Question

find the least number when divided by 35,56 and 91 leave the same remainder 7 in each case

Answer 1

The least number is 7
As if 35 is devided by 7
Then 7*4 = 28, balance is 7
In case of 56
7*7 = 49, balance is 7
In case of 91
7*12 = 84, balance is 7
So it is clear that 7 is the least no. With given requirement

Answer 2

The least number is 3647

Given:

The number leaves 7 as remainder in each case when divided by 35, 56 and 91

To find:  

The “Least number”

Solution:

Now, find the L.C.M of the three numbers 35, 56, 91

$\begin{array}{c}{35=7 \times 5} \\ {56=7 \times 2 \times 2 \times 2} \\ {91=7 \times 13}\end{array}$

LCM of 35 , 56 , 91 $=7 \times 5 \times 2 \times 2 \times 2 \times 13=3640$

Since the number leaves a remainder add 7 to the LCM of numbers obtained,  

7,3640 + 7 = 3647.