Question

What is the smallest
number that when divided by
35, 56, and 91 leaves
remainder of 7 in each case.
Step by step explanation

Answer 1

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Answer 2

Here,

$35 = 5 \times 7 \\ 56 = 2 \times 2 \times 2 \times 7 \\ 91 = 7 \times 13$

• Smallest number dividing each of 35, 56, & 91

$= > 5 \times 7 \times 2 \times 2 \times 2 \times 13 \\ = > 3640$

So, 3640 is divided by 35, 56, & 91 without leaving a remainder.

Thus,

$3640 + 7 = \pink{ 3647}$

The number obtained⬆️⬆️ i.e. 3647 is that required number which will leave 7 as remainder when divided by 35, 56, and 91.