# What is the smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case?

By BYJU'S Exam Prep

Updated on: September 25th, 2023

The smallest number that, when divided by 35, 56 and 91 leaves the remainder of 7 in each case is 3647. to find this solution, we will factories the numbers provided and then calculate their LCM. We will then add the LCM to the remainder. Through this process, we will find out the answer as to which is the smallest number providing the remainder 7 when divided by 35, 56 and 91 every time.

## Smallest Number Divided by 35, 56 and 91 Leaving 7 as remainder

To find out the smallest number divisible by 35, 56 and 91 that leaves 7 as a remainder, we will first find out the factors of all these numbers.

35 = 5 x 7
56 = 2 x 2 x 2 x 7
91 = 7 x 13

Now, taking the prime factors of all three numbers, we will now find the LCM of these numbers:

LCM = 7 x 5 x 2 x 2 x 2 x 13 = 3640

Now, that we have found the LCM, we will add 7 to this number to get our final answer,

3640 + 7 = 3647

Thus, the smallest number that, when divided by 35,56 and 91 leaves remainder of 7 in each case is 3647.

## What is LCM?

The full form of LCM is the Least Common Multiple. A common multiple refers to the number which is a multiple of two or more numbers. The LCM helps us find the smallest such multiple. It can be found via three methods, which include, the prime factorization method, division method, and listing multiples method. Another name for LCM is the Least Common Divisor.

Summary:

## What is the smallest number that, when divided by 35,56 and 91 leaves remainder of 7 in each case?

3674 is the smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case. to reach this solution, we first found the factors of the three numbers provided and then found their LCM. After adding the remainder 7 to it, we got our answer.

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