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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?
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.
How many factors of 2^{5} * 3^{6} * 5^{2} are perfect squares?
Any factor of this number should be of the form 2^{a} × 3^{b} × 5^{c}
For the factor to be a perfect square a, b, c has to be even.
⇒ a can take values 0, 2, 4
⇒ b can take values 0, 2, 4, 6
⇒ c can take values 0, 2
So, total number of perfect squares = 3 × 4 × 2 = 24
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?
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.
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
Which of the following options are true?
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 .
The sum of the first 100 natural numbers, 1 to 100 is divisible by:
The sum of the first 100 natural numbers is:
= (n * (n + 1)) / 2
= (100 * 101) / 2
= 50 * 101
101 is an odd number and 50 is divisible by 2.
Hence, 50 * 101 will be divisible by 2.
In a fourdigit 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?
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.
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?
Let's assume that the total number of pages in the book is n. The sum of the page numbers from 1 to n is given by the formula (n*(n+1))/2.
If we call the repeated page number as x, the sum of all the page numbers from 1 to n with x being added twice is (n*(n+1))/2 + x + x = 1000.
Expanding this equation we get: n(n+1) + 2x = 2000
n^2 + n + 2x = 2000
n^2 + n = 2000  2x
(n+1)(n) = 2000  2x
Now we can find the value of n by factoring it, n(n+1) = (44)(46) = 2004
n = 44 or 45
Now we know that one of the repeated page number is either 44 or 45, so the repeated page number is x = 44 or x = 45. Out of the options given, the only number that is repeated is 45, So the correct answer is 3.
The integers 34041 and 32506 when divided by a threedigit integer n leave the same remainder. What is n?
The difference of the numbers = 34041 – 32506 = 1535
The number that divides both these numbers must be a factor of 1535.
307 is the only 3 digit integer that divides 1535.
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?
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 x 2 x 3
⇒ 15 = 3 x 5
⇒ 18 = 2 x 3 x 3
⇒ 20 = 2 x 2 x 5
∴ LCM = 2 x 2 x 3 x 5 x 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 x 2 x 3 x 3 x 5 x 5
= 900
The factors of 1080 which are perfect square:
1080 → 2^{3} × 3^{3} × 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
So, the factors which are perfect square are 1, 4, 9, 36.
Hence, Option B is correct.
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 is 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 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?
The size of each row would be the HCF of 363, 429 and 693. Difference between 363 and 429 =66.
Factors of 66 are 66, 33, 22, 11, 6, 3, 2, 1.
66 need not to be checked as it is even and 363 is odd. 33 divides 363, hence would automatically divide
429 and also divides 693. Hence, 33 is the correct answer for the size of each row.
For how many rows would be required we need to follow the following process:
Minimum number of rows required = 363/33 + 429/33 + 693/33 = 11 + 13 + 21 = 45 rows.
Therefore, the correct answer is A
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?
Solve this question step by step:
Hence it is not a perfect square for any arrangement.
Find the remainder when 73 * 75 * 78 * 57 * 197 * 37 is divided by 34.
Remainder = (73 x 75 x 78 x 57 x 197 x 37) / 34
= (5 x 7 x 10 x 23 x 27 x 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 x 7 x 10 x 23 x 27 x 3) / 34
= (35 x 30 x 23 x 27) / 34
= (1 x (4) x (11) x (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 the negative or positive remainders at any time.)
= (1 x 4 x 11 x 7) / 34
= (28 x 11) / 34
= (6 x 11) / 34
= 66 / 34
∴ R = 32
There are two integers 34041 and 32506, when divided by a threedigit integer n, leave the same remainder. What is the value of n?
Find the difference between both 34041 and 32506.
34041  32506 = 1535
So, N should be a factor of 1535. Factors of 1535:
1, 5, 307, 1535
Here, 307 is a threedigit factor of 1535. So, N = 307.
34041 and 32506 divided by 307 leaves remainder 271. So, 307 is the 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?
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.
Therefore, the correct answer is Option D.
Teacher said that there were 100 students in his class, 24 of whom were boys and 32 were girls. Which base system did the teacher use in this statement?
We are provided with the equation (32) + (24) = (100). Let us assume our base be 'b'
Then,we can say:
⇒ 32 = 3 x b^{1 }+ 2 x b^{0} = 3b+2
⇒ 24 = 2 x b^{1}+ 4 x b^{0 }= 2b+4
⇒ 100 = 1 x b^{2 }+ 0 x b^{1 }+ 0 x b^{0} = b^{2}
Now, according to our question:
⇒ 32 + 24=100
⇒ (3b + 2) + (2b + 4) = (b^{2})
⇒ 5b + 6 = b^{2}
⇒ b^{2 } 5b  6 = 0
⇒ b^{2 } 6b + b  6 = 0
⇒ b(b  6) + 1(b  6) = 0
⇒ (b  6) * (b + 1) = 0
⇒ b = 6, 1
Base can't be negative. Hence b = 6.
∴ Base assumed in the asked question must be 6.
24 = 8 × 3
Therefore, we need to find the highest power of 8 and 3 in 150!
8 = 2^{3}
Highest power of 8 in 150! is:
= [(150 / 2) + (150 / 4) + (150 / 8) + (150 / 16) + (150 / 32) + (150 / 64) +(150 / 128)] / 3
= 48
Highest power of 3 in 150! is:
= [150 / 3] + [150 / 9] + [150 / 27] + [150 / 81]
= 72
As the powers of 8 are less, powers of 24 in 150! = 48
If a three digit number ‘abc’ has 2 factors (where a, b, c are digits), how many factors does the 6digit number ‘abcabc’ have?
Factors: Beauty of the number 1001. This number is not prime, is a product of three distinct primes and does wonderful things to threedigit numbers when multiplied to them.
To start with ‘abcabc’ = ‘abc’ * 1001 or abc * 7 * 11 * 13 (This is a critical idea to remember).
‘abc’ has only two factors. Or, ‘abc’ has to be prime. Only a prime number can have exactly two factors. (This is in fact the definition of a prime number)
So, ‘abcabc’ is a number like 101101 or 103103.
’abcabc’ can be broken as ‘abc’ * 7 * 11 * 13. Or, a p * 7 * 11 * 13 where p is a prime.
As we have already seen, any number of the form paqbrc will have (a + 1) (b + 1) (c + 1) factors, where p, q, r are prime.
So, p * 7 * 11 * 13 will have = (1 + 1) * (1 + 1) * (1 + 1) * (1 + 1) = 16 factors
The question is "How many factors does the 6digit number ‘abcabc’ have?"
Hence the answer is 16 factors.
Choice A is the correct answer.
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