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Question

What least number should be added to 3500 to make it exactly divisible by 42, 49, 56 and 63?

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
28

Finding Least Number for Divisibility

The problem asks for the smallest number that should be added to 3500 to make it exactly divisible by 42, 49, 56, and 63. This means we need to find a number $N$ such that $3500 + N$ is a common multiple of these numbers. We need the least such positive $N$. The first step is to find the Least Common Multiple (LCM) of 42, 49, 56, and 63.

Calculating the LCM

First, find the prime factorization of each number:

  • $42 = 2 \times 3 \times 7$
  • $49 = 7^2$
  • $56 = 2^3 \times 7$
  • $63 = 3^2 \times 7$

The LCM is found by taking the highest power of each prime factor present:

  • LCM$(42, 49, 56, 63) = 2^3 \times 3^2 \times 7^2$
  • LCM $= 8 \times 9 \times 49$
  • LCM $= 72 \times 49$
  • LCM $= 3528$

The LCM is 3528. This is the smallest number that is exactly divisible by 42, 49, 56, and 63.

Determining the Number to Add

We need to find the least number to add to 3500 to reach the next multiple of the LCM (3528). Since 3500 is less than 3528, the smallest multiple of 3528 that is greater than or equal to 3500 is 3528 itself.

Let the number to be added be $N$. Then:

$3500 + N = 3528$

Solving for $N$:

$N = 3528 - 3500$ $N = 28$

Therefore, the least number that should be added to 3500 to make it exactly divisible by 42, 49, 56, and 63 is 28.

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