Which is the smallest multiple of 7, which leaves 5 as the remainder in each case. When divided by 8, 9, 12 and 15?
Correct Answer :
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 , such that:
1. is a multiple of 7 (i.e., is completely divisible by 7).
2. When 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:
where is a positive integer.
Let us find the LCM of 8, 9, 12, and 15 using prime factorization:
- Prime factorization of 8:
- Prime factorization of 9:
- Prime factorization of 12:
- Prime factorization of 15:
The LCM is the product of the highest power of each prime factor present in these numbers:
So, the required number must be of the form:
Step 3: Find the value of such that is divisible by 7.
We need to find the smallest integer value of for which is divisible by 7.
Let us express 360 in terms of a multiple of 7:
Therefore, we can rewrite the expression as:
For the number to be divisible by 7, the term must be divisible by 7.
Let us test positive integer values of starting from 1:
- If : (not divisible by 7)
- If : (not divisible by 7)
- If : (which is divisible by 7 since )
Thus, the smallest integer value is .
Step 4: Calculate the final number.
Substitute back into our expression for :
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.
Access expert-curated educational resources and study materials—completely free.
Create, conduct, and manage professional online assessments with Mindyard. Perfect for teachers and institutes.
Copyright © 2026 Mindyard. All Rights Reserved.