CAT Past Year Questions: Factors & Divisibility

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.

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

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

(√2)¹⁹ 3⁴ 4² 9^m 8ⁿ = 3ⁿ 16^m (∜64)

⇒ 2^19/2 × 3⁴ × 2⁴ × 3^2m × 2³ⁿ = 3ⁿ × 2^4m × 2^3/2

⇒ 2^{(19/2 + 4 + 3n)} × 3^{(4 + 2m)} = 3^{n×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

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

⇒ 0.25 < 2^x  < 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

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

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.

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.

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).

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.

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.

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.

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.

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.

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.

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.

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