Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:How many factors of 24 × 35 × 104 are perfect squares which are greater than 1?
[2019]
Explanation
24 × 35 × 104 = 28 × 35 × 54
For perfect squares, we have to take only even powers of the prime factors of the number
The number of ways 2’s can be used is 5 i.e. 20, 22, 24, 26, 28
The number of ways 3’s can be used is 3 i.e. 30, 32, 34
The number of ways 5’s can be used is 3 i.e. 50, 52, 54
Therefore, the total number of factors which are perfect squares = 5 × 3 × 3 = 45
But this also includes the number 1.
Hence excluding 1, the required number is 45 – 1 = 44.
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:How many pairs (m, n) of positive integers satisfy the equation m2 + 105 = n2?
[2019]
Explanation
m2 + 105 = n2 ⇒ n2 – m2 = 105
⇒ (n – m) (n + m) = 105
Since m an d n are positive integers (n – m) < (n + m), then by splitting 105 in two factors, we get
⇒ (n – m) (n + m) = 1 × 105
For (n – m) = 1 and (n + m) = 105, (m, n) = (52, 53)
⇒ (n – m) (n + m) = 3 × 35
For (n – m) = 3 and (n + m) = 35 (m, n) = (16, 19)
⇒ (n – m) (n + m) = 5 × 21
For (n – m) = 5 and (n + m) = 21 (m, n) = (8, 13)
⇒ (n – m) (n + m) = 7 × 15
For (n – m) = 7 and (n + m) = 15, (m, n) = (4, 11)
Hence, there are four required pairs.
Shortcut approach :
Number of pairs =
105 = 3 × 5 × 7
Factors of 105 are 1, 3, 5, 7, 15, 21, 35 and 105
Number of factors of 105 = 8
Hence, required number of pairs = 8 / 2 = 4
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:In a six-digit number, the sixth, that is, the rightmost, digit is the sum of the first three digits, the fifth digit is the sum of first two digits, the third digit is equal to the first digit, the second digit is twice the first digit and the fourth digit is the sum of fifth and sixth digits. Then, the largest possible value of the fourth digit is?
[2019]
Explanation
Let the number be ABCDEF, where A, B, C, D, E and F be the digits
Now, C = A, B = 2A
F = A + B + C = A + 2A + A = 4A, E = A + B = A + 2A = 3A
and D = E + F = 3A + 4A = 7A
Since A an d D both are digit, the maximum possible value of A = 1. Therefore, the maximum value of D = 7
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:If m and n are integers such that (√2)19 34 42 9m 8n = 3n 16m (∜64) then m is
[2019]
Explanation
(√2)19 34 42 9m 8n = 3n 16m (∜64)
⇒ 219/2 × 34 × 24 × 32m × 23n = 3n × 24m × 23/2
⇒ 2(19/2 + 4 + 3n) × 3(4 + 2m) = 3n×2((4m+6/4))
On comparing the powers of same bases from both side, we get
19 / 2 + 4 + 3n = 4m + 3 / 2
⇒ 3n = 4m + - 4
⇒ 3n = 4m +
⇒ n =
⇒ n = ...(i)
4 + 2m = n ...(ii)
From equation (i) and (ii), we get
⇒ 2m = –24
∴ m = –12
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157 : 3, then the sum of the two numbers is
[2019]
Explanation
Let the two numbers be x and y
∴ x × y = 616
and
Now, let x3 – y3 = 157k and (x – y)3 = 3k
Since, (x – y)3 = x3 – y3 – 3xy(x – y)
∴ 3k = 157k – 3 × 616(3k)1/3
⇒ 154k = 3 × 616 × (3k)1/3 ⇒ k =
⇒ k = 12 × (3k)1/3 ⇒ k3 = 123 × 3 × k
⇒ k(k2 – 123 × 3) = 0
∵ k ≠ 0
∴ k2 = 3 × 123
∴ k = 72
⇒ x – y = (3k)1/3 = (3 × 72)1/3 = 6
∵ (x + y)2 = (x – y)2 + 4xy
∴ (x + y)2 = 62 + 4 × 616 = 2500
⇒ x + y = 50
Hence, sum of the two numbers = 50
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:The number of integers x such that 0.25 < 2x < 200, and 2x + 2 is perfectly divisible by either 3 or 4, is
[2018]
Explanation
⇒ 0.25 < 2x < 200
⇒ 0.25 = 1/4 = 2-2
⇒ 200 = 28
value of x varies between = {-1, 0, 1, 2, 3, 4, 5, 6, 7}
When x = 1, is it divisible by 4.
When x = multiple of 2 as 2 and 4 is divisible y 3
∴ Total number of numbers = {0, 1, 2, 4, 6} = 5
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:A sequence of 4 digits, when considered as a number in base 10 is four times the number it represents in base 6. What is the sum of the digits of the sequence?
[2016]
Explanation
Let the 4-digit sequence be abcd.
In base 6, this represents 216a + 36b + 6c + d and each of a, b, c, d is less than 6.
In base 10, it represents 1000a + 100b + 10c + d.
Given 4(216a + 36b + 6c + d)
= 1000a + 100b + 10c + d
⇒ 136a = 44b + 14c + 3d ...(A)
By trial a = 1, b = 2, c = 3, d = 2
If a = 2, the LHS = 272
[If we consider b = 5, we need 272 – 220 or 52 from 14c + 3d (c, d)=(2, 8) but 8 is not a proper digit in base 6.
If a= 3, the LHS = 408, while 44b + 14c + 3d can at the most be (44 + 14 + 3) 5 or 305.
∴ There are no other possible values that satisfy (A)]
∴ abcd = 1232 and a + b + c + d = 8
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:Which of the following will completely divide (10690 – 4990)?
[2015]
Explanation
x = 10690 – 4990
∵ (xn – an) is divisible by both (x – a) and (x + a) whenever n is even
⇒ (10690 – 4990) is divisible by both 57 and 155
57 = 19 × 3
155 = 31 × 5
Therefore, (10690 – 4990) will be divisible by (19 × 31) = 589 as well.
Also, note that (10690– 4990) will be odd and options (b) and (c) are even. Hence, they can be rejected.
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:Let P be the set of all odd positive integers such that every element in P satisfies the following conditions.
I. 100 ≤ n < 1000
II. The digit at the hundred’s place is never greater than the digit at tens place and also never less than the digit at units place.
How many elements are there in P?
[2015]
Explanation
Let n be xyz and since n is odd z can take only odd values i.e. 1, 3, 5 and 9 Now, x ≤ y and x ≥ z
∴ Total number of elemenls in P = 95.
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:A four-digit number is divisible by the sum of its digits. Also, the sum of these four digits equals the product of the digits. What could be the product of the digits of such a number?
[2015]
Explanation
Solve by option.
Option (a): If the product of the digits is 6. then the factors of 6 are 1,2, 3 and 6. This combination of digits is not suitable. So it is not the answer.
Option (b): If the product of the digits is 8, then the factors of 8 are 1, 2, 4 and 8. So only possible combination is 1, 1, 2, 4.
Hence, the number is 4112. It is suitable for answer.
Similarly, we can check options (c) and (d).
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:P is the product of the first 100 multiples of 15 and Q is the product of the first 50 multiples of 2520. Find the number of consecutive zeroes at the end of P2/Q × 101767
[2015]
Explanation
P = 15100 (1 × 2 × 3 × .... × 100)
= 15100 × 100!
Highest power of 2 in P = 97 (2 will be deciding factor for number of zeroes because number of lives will be greater than number of zeroes in this number) Q = 2520×50 (1 × 2 × 3x .... × 50) = 52000 × 50!
Highest power of 2 in θ = 47
So Highest power of 2 in
P2/Q × 101767 = 2 × 97 + 1767 – 47 = 1914
Hence, number of zeroes = 1914.
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:The number of factors of the square of a natural number is 105. The number of factors of the cube of the same number is ‘F’. Find the maximum possible value of ‘F’.
[2013]
Explanation
Let the number be N.
In order to maximize the number of factors of N3, N2 must be expressed as a product of as many prime factors as possible.
No. of factors of N2 = 105 = 3 × 5 × 7
where a = 2 b = 4 c = 6
then power original number = (2 + 1) (4 + 1) (6 + 1)
∴ N2 = (a)2 (b)4 (c)6, where a, b and c are prime numbers.
∴ N3 = (a)3 (b)6 (c)9
Where N = ap bq cr no = (p + 1) (q + 1) (r + 1)
Hence, the number of factors of N3
= (3 + 1) × (6 + 1) × (9 × 1) = 4 × 7 × 10 = 280.
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:‘ab’ is a two-digit prime number such that one of its digits is 3. If the absolute difference between the digits of the number is nota factor of 2, then how many values can ‘ab’ assume?
[2013]
Explanation
Since 'ab' is a two - digit prime number and one of its digit is 3, it can assume any of the values among 13, 23, 31, 37,43, 53, 73 and 83.
As the absolute difference between the digits of the number is not a factor of 2, the number among the obtained numbers that satisfy the aforementioned condition are 37, 73 and 83. Hence, the number of values that 'ab' can assume is 3.
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:If E = 3 + 8 + 15 + 24 + … + 195, then what is the sum of the prime factors of E?
[2013]
Explanation
E = 3 + 8 + 15 + 24 + .... + 195 = 1 × 3 + 2 × 4 + 3 × 5 + 4 × 6 + .... + 13 × 15
∴ Tn = n (n + 2) and n = 13
= 7 × 11 × 13
Hence the sum of the prime factors of E
= 7 + 11 + 13 = 31.
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:The number 44 is written as a product of 5 distinct integers. If ‘n’ is the sum of these five integers then what is the sum of all the possible values of n?
[2012]
Explanation
Prime factorization of 44 is = 2 × 2 × 11
To express 44 as product of five distinct integers
So, we'll have to put 1 and –1.
The only possible way comes out to be:
44 = 2 × (–2) × 11 × 1 × (–1)
In this case the value of n would be 11 which is also the only possible value.
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:All the two-digit natural numbers whose unit digit is greater than their ten’s digit are selected. If all these numbers are written one after the other in a series, how many digits are there in the resulting number?
[2012]
Explanation
Here find the number of two–digit natural numbers such that unit digit is greater than their ten's digit.
In such natural numbers, we cannot take 0 or 1 in units place.
When we take 2 at unit's place, we obtain only 1 Such number is 12.
When we take 3 at unit's place, we obtain 2 such numbes are 13 and 23.
When we take 9 at unit's place, we obtain 8 such numbers.
So, number of such numbers is (1 + 2 + 3 + .... + 8) = 36
Hence, the required number has 72 digits.
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:A positive integer is equal to the square of the number of factors it has. How many such integers are there?
[2011]
Explanation
One such number is 1 which has no factor other than itself.
If the number has only one prime factor i.e. it is of the form pa where p is a prime number and a is a natural number, then according to the question: (a + 1)2 = pa
This is possible only if a = 2 and p = 3. So the number is 9.
If the number has two prime factors then it would be of the type pa × qb, where p and q are two distinct prime numbers. Then according to the question:
(a + 1)2 (b + 1)2 = pa × qb
This is possible only if p and q are both 3. Since they are different, this is not a valid case. So there would no such case with two or more prime factors.
So there are only two such integers - 1 and 9.
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Question for CAT Past Year Questions: Factors & Divisibility
Try yourself:If ‘a’ is one of the roots of x5 – 1 = 0 and a ≠ 1, then what is the value of a15 + a16 + a17 +.......a50?
[2010]
Explanation
a15 + a16 + a17 + .... + a50|
Sum = a15 {1 + a + a2 + ..... a35}
Since a is the root of equation x5 – 1 = 0,
a5 – 1 = 0 ⇒ a5 = 1
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