Question Details

Which is the smallest multiple of 7, which leaves 5 as the remainder in each case. When divided by 8, 9, 12 and 15?

Options

A

365

B

1,085

C

2,525

D

725

Show Answer

Correct Answer :

Option B

1,085

1,085

Solution :

The correct option is 1,085.

Let us find the solution step-by-step.

Step 1: Understand the requirements of the problem.
We need to find the smallest number, say N, such that:
1. N is a multiple of 7 (i.e., N is completely divisible by 7).
2. When N is divided by 8, 9, 12, and 15, it leaves a remainder of 5 in each case.

Step 2: Find the Least Common Multiple (LCM) of the divisors.
Any number that leaves a remainder of 5 when divided by 8, 9, 12, and 15 must be of the form:
N=(LCM of 8, 9, 12, 15)×k+5
where k is a positive integer.

Let us find the LCM of 8, 9, 12, and 15 using prime factorization:
- Prime factorization of 8: 23
- Prime factorization of 9: 32
- Prime factorization of 12: 22×3
- Prime factorization of 15: 3×5

The LCM is the product of the highest power of each prime factor present in these numbers:
LCM=23×32×5
LCM=8×9×5=360

So, the required number must be of the form:
N=360k+5

Step 3: Find the value of k such that N is divisible by 7.
We need to find the smallest integer value of k for which 360k+5 is divisible by 7.
Let us express 360 in terms of a multiple of 7:
360=7×51+3
Therefore, we can rewrite the expression as:
360k+5=(7×51+3)k+5
360k+5=7×51k+(3k+5)

For the number to be divisible by 7, the term 3k+5 must be divisible by 7.
Let us test positive integer values of k starting from 1:
- If k=1: 3(1)+5=8 (not divisible by 7)
- If k=2: 3(2)+5=11 (not divisible by 7)
- If k=3: 3(3)+5=14 (which is divisible by 7 since 14/7=2)

Thus, the smallest integer value is k=3.

Step 4: Calculate the final number.
Substitute k=3 back into our expression for N:
N=360×3+5
N=1080+5=1085

Thus, the smallest multiple of 7 which leaves a remainder of 5 in each case when divided by 8, 9, 12, and 15 is 1,085.

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