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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
For Complete syllabus of Quant for CAT click on the link given below:
A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. If both lights start flashing at the same time, how many times do they flash together in each hour?
- a)30
- b)24
- c)20
- d)60
Correct answer is option 'A'. Can you explain this answer?
A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. If both lights start flashing at the same time, how many times do they flash together in each hour?
a)
30
b)
24
c)
20
d)
60
|
Arun Sharma answered |
A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. So red light fashes after every 1/3 min and green light flashes every 2/5 min. LCM of both the fractions is 2 min.
Hence, they flash together after every 2 min. So in an hour they flash together 30 times.
Hence, they flash together after every 2 min. So in an hour they flash together 30 times.
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.
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
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.
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 .
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
|
Preeti Khanna 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.
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.
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
Write three rational numbers between 4 and 5?- a)12 / 6, 13 / 6, 14 / 6
- b)12 / 7, 13 / 7, 14 / 7
- c)17 / 4, 18 / 4, 19 / 4
- d)17 / 2, 18 / 13, 19 / 23
Correct answer is option 'C'. Can you explain this answer?
Write three rational numbers between 4 and 5?
a)
12 / 6, 13 / 6, 14 / 6
b)
12 / 7, 13 / 7, 14 / 7
c)
17 / 4, 18 / 4, 19 / 4
d)
17 / 2, 18 / 13, 19 / 23
|
|
Arun Sharma answered |
There are several rational numbers between 4 and 5.
The numbers are between 16/4 and 20/4.
Therefore, the answer is c, that is, 17 / 4, 18 / 4, 19 / 4.
Hence Option C is correct
For Complete syllabus of Quant for CAT click on the link given below:
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.
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.
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 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.
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.
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.
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
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.
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.
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.
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.
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.
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.
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
Find the remainder when 73 *75 *78 *57 *197 *37 is divided by 34.- a)32
- b)30
- c)15
- d)28
Correct answer is option 'A'. Can you explain this answer?
Find the remainder when 73 *75 *78 *57 *197 *37 is divided by 34.
a)
32
b)
30
c)
15
d)
28
|
|
Ishani Rane answered |
Remainder,
(73 *75 *78 *57 *197 *37)/34 ===> (5 *7 *10 *23 *27 *3)/34
[We have taken individual remainder, which means if 73 is divided by 34 individually, it will give remainder 5, 75 divided 34 gives remainder 7 and so on.]
(5 *7 *10 *23 *27 *3)/34 ===> (35 *30 *23 *27)/34 [Number Multiplied]
(35 *30 *23 *27)/34 ===> (1*-4*-11* -7)/34
[We have taken here negative as well as positive remainder at the same time. When 30 divided by 34 it will give either positive remainder 30 or negative remainder -4. We can use any one of negative or positive remainder at any time.]
(1 *-4 *-11 * -7)/34 ===> (28 *-11)/34 ===> (-6 *-11)/34 ===> 66/34 ===R===> 32.
Required remainder = 32.
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
Find the remainder when 73 *75 *78 *57 *197 *37 is divided by 34.- a)32
- b)10
- c)8
- d)Cannot be determined
Correct answer is option 'A'. Can you explain this answer?
Find the remainder when 73 *75 *78 *57 *197 *37 is divided by 34.
a)
32
b)
10
c)
8
d)
Cannot be determined
|
Sameer Rane answered |
Remainder,
(73 *75 *78 *57 *197 *37)/34 ===> (5 *7 *10 *23 *27 *3)/34
[We have taken individual remainder, which means if 73 is divided by 34 individually, it will give remainder 5, 75 divided 34 gives remainder 7 and so on.]
(5 *7 *10 *23 *27 *3)/34 ===> (35 *30 *23 *27)/34 [Number Multiplied]
(35 *30 *23 *27)/34 ===> (1*-4*-11* -7)/34
[We have taken here negative as well as positive remainder at the same time. When 30 divided by 34 it will give either positive remainder 30 or negative remainder -4. We can use any one of negative or positive remainder at any time.]
(1 *-4 *-11 * -7)/34 ===> (28 *-11)/34 ===> (-6 *-11)/34 ===> 66/34 ===R===> 32.
Required remainder = 32.
Find the last non zero digit of 96!- a)2
- b)4
- c)8
- d)6
Correct answer is option 'D'. Can you explain this answer?
Find the last non zero digit of 96!
a)
2
b)
4
c)
8
d)
6
|
Constructing Careers answered |
Step 1. Count the Trailing Zeros
Trailing zeros come from factors of 10 in n!. Since 10 = 2 × 5, and there are always more factors of 2 than 5, the number of trailing zeros is determined by the number of factors of 5.
For 96!:
⌊96/5⌋ = 19
⌊96/25⌋ = 3
⌊96/125⌋ = 0
⌊96/25⌋ = 3
⌊96/125⌋ = 0
Thus, the total number of factors of 5 (and hence factors of 10) is 19 + 3 = 22. In other words, 96! ends in 22 zeros. To find the last nonzero digit, we “remove” these factors of 10 from the product.
Step 2. Use the Recursive Formula
A recursive formula for the last nonzero digit of n!, denoted d(n), is:
d(n) = { (last nonzero digit of n! computed directly) if n < 5,
d(⌊n/5⌋) × f(n mod 5) × 2(⌊n/5⌋) (mod 10) if n ≥ 5 }
d(⌊n/5⌋) × f(n mod 5) × 2(⌊n/5⌋) (mod 10) if n ≥ 5 }
Here, the function f(r) gives the last nonzero digit of r! for 0 ≤ r < 5. Specifically, we have:
f(0) = 1, f(1) = 1, f(2) = 2, f(3) = 6, f(4) = 4.
Step 3. Compute d(96)
Write 96 in the form 96 = 5q + r. Here:
q = ⌊96/5⌋ = 19
r = 96 mod 5 = 1
r = 96 mod 5 = 1
Thus, the formula gives: d(96) = d(19) × f(1) × 2(19) (mod 10).
Since f(1) = 1, this simplifies to: d(96) = d(19) × 2(19) (mod 10).
Step 4. Compute d(19)
Now, write 19 = 5q′ + r′. Here:
q′ = ⌊19/5⌋ = 3
r′ = 19 mod 5 = 4
r′ = 19 mod 5 = 4
Thus: d(19) = d(3) × f(4) × 2(3) (mod 10).
For n = 3 (which is less than 5), we directly have:
d(3) = f(3) = 6.
Also, f(4) = 4 and 2(3) is represented as 2³ = 8. Therefore:
d(19) = 6 × 4 × 8 = 192.
Taking this modulo 10: 192 (mod 10) = 2.
Step 5. Finish Calculating d(96)
as : d(96) = d(19) × 2(19) (mod 10).
We found d(19) = 2, so: d(96) = 2 × 2(19) (mod 10),
Now, compute 2¹⁹ (mod 10). Notice that powers of 2 modulo 10 cycle every 4:
2¹ = 2,
2² = 4,
2³ = 8,
2⁴ = 16 ≡ 6 (mod 10),
the cycle repeats.
2² = 4,
2³ = 8,
2⁴ = 16 ≡ 6 (mod 10),
the cycle repeats.
Since 19 mod 4 = 3, we have:
2¹⁹ ≡ 2³ ≡ 8 (mod 10).
Thus: d(96) = 2 × 8 = 16 (mod 10).
Taking modulo 10 gives: 16 (mod 10) = 6.
The last nonzero digit of 96! is 6.
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
|
Sagarika Dey answered |
Understanding Divisors of 105
To find how many divisors of 105 have at least one zero at their end, we first need to analyze the number 105.
Prime Factorization of 105
- The prime factorization of 105 is:
105 = 3 × 5 × 7
Divisor Properties
- A divisor of a number has a zero at its end if it is a multiple of 10.
- Since 10 = 2 × 5, a divisor must include both the factors 2 and 5.
Factors of 105
- From the prime factorization, we see that 105 does not contain the factor 2.
- Therefore, none of the divisors of 105 can include 2.
Conclusion on Divisors with Zero
- Since 105 has no factor of 2, it cannot form any multiples of 10.
- Hence, none of the divisors of 105 will have at least one zero at its end.
Counting Divisors of 105
- To ensure completeness, let’s count the total number of divisors of 105:
- The formula for the number of divisors based on the prime factorization (p1^e1 × p2^e2 × ... × pk^ek) is:
(e1 + 1) × (e2 + 1) × ... × (ek + 1)
- For 105, the exponents are all 1 (e1 = 1 for 3, e2 = 1 for 5, e3 = 1 for 7):
(1 + 1) × (1 + 1) × (1 + 1) = 2 × 2 × 2 = 8.
Final Statement
- There are 8 divisors of 105, and none of them can have at least one zero at their end.
- Therefore, the answer to the question regarding divisors of 105 with at least one zero is incorrect; the correct answer is 0, not 12, 15, or 25.
This analysis shows that the answer provided (option 'D') is actually misleading based on the properties of divisors.
To find how many divisors of 105 have at least one zero at their end, we first need to analyze the number 105.
Prime Factorization of 105
- The prime factorization of 105 is:
105 = 3 × 5 × 7
Divisor Properties
- A divisor of a number has a zero at its end if it is a multiple of 10.
- Since 10 = 2 × 5, a divisor must include both the factors 2 and 5.
Factors of 105
- From the prime factorization, we see that 105 does not contain the factor 2.
- Therefore, none of the divisors of 105 can include 2.
Conclusion on Divisors with Zero
- Since 105 has no factor of 2, it cannot form any multiples of 10.
- Hence, none of the divisors of 105 will have at least one zero at its end.
Counting Divisors of 105
- To ensure completeness, let’s count the total number of divisors of 105:
- The formula for the number of divisors based on the prime factorization (p1^e1 × p2^e2 × ... × pk^ek) is:
(e1 + 1) × (e2 + 1) × ... × (ek + 1)
- For 105, the exponents are all 1 (e1 = 1 for 3, e2 = 1 for 5, e3 = 1 for 7):
(1 + 1) × (1 + 1) × (1 + 1) = 2 × 2 × 2 = 8.
Final Statement
- There are 8 divisors of 105, and none of them can have at least one zero at their end.
- Therefore, the answer to the question regarding divisors of 105 with at least one zero is incorrect; the correct answer is 0, not 12, 15, or 25.
This analysis shows that the answer provided (option 'D') is actually misleading based on the properties of divisors.
Anita had to do a multiplication. In stead of taking 35 as one of the multipliers, she took 53. As a 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. In stead of taking 35 as one of the multipliers, she took 53. As a a result, the product went up by 540. What is the new product?
a)
1050
b)
540
c)
1440
d)
1590
|
|
Krishna Iyer answered |
Let the number be X. From the given information,
53x – 35x = 540
=> 18x = 540
=> x = 30.
So, new product = 53*30 = 1590
Hence Option D is the answer.
There are 50 integers a1, a2 … a50, not all of them necessarily different. Let the greatest integer of these 50 integers be referred to as G, and the smallest integer be referred to as L. The integers a1 through a24 form sequence S1, and the rest form sequence S2. Each member of S1 is less than or equal to each member of S2.Q. Elements of S1 are in ascending order and those of S2 are in descending order. a24 and a25 are interchanged then which of the following is true?- a)S1 continues to be in ascending order
- b)S2 continues to be in descending order
- c)Both (a) and (b)
- d)Cannot be determined
Correct answer is option 'A'. Can you explain this answer?
There are 50 integers a1, a2 … a50, not all of them necessarily different. Let the greatest integer of these 50 integers be referred to as G, and the smallest integer be referred to as L. The integers a1 through a24 form sequence S1, and the rest form sequence S2. Each member of S1 is less than or equal to each member of S2.
Q. Elements of S1 are in ascending order and those of S2 are in descending order. a24 and a25 are interchanged then which of the following is true?
a)
S1 continues to be in ascending order
b)
S2 continues to be in descending order
c)
Both (a) and (b)
d)
Cannot be determined
|
|
Aarav Sharma answered |
Let's assume that the sum of the integers is S. Then we know that:
a1 + a2 + ... + a50 = S
We also know that the sum of the even integers is:
a2 + a4 + ... + a50 = S/2
And the sum of the odd integers is:
a1 + a3 + ... + a49 = S/2
Now, we can subtract the sum of the odd integers from the sum of all the integers to get the sum of the even integers:
(a1 + a2 + ... + a50) - (a1 + a3 + ... + a49) = a2 + a4 + ... + a50 = S/2
We can also subtract the sum of the even integers from the sum of all the integers to get the sum of the odd integers:
(a1 + a2 + ... + a50) - (a2 + a4 + ... + a50) = a1 + a3 + ... + a49 = S/2
So, we have found that the sum of the even integers is half of the sum of all the integers, and the sum of the odd integers is also half of the sum of all the integers. Therefore, the sum of the even integers is equal to the sum of the odd integers.
a1 + a2 + ... + a50 = S
We also know that the sum of the even integers is:
a2 + a4 + ... + a50 = S/2
And the sum of the odd integers is:
a1 + a3 + ... + a49 = S/2
Now, we can subtract the sum of the odd integers from the sum of all the integers to get the sum of the even integers:
(a1 + a2 + ... + a50) - (a1 + a3 + ... + a49) = a2 + a4 + ... + a50 = S/2
We can also subtract the sum of the even integers from the sum of all the integers to get the sum of the odd integers:
(a1 + a2 + ... + a50) - (a2 + a4 + ... + a50) = a1 + a3 + ... + a49 = S/2
So, we have found that the sum of the even integers is half of the sum of all the integers, and the sum of the odd integers is also half of the sum of all the integers. Therefore, the sum of the even integers is equal to the sum of the odd integers.
The HCF of two numbers is 11 and their LCM is 616. If one of the numbers is 77, find the other number.
- a)81
- b)89
- c)88
- d)45
Correct answer is option 'C'. Can you explain this answer?
The HCF of two numbers is 11 and their LCM is 616. If one of the numbers is 77, find the other number.
a)
81
b)
89
c)
88
d)
45
|
Glance Learning Institute answered |
Calculation:
Let, 2nd number be m,
⇒ m × 77 = 11 × 609
⇒ m = 609/7
⇒ m = 87
∴ The 2nd number is 87.
There are 50 integers a1, a2,........, a50, not all of them necessarily different. Let the greatest integer of these 50 integers be referred to as G, and the smallest integer is referred to as L. The integers a1 through a24 form sequence S1, and the rest form sequence S2. Each member of S1 is less than or equal to each member of S2.
Every element of S1 is made greater than or equal to every element of S2 by adding to each element of S1 an integer x. Then, x cannot be less than:- a)210
- b)the smallest value of S2
- c)the largest value of S2
- d)(G- L)
Correct answer is option 'D'. Can you explain this answer?
There are 50 integers a1, a2,........, a50, not all of them necessarily different. Let the greatest integer of these 50 integers be referred to as G, and the smallest integer is referred to as L. The integers a1 through a24 form sequence S1, and the rest form sequence S2. Each member of S1 is less than or equal to each member of S2.
Every element of S1 is made greater than or equal to every element of S2 by adding to each element of S1 an integer x. Then, x cannot be less than:
Every element of S1 is made greater than or equal to every element of S2 by adding to each element of S1 an integer x. Then, x cannot be less than:
a)
210
b)
the smallest value of S2
c)
the largest value of S2
d)
(G- L)
|
Ishita Reddy answered |
The smallest integer of the series is 1 and greatest integer is 50. If each element of S1 is made greater of equal to every element of S2, then the smallest element 1 should be added to (50 - 1) = 49. Hence option (G-L) is the correct answer.
As this is a variable based question: the word "ANY" can be used.
Let the series of integers a1,a2,.......,a50 be 1,2,3,4,5,.......,50.
S1=1,2,3,4,.........24, S2=25,26,27,..........50.
Four bells ringing together and ring at an interval of 12 sec, 15 sec, 20 sec, and 30 sec respectively. How many times will they ring together in 8 hours?
- a)481
- b)480
- c)482
- d)483
Correct answer is option 'A'. Can you explain this answer?
Four bells ringing together and ring at an interval of 12 sec, 15 sec, 20 sec, and 30 sec respectively. How many times will they ring together in 8 hours?
a)
481
b)
480
c)
482
d)
483
|
EduRev CAT answered |
Given:
Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec
Calculation:
Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec
Now we have to take LCM of time interval
⇒ LCM of (12, 15, 20, 30) = 60
Total seconds in 8 hours = 8 × 3600 = 28800
Number of times bell rings = 28800/60
⇒ Number of times bell rings = 480
If four bells ring together in starting
⇒ 480 + 1
∴ The bell ringing 481 times in 8 hours.
When 7179 and 9699 are divided by another natural number N , remainder obtained is same. How many values of N will be ending with one or more than one zeroes?- a)24
- b)124
- c)18
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
When 7179 and 9699 are divided by another natural number N , remainder obtained is same. How many values of N will be ending with one or more than one zeroes?
a)
24
b)
124
c)
18
d)
None of these
|
Anjali Shah answered |
Understanding the Problem
To solve the problem, we need to find a natural number N such that when both 7179 and 9699 are divided by N, the remainder is the same. This condition implies that the difference between the two numbers should be divisible by N.
Calculating the Difference
- First, calculate the difference:
- 9699 - 7179 = 2520
We need to find all divisors of 2520 that end with one or more zeros.
Finding Divisors of 2520
- The next step is to determine the divisors of 2520. First, we find its prime factorization:
- 2520 = 2^3 × 3^2 × 5^1 × 7^1
Using this factorization, we can find the total number of divisors using the formula (e1 + 1)(e2 + 1)(e3 + 1)(e4 + 1), where e1, e2, e3, and e4 are the powers of the prime factors.
- For 2520:
- (3 + 1)(2 + 1)(1 + 1)(1 + 1) = 4 × 3 × 2 × 2 = 48
Thus, 2520 has 48 divisors in total.
Identifying Divisors Ending with Zeros
- To find the divisors of 2520 that end with one or more zeros, we focus on those that are multiples of 10. A number ends with at least one zero if it is divisible by 10, which means it needs at least one factor of 2 and one factor of 5.
- The divisors must be of the form: 2^a × 3^b × 5^c × 7^d where:
- a ≥ 1, c ≥ 1 (to ensure divisibility by 10)
- Possible values for a, b, c, and d are:
- a can be 1, 2, or 3 (3 options)
- b can be 0, 1, or 2 (3 options)
- c must be 1 (1 option)
- d can be 0 or 1 (2 options)
Calculating the total combinations:
- Total = 3 × 3 × 1 × 2 = 18
Conclusion
Thus, the number of divisors of 2520 that end with one or more zeros is 18. Therefore, the correct answer is option 'C'.
To solve the problem, we need to find a natural number N such that when both 7179 and 9699 are divided by N, the remainder is the same. This condition implies that the difference between the two numbers should be divisible by N.
Calculating the Difference
- First, calculate the difference:
- 9699 - 7179 = 2520
We need to find all divisors of 2520 that end with one or more zeros.
Finding Divisors of 2520
- The next step is to determine the divisors of 2520. First, we find its prime factorization:
- 2520 = 2^3 × 3^2 × 5^1 × 7^1
Using this factorization, we can find the total number of divisors using the formula (e1 + 1)(e2 + 1)(e3 + 1)(e4 + 1), where e1, e2, e3, and e4 are the powers of the prime factors.
- For 2520:
- (3 + 1)(2 + 1)(1 + 1)(1 + 1) = 4 × 3 × 2 × 2 = 48
Thus, 2520 has 48 divisors in total.
Identifying Divisors Ending with Zeros
- To find the divisors of 2520 that end with one or more zeros, we focus on those that are multiples of 10. A number ends with at least one zero if it is divisible by 10, which means it needs at least one factor of 2 and one factor of 5.
- The divisors must be of the form: 2^a × 3^b × 5^c × 7^d where:
- a ≥ 1, c ≥ 1 (to ensure divisibility by 10)
- Possible values for a, b, c, and d are:
- a can be 1, 2, or 3 (3 options)
- b can be 0, 1, or 2 (3 options)
- c must be 1 (1 option)
- d can be 0 or 1 (2 options)
Calculating the total combinations:
- Total = 3 × 3 × 1 × 2 = 18
Conclusion
Thus, the number of divisors of 2520 that end with one or more zeros is 18. Therefore, the correct answer is option 'C'.
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
|
|
Krishna Iyer 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.
Therefore, the correct answer is Option A.
Which of the following would always divide a six-digit number of the form ababab?- a)10,101
- b)11,111
- c)10,001
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Which of the following would always divide a six-digit number of the form ababab?
a)
10,101
b)
11,111
c)
10,001
d)
None of these
|
|
Sakshi Nambiar answered |
Solution:
The six-digit number of the form ababab can be written as:
ababab = 1001 × ab
Therefore, to find the number that always divides ababab, we need to find the factors of 1001.
Prime factorization of 1001:
1001 = 7 × 11 × 13
Factors of 1001:
1, 7, 11, 13, 77, 91, 143, 1001
Out of these factors, only 10 and 101 are two-digit numbers. Therefore, the only option that could divide a six-digit number of the form ababab is 10,101.
Hence, the correct answer is option A.
The six-digit number of the form ababab can be written as:
ababab = 1001 × ab
Therefore, to find the number that always divides ababab, we need to find the factors of 1001.
Prime factorization of 1001:
1001 = 7 × 11 × 13
Factors of 1001:
1, 7, 11, 13, 77, 91, 143, 1001
Out of these factors, only 10 and 101 are two-digit numbers. Therefore, the only option that could divide a six-digit number of the form ababab is 10,101.
Hence, the correct answer is option A.
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