*5;
18 = 2*3*3;
20 = 2*2*5;
Hence, LCM = 2*2*3*5*3
Since, the soldiers are in the form of a solid square.
Hence, LCM must be a perfect square. To make the LCM a perfect square, We have to multiply it by 5, hence, the required number of soldiers = 2*2*3*3*5*5 = 900.
LCM of 12, 15, 18 and 20;
12 = 2*2*3;
15 = 3*5;
18 = 2*3*3;
20 = 2*2*5;
Hence, LCM = 2*2*3*5*3
Since, the soldiers are in the form of a solid square.
Hence, LCM must be a perfect square. To make the LCM a perfect square, We have to multiply it by 5, hence, the required number of soldiers = 2*2*3*3*5*5 = 900.
The remainder when 1010+10100+101000+....+1010000000000 is divided by 7 is- a)1
- b)2
- c)5
- d)7
Correct answer is option 'C'. Can you explain this answer?
The remainder when 1010+10100+101000+....+1010000000000 is divided by 7 is
a)
1
b)
2
c)
5
d)
7
|
|
Anjana Kaur answered |
Solution:
The given number is 1010 10100 101000 .... 1010000000000.
We observe that the number is formed by concatenating the numbers 10, 100, 1000, ... up to 1000000000000.
Let S be the sum of these numbers: S = 10 + 100 + 1000 + ... + 1000000000000.
We can write S as follows:
S = 10(1 + 10 + 100 + ... + 100000000000) = 10(111111111111) = 1111111111110.
Now, let's consider the remainders when the powers of 10 are divided by 7:
10^1 ≡ 3 (mod 7)
10^2 ≡ 2 (mod 7)
10^3 ≡ 6 (mod 7)
10^4 ≡ 4 (mod 7)
10^5 ≡ 5 (mod 7)
10^6 ≡ 1 (mod 7)
10^7 ≡ 3 (mod 7)
10^8 ≡ 2 (mod 7)
10^9 ≡ 6 (mod 7)
10^10 ≡ 4 (mod 7)
10^11 ≡ 5 (mod 7)
10^12 ≡ 1 (mod 7)
We observe that the remainders repeat in cycles of length 6.
Therefore, we can write the remainder when S is divided by 7 as follows:
S ≡ (3 + 2 + 6 + 4 + 5 + 1) + (3 + 2 + 6 + 4 + 5 + 1) + ... (mod 7)
There are 11 numbers in the sum, so we have:
S ≡ 11(21) ≡ 2 (mod 7)
Therefore, the remainder when 1010 10100 101000 .... 1010000000000 is divided by 7 is 5.
Hence, the correct answer is option (c) 5.
The given number is 1010 10100 101000 .... 1010000000000.
We observe that the number is formed by concatenating the numbers 10, 100, 1000, ... up to 1000000000000.
Let S be the sum of these numbers: S = 10 + 100 + 1000 + ... + 1000000000000.
We can write S as follows:
S = 10(1 + 10 + 100 + ... + 100000000000) = 10(111111111111) = 1111111111110.
Now, let's consider the remainders when the powers of 10 are divided by 7:
10^1 ≡ 3 (mod 7)
10^2 ≡ 2 (mod 7)
10^3 ≡ 6 (mod 7)
10^4 ≡ 4 (mod 7)
10^5 ≡ 5 (mod 7)
10^6 ≡ 1 (mod 7)
10^7 ≡ 3 (mod 7)
10^8 ≡ 2 (mod 7)
10^9 ≡ 6 (mod 7)
10^10 ≡ 4 (mod 7)
10^11 ≡ 5 (mod 7)
10^12 ≡ 1 (mod 7)
We observe that the remainders repeat in cycles of length 6.
Therefore, we can write the remainder when S is divided by 7 as follows:
S ≡ (3 + 2 + 6 + 4 + 5 + 1) + (3 + 2 + 6 + 4 + 5 + 1) + ... (mod 7)
There are 11 numbers in the sum, so we have:
S ≡ 11(21) ≡ 2 (mod 7)
Therefore, the remainder when 1010 10100 101000 .... 1010000000000 is divided by 7 is 5.
Hence, the correct answer is option (c) 5.
The integers 34041 and 32506 when divided by a three-digit integer n leave the same remainder. What is n?
- a)289
- b)367
- c)453
- d)307
Correct answer is option 'D'. Can you explain this answer?
The integers 34041 and 32506 when divided by a three-digit integer n leave the same remainder. What is n?
a)
289
b)
367
c)
453
d)
307
|
|
Arun Sharma answered |
Let the common remainder be x.
32506 – x is divisible by n.
34041 – x is divisible by n.
Difference of (32506 – x) and (34041 – x) = (32506 – x) – (34041 – x)
⇒ 32506 – x – 34041 + x
⇒ 32506 – 34041
⇒ 1535
Factors of 1535 = 1 × 5 × 307 × 1535
3-digit number = 307
⇒ n = 307
∴ The value of n is 307.
16 students were writing a test in a class. Rahul made 14 mistakes in the paper, which was the highest number of mistakes made by any student. Which of the following statements is definitely true?
- a)At least two students made the same number of mistakes
- b)Exactly two students made the same number of mistakes
- c)At most two students made the same number of mistakes
- d)All students made different number of mistakes.
Correct answer is option 'A'. Can you explain this answer?
16 students were writing a test in a class. Rahul made 14 mistakes in the paper, which was the highest number of mistakes made by any student. Which of the following statements is definitely true?
a)
At least two students made the same number of mistakes
b)
Exactly two students made the same number of mistakes
c)
At most two students made the same number of mistakes
d)
All students made different number of mistakes.
|
|
Alok Verma answered |
The number of mistakes made by all the students will be between 0 and 14, i.e., students are having a total of 15 options to make mistakes. Since the number of students = 16, at least two students will have the same number of mistakes (that can be zero also, i.e., two students are making no mistakes). Hence, option 1 is the answer.
A number when divided by 703 gives a remainder of 75. What will be the remainder when we divide the same number by 37?- a)1
- b)2
- c)5
- d)7
Correct answer is option 'A'. Can you explain this answer?
A number when divided by 703 gives a remainder of 75. What will be the remainder when we divide the same number by 37?
a)
1
b)
2
c)
5
d)
7
|
|
Preeti Khanna answered |
Let the number be N and its quotient be k.
Then the number N can be written in the form of:
N = 703k + 75
Now, we have to find out the what will be the remainder when it is divided by 37.
The number is (703k + 75)
Let’s divide it by 37
(703k + 75)/ 37
703 is divisible by 37 hence, remainder will be 0 whereas, 75 when divided by 37 leaves remainder 1.
Therefore, the remainder when the number N is divided by 37 will be (0+1) i.e. 1.
Then the number N can be written in the form of:
N = 703k + 75
Now, we have to find out the what will be the remainder when it is divided by 37.
The number is (703k + 75)
Let’s divide it by 37
(703k + 75)/ 37
703 is divisible by 37 hence, remainder will be 0 whereas, 75 when divided by 37 leaves remainder 1.
Therefore, the remainder when the number N is divided by 37 will be (0+1) i.e. 1.
Find the unit digit:
(17) (19) (13)- a)2
- b)3
- c)7
- d)9
Correct answer is option 'B'. Can you explain this answer?
Find the unit digit:
(17) (19) (13)
(17) (19) (13)
a)
2
b)
3
c)
7
d)
9
|
|
Anaya Patel answered |
17 is raised to the power of 19 and 19 is raised to the power of 13.
To find the last digit of the number of this kind we will start with the base, and the base here is 17.
To get the unit digit of a number our only concern is the digit at the unit place i.e.7.
The cyclicity of 7 is 4.
Dividing 1913 by 4.
Remainder will be 3.
7 raised to power 3 (73), the unit digit of this number will be 3.
To find the last digit of the number of this kind we will start with the base, and the base here is 17.
To get the unit digit of a number our only concern is the digit at the unit place i.e.7.
The cyclicity of 7 is 4.
Dividing 1913 by 4.
Remainder will be 3.
7 raised to power 3 (73), the unit digit of this number will be 3.
Find the unit digit:
346 765 * 768 983 * 987 599
- a)3
- b)5
- c)6
- d)9
Correct answer is option 'C'. Can you explain this answer?
Find the unit digit:
346 765 * 768 983 * 987 599
346 765 * 768 983 * 987 599
a)
3
b)
5
c)
6
d)
9
|
|
Anaya Patel answered |
In this type of problem
Step 1: we find the unit digit of each term
Step 2: we find the product of the unit digits of each term
Step 3: The unit digit of the product will be the product of whole number
The unit digit of 346 765 = 6
The unit digit of 768 983 = 2 (for unit digit remainder of (power)/4 is checked and periodicity is checked as per base no ) like r(remainder)=983/4 is 3 so 83 unit digit is 2
The unit digit of 987599 = 3 (for unit digit remainder of (power)/4 is checked and periodicity is checked as per base no ) like r(remainder)=983/4 is 3 so 73 unit digit is 3
6 * 2 * 3 = 36
Hence, the unit digit is 6.
Step 1: we find the unit digit of each term
Step 2: we find the product of the unit digits of each term
Step 3: The unit digit of the product will be the product of whole number
The unit digit of 346 765 = 6
The unit digit of 768 983 = 2 (for unit digit remainder of (power)/4 is checked and periodicity is checked as per base no ) like r(remainder)=983/4 is 3 so 83 unit digit is 2
The unit digit of 987599 = 3 (for unit digit remainder of (power)/4 is checked and periodicity is checked as per base no ) like r(remainder)=983/4 is 3 so 73 unit digit is 3
6 * 2 * 3 = 36
Hence, the unit digit is 6.
1 ’s are given 100 times, 2 ’s are given 100 times and 3’s are given 100 times. Now numbers are made by arranging these 300 digits in all possible ways. How many of these numbers will be perfect squares?
- a)10101
- b)1001001
- c)100010001
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
1 ’s are given 100 times, 2 ’s are given 100 times and 3’s are given 100 times. Now numbers are made by arranging these 300 digits in all possible ways. How many of these numbers will be perfect squares?
a)
10101
b)
1001001
c)
100010001
d)
None of these
|
Dhruv Mehra answered |
Solve this question step by step:
- Any number formed by this method is clearly divisible by 3.
- Since it needs to be a square, it should be divisible by (3)[2*k]. k varies over the natural numbers.
- Now consider the original number. It has hundred 1’s, hundred 2’s and hundred 3’s. Sum of these digits is 600. This is not divisible by 9. Hence number is not divisible by 9.
- If a number is divisible by (3)[2*k], it is divisible by 3k.
- This number is not divisible by 3k for any k > 1.
Hence it is not a perfect square for any arrangement.
The sum of first five, 3 digit prime numbers is:- a)547
- b)533
- c)529
- d)555
Correct answer is option 'B'. Can you explain this answer?
The sum of first five, 3 digit prime numbers is:
a)
547
b)
533
c)
529
d)
555
|
|
Krishna Iyer answered |
► First 5, 3 digit prime numbers are listed below:
► Their Sum is 533
► Their Sum is 533
The sum of the factorials of the three-digits of a 3-digit number is equal to the three-digit number formed by these three digits, taken in the same order. Which of the following is true of the number of such three-digit numbers, if no digit occurs more than once?- a)No such number exists
- b)Exactly one such number exists
- c)There is more than one such number, but they are finite in number
- d)There are infinite such numbers
Correct answer is 'B'. Can you explain this answer?
The sum of the factorials of the three-digits of a 3-digit number is equal to the three-digit number formed by these three digits, taken in the same order. Which of the following is true of the number of such three-digit numbers, if no digit occurs more than once?
a)
No such number exists
b)
Exactly one such number exists
c)
There is more than one such number, but they are finite in number
d)
There are infinite such numbers
|
Sagar Bavliya answered |
How??? Please explain this concept dear
Srini wrote his class 10th board examination this year. When the result came out he searched for his hall ticket to see his roll number but could not trace it. He could remember only the first three digits of the 6 digit number as 267. His father, however, remembered that the number was divisible by 11. His mother gave the information that the number was also divisible by 13. They tried to recollect the number when all of a sudden Srini told that the number was a multiple of 7. What was the unit digits of the number?- a)5
- b)7
- c)2
- d)Cannot be determined
Correct answer is option 'B'. Can you explain this answer?
Srini wrote his class 10th board examination this year. When the result came out he searched for his hall ticket to see his roll number but could not trace it. He could remember only the first three digits of the 6 digit number as 267. His father, however, remembered that the number was divisible by 11. His mother gave the information that the number was also divisible by 13. They tried to recollect the number when all of a sudden Srini told that the number was a multiple of 7. What was the unit digits of the number?
a)
5
b)
7
c)
2
d)
Cannot be determined
|
Manoj Ghosh answered |
His roll no. is divisible by 1001 (13x11x7)
and 1st three digit are 267.
Hence, the Last three digits will also be 267.
What is the remainder when (103 + 93)752 is divided by 123?- a)10
- b)729
- c)752
- d)1000
Correct answer is option 'A'. Can you explain this answer?
What is the remainder when (103 + 93)752 is divided by 123?
a)
10
b)
729
c)
752
d)
1000
|
|
Palak Dasgupta answered |
A remainder can never be greater than the no. which is the dividing factor.
hence, the remainder<123
which leaves us with only one option, i.e. (a)
Find the gcd (111....11 hundred ones; 11....11 sixty ones).- a)111....forty ones
- b)111.....twenty five ones
- c) 111..... twenty ones
- d)111.... sixty ones
Correct answer is option 'C'. Can you explain this answer?
Find the gcd (111....11 hundred ones; 11....11 sixty ones).
a)
111....forty ones
b)
111.....twenty five ones
c)
111..... twenty ones
d)
111.... sixty ones
|
Shanaya Yadav answered |
Solution:
To solve this problem we need to use the concept of divisibility rules.
Divisibility rule of 11 states that if the difference between the sum of digits at odd places and the sum of digits at even places is either 0 or a multiple of 11, then the number is divisible by 11.
In this case, both the numbers have only one digit repeated multiple times. Therefore, the sum of digits at odd places and even places will be the same for both numbers.
Hence, we can conclude that the gcd of both numbers will be the smaller number.
a) 111....forty ones
The number of ones is even. Therefore, the sum of digits at odd places and even places will be the same. Hence, the number is divisible by 11.
b) 111.....twenty five ones
The number of ones is odd. Therefore, the sum of digits at odd places and even places will be different. Hence, the number is not divisible by 11.
c) 111..... twenty ones
The number of ones is even. Therefore, the sum of digits at odd places and even places will be the same. Hence, the number is divisible by 11.
d) 111.... sixty ones
The number of ones is even. Therefore, the sum of digits at odd places and even places will be the same. Hence, the number is divisible by 11.
Hence, the gcd of 111....20 ones and 111....60 ones is 111....20 ones.
To solve this problem we need to use the concept of divisibility rules.
Divisibility rule of 11 states that if the difference between the sum of digits at odd places and the sum of digits at even places is either 0 or a multiple of 11, then the number is divisible by 11.
In this case, both the numbers have only one digit repeated multiple times. Therefore, the sum of digits at odd places and even places will be the same for both numbers.
Hence, we can conclude that the gcd of both numbers will be the smaller number.
a) 111....forty ones
The number of ones is even. Therefore, the sum of digits at odd places and even places will be the same. Hence, the number is divisible by 11.
b) 111.....twenty five ones
The number of ones is odd. Therefore, the sum of digits at odd places and even places will be different. Hence, the number is not divisible by 11.
c) 111..... twenty ones
The number of ones is even. Therefore, the sum of digits at odd places and even places will be the same. Hence, the number is divisible by 11.
d) 111.... sixty ones
The number of ones is even. Therefore, the sum of digits at odd places and even places will be the same. Hence, the number is divisible by 11.
Hence, the gcd of 111....20 ones and 111....60 ones is 111....20 ones.
If (6a + 12) is odd then 'a' would be, a is an Integer- a)Odd
- b)Even
- c)Cannot be determined
- d)Inconsistent information
Correct answer is option 'D'. Can you explain this answer?
If (6a + 12) is odd then 'a' would be, a is an Integer
a)
Odd
b)
Even
c)
Cannot be determined
d)
Inconsistent information
|
|
Aarav Chavan answered |
► There are two possibilities for a
► Case 1 : a is even
• 6a = Even x Even = Even
• 6a + 12 = Even + Even = Even
► Case 1 : a is even
• 6a = Even x Even = Even
• 6a + 12 = Even + Even = Even
► Case 2 : a is odd
• 6a = Even x Odd = Even
• 6a + 12 = Even + Even = Even
• 6a = Even x Odd = Even
• 6a + 12 = Even + Even = Even
► In both the cases the answer is even, so the information provided is inconsistent.
What will be remainder when 1212121212... 300 times, is being divided by 99 ?- a)18
- b)81
- c)54
- d)36
Correct answer is option 'A'. Can you explain this answer?
What will be remainder when 1212121212... 300 times, is being divided by 99 ?
a)
18
b)
81
c)
54
d)
36
|
Shalini Kumar answered |
This number 1212121212... 300 times is divisible by 9. So, we can write 1212121212...300 times = 9 N, where N is the quotient obtained when divided by 9. Now this question is like -
Now we will have to find the reminder obtained when 134680134680.. . 50 times is divided by 11.
For this, we are supposed to use the divisibility rule of 11 from right hand side. [Using the divisibility rule from left hand side might give us the wrong remainder, like if we find out the remainder obtained when 12 is divided by 11, remainder = 1 = (2-1)≠(1 - 2)]
For this, we are supposed to use the divisibility rule of 11 from right hand side. [Using the divisibility rule from left hand side might give us the wrong remainder, like if we find out the remainder obtained when 12 is divided by 11, remainder = 1 = (2-1)≠(1 - 2)]
Alternatively, divisibility rule of 10" - 1, n = 2 can be used to find the remainder in this case.
The History teacher was referring to a year in the 19th century. Rohan found an easy way to remember the year. He found that the number, when viewed in a mirror, increased 4.5 times. Which year was the teacher referring to?- a)1,801
- b)1,810
- c)1,818
- d)More than one value
Correct answer is option 'C'. Can you explain this answer?
The History teacher was referring to a year in the 19th century. Rohan found an easy way to remember the year. He found that the number, when viewed in a mirror, increased 4.5 times. Which year was the teacher referring to?
a)
1,801
b)
1,810
c)
1,818
d)
More than one value
|
|
Ravi Singh answered |
Going through the options, 8181/1818 = 4.5
Find the unit digit:
(76476756749)8754874878
- a)1
- b)2
- c)3
- d)4
Correct answer is option 'A'. Can you explain this answer?
Find the unit digit:
(76476756749)8754874878
(76476756749)8754874878
a)
1
b)
2
c)
3
d)
4
|
|
Preeti Khanna answered |
Correct Answer :- a
Explanation : The unit digit of the number will depend on the last digit.
As we know that 91 = 9
92 = 81
93 = 729
94 = 6561
The unit digit of the number is 1 and 9, from the options we can pick the answer
Hence option a) is correct
a, b, and c are 3 single-digit numbers such that 0.abcabcabc..... = 51/111.
Find the value of a + b + c. - a)10
- b)18
- c)27
- d)9
Correct answer is option 'B'. Can you explain this answer?
a, b, and c are 3 single-digit numbers such that 0.abcabcabc..... = 51/111.
Find the value of a + b + c.
Find the value of a + b + c.
a)
10
b)
18
c)
27
d)
9
|
|
Saptarshi Saini answered |
► 0.abcabc.....is a non terminating recurring decimal number
► so as per the formula we discussed
• 0.abcabc .... = abc/999 = 51/111,
• abc = 51 x 9 = 459
► Since a, b and c are individual digits , a = 4; b = 5; and c = 9
• a + b + c = 18
► so as per the formula we discussed
• 0.abcabc .... = abc/999 = 51/111,
• abc = 51 x 9 = 459
► Since a, b and c are individual digits , a = 4; b = 5; and c = 9
• a + b + c = 18
What would be the greatest number that divides 14, 20, and 32 leaving the same remainder?
- a)3
- b)6
- c)12
- d)14
Correct answer is option 'B'. Can you explain this answer?
What would be the greatest number that divides 14, 20, and 32 leaving the same remainder?
a)
3
b)
6
c)
12
d)
14
|
Yash Patel answered |
Here, the number which divides 14, 20, and 32 leaves the same remainder.
∴ We will be using HCF model 2
The required number will be the HCF of (20 - 14), (32 - 20), and (32 - 14).
i.e. HCF (6, 12, 18)
which will be 6.
Therefore, the required number is 6.
∴ We will be using HCF model 2
The required number will be the HCF of (20 - 14), (32 - 20), and (32 - 14).
i.e. HCF (6, 12, 18)
which will be 6.
Therefore, the required number is 6.
What is the difference between the largest four digit number and the smallest four digit number, both written in hexadecimal system?- a)FFFF
- b)EFFF
- c)EEEE
- d)EFEF
Correct answer is option 'A'. Can you explain this answer?
What is the difference between the largest four digit number and the smallest four digit number, both written in hexadecimal system?
a)
FFFF
b)
EFFF
c)
EEEE
d)
EFEF
|
Ramit Mitra answered |
For hexa, we have 0-9 & A-F,
hence largest 4 digit number is FFFF, smallest 4 digit is 1000, when we substract,
FFFF - 1000 = EFFF
So, I think, the question has a bug and the answer should be b) EFFF
hence largest 4 digit number is FFFF, smallest 4 digit is 1000, when we substract,
FFFF - 1000 = EFFF
So, I think, the question has a bug and the answer should be b) EFFF
A rectangular floor is fully covered with square tiles of identical size. The tiles on the edges are white and the tiles in the interior are red. The number of white tiles is the same as the number of red tiles. A possible value of the number of tiles along one edge of the floor is :- a)10
- b)16
- c)14
- d)12
Correct answer is option 'D'. Can you explain this answer?
A rectangular floor is fully covered with square tiles of identical size. The tiles on the edges are white and the tiles in the interior are red. The number of white tiles is the same as the number of red tiles. A possible value of the number of tiles along one edge of the floor is :
a)
10
b)
16
c)
14
d)
12
|
UPSC Achievers answered |
Let the rectangle has x and y tiles along its length and breadth respectively.
The number of white tiles
W = 2x + 2(y – 2) = 2 (x + y – 2)
And the number of red tiles = R = xy – 2 (x + y – 2)
Given that the number of white tiles is the same as the number of red tiles
⇒ 2 (x + y – 2) = xy – 2 (x + y – 2)
⇒ 4 (x + y – 2) = xy
⇒ xy – 4x – 4y = –8
⇒ (x – 4) (y – 4) = 8 = 8 �1 or 4 � 2
⇒ m – 4 = 8 or 4
⇒m = 12 or 8
Therefore, the number of tiles along one edge of the floor can be 12
The number of white tiles
W = 2x + 2(y – 2) = 2 (x + y – 2)
And the number of red tiles = R = xy – 2 (x + y – 2)
Given that the number of white tiles is the same as the number of red tiles
⇒ 2 (x + y – 2) = xy – 2 (x + y – 2)
⇒ 4 (x + y – 2) = xy
⇒ xy – 4x – 4y = –8
⇒ (x – 4) (y – 4) = 8 = 8 �1 or 4 � 2
⇒ m – 4 = 8 or 4
⇒m = 12 or 8
Therefore, the number of tiles along one edge of the floor can be 12
Hence Option D is correct
For Complete syllabus of Quant for CAT click on the link given below:
Which of the following is a Real Number but not a Rational Number?- a)−√4
- b)−√3
- c)√−4
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Which of the following is a Real Number but not a Rational Number?
a)
−√4
b)
−√3
c)
√−4
d)
None of these
|
Ishita Reddy answered |
(a) −√4 = −2 is a Real Number as well as a Rational Number
(b) −√3 is a Real number and an irrational number
(c) √−4 is an imaginary number
(b) −√3 is a Real number and an irrational number
(c) √−4 is an imaginary number
Which is the largest Even Prime number?- a)109+ 2
- b)2
- c)Cannot be determined.
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
Which is the largest Even Prime number?
a)
109+ 2
b)
2
c)
Cannot be determined.
d)
None of the above
|
Rajat Mehra answered |
Understanding Prime Numbers
Prime numbers are integers greater than 1 that have no positive divisors other than 1 and themselves. This means a prime number can only be divided evenly by 1 and the number itself.
Identifying Even Prime Numbers
- An even number is any integer that is divisible by 2.
- The only even prime number is 2, as it can only be divided by 1 and 2 itself.
- All other even numbers (like 4, 6, 8, etc.) can be divided by 2, making them non-prime.
Evaluating the Options
- Option A: 109 + 2
- This equals 111, which is not a prime number because it can be divided by 3 and 37.
- Option B: 2
- As discussed, 2 is the only even prime number.
- Option C: Cannot be determined
- This option is incorrect as we can definitively determine the largest even prime.
- Option D: None of the above
- This also does not apply as option B is correct.
Conclusion
In summary, the correct answer to the question of the largest even prime number is option B: 2. It is the only even prime number, as all other even numbers are composite. Thus, option B is the only valid choice among those listed.
Prime numbers are integers greater than 1 that have no positive divisors other than 1 and themselves. This means a prime number can only be divided evenly by 1 and the number itself.
Identifying Even Prime Numbers
- An even number is any integer that is divisible by 2.
- The only even prime number is 2, as it can only be divided by 1 and 2 itself.
- All other even numbers (like 4, 6, 8, etc.) can be divided by 2, making them non-prime.
Evaluating the Options
- Option A: 109 + 2
- This equals 111, which is not a prime number because it can be divided by 3 and 37.
- Option B: 2
- As discussed, 2 is the only even prime number.
- Option C: Cannot be determined
- This option is incorrect as we can definitively determine the largest even prime.
- Option D: None of the above
- This also does not apply as option B is correct.
Conclusion
In summary, the correct answer to the question of the largest even prime number is option B: 2. It is the only even prime number, as all other even numbers are composite. Thus, option B is the only valid choice among those listed.
How many different four digit numbers are there in the octal (Base 8) system, expressed in that system?- a)3584
- b)2058
- c)6000
- d)7000
Correct answer is option 'D'. Can you explain this answer?
How many different four digit numbers are there in the octal (Base 8) system, expressed in that system?
a)
3584
b)
2058
c)
6000
d)
7000
|
Pallabi Kaur answered |
The total number of numbers of four digits in octal system = 7 x 8 x 8 x 8 = 3584 When we convert this number into octal system, this is equal to 7000.
How many divisors of 105 will have at least one zero at its end?- a)9
- b)12
- c)15
- d)25
Correct answer is option 'D'. Can you explain this answer?
How many divisors of 105 will have at least one zero at its end?
a)
9
b)
12
c)
15
d)
25
|
Prashanth Sachin answered |
FIRST SIMPLY FIND TOTAL NUMBER OF FACTORS OF 10 POWER 5
10^5 can be written as 2^5 × 5^5
(5+1) × (5+1)
6×6 = 36
TOTAL 36 FACTORS
from this we have to subtract the factors which containing zero
36/5 = 7
36/2 = 18
18 - 7 = 11 ( it means there are 11 factors which containing zero)
simply subtract 36-11 is equal to 25
so ANS = 25
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