Arun Sharma Test: HCF & LCM - Free MCQ with solutions for CAT Exam

First, we calculate the least common multiple (LCM) of these numbers. The LCM of 5, 6, 4, and 3 is 60. We need to determine the smallest number to add to 2497 such that the result is divisible by 60.

When 2497 is divided by 60, the remainder is 37 (since 2497 divided by 60 equals 41 with a remainder of 37). To achieve divisibility, we must add 23 (60 - 37) to 2497, making the correct answer option C (23).

We need to find the least number that, when divided by 5, 6, 7, and 8, leaves a remainder of 3, and when divided by 9 leaves no remainder (i.e., is divisible by 9).

First, the condition "divided by 5, 6, 7, and 8 leaves a remainder of 3" means the number can be expressed as: Number = LCM of (5, 6, 7, 8) * k + 3, where k is a positive integer.

Calculate the LCM of 5, 6, 7, and 8:

• Prime factors: 5 = 5, 6 = 2 * 3, 7 = 7, 8 = 2^3
• LCM = 2^3 * 3 * 5 * 7 = 8 * 3 * 5 * 7 = 840

So, the number can be written as: Number = 840k + 3

Additionally, the number must be divisible by 9 (remainder 0), so: 840k + 3 ≡ 0 (mod 9)

First, find 840 mod 9: 8 + 4 + 0 = 12, 1 + 2 = 3, so 840 ≡ 3 (mod 9)

Now substitute: 3k + 3 ≡ 0 (mod 9) 3k ≡ -3 (mod 9) Since -3 + 9 = 6, this is: 3k ≡ 6 (mod 9)

Divide by 3 (note that 3 and 9 are not coprime, so we check values): k ≡ 2 (mod 3) (because 3 * 2 = 6, and 6 mod 9 = 6)

So, k = 3m + 2 (where m is a non-negative integer).

Substitute k = 3m + 2 into the number: Number = 840 * (3m + 2) + 3 = 2520m + 1680 + 3 = 2520m + 1683

The smallest positive value occurs when m = 0: Number = 1683

Verify:

• 1683 ÷ 5 = 336 remainder 3 (1680 is divisible by 5, 1683 - 1680 = 3)
• 1683 ÷ 6 = 280 remainder 3 (1674 is divisible by 6, 1683 - 1674 = 9, but adjust: 1680 ÷ 6 = 280, 1683 - 1680 = 3)
• 1683 ÷ 7 = 240 remainder 3 (1680 ÷ 7 = 240, 1683 - 1680 = 3)
• 1683 ÷ 8 = 210 remainder 3 (1680 ÷ 8 = 210, 1683 - 1680 = 3)
• 1683 ÷ 9 = 187 remainder 0 (exact division)

All conditions are satisfied.

Answer: b) 1683

L.C.M. of 252, 308 and 198 = 2772.

So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.

So, the numbers 12 and 16.

L.C.M. of 12 and 16 = 48.

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7

   = 1008 + 7

   = 1015

The prime factorization of 252 is calculated as follows:

• 252 ÷ 2 = 126 (One 2)
• 126 ÷ 2 = 63 (Two 2s)
• 63 ÷ 3 = 21 (One 3)
• 21 ÷ 3 = 7 (Two 3s)
• 7 is a prime number (One 7)

Thus, 252 = 2 × 2 × 3 × 3 × 7. Therefore, the correct answer is A.