Question

Find the smallest no. That when divided by 35;56;91 leaves remainders of 7 in each case

Answer 1

Answer:

The smallest number is 7.

The next smallest number is 3647.

Step-by-step explanation:

7 has that property!

• 7 divided by 35 is 0 with a remainder of 7
• 7 divided by 56 is 0 with a remainder of 7
• 7 divided by 91 is 0 with a remainder of 7

But what about the next smallest number?

We just need to add the lcm of 35, 56 and 91.

The factorizations are:  35 = 5×7, 56 = 7×8, 91 = 7×13.

So their lcm is 5×7×8×13 = 3640.

So the next smallest number that leaves a remainder of 7 each time is 3640+7 = 3647.

Answer 2

$\sf\underline{Step-by-step \: explanation}$

Step - 1 : Find Lcm of 35 , 56 and 91

7 | 35 , 56 , 91

2 | 5 , 8 , 13

2 | 5 , 4 , 13

2 | 5 , 2 , 13

5 | 5 , 1 , 13

13 | 1 , 1 , 13

1 | 1 , 1 , 1

Lcm = 7 × 2 × 2 × 5 × 13

Lcm = 3640

Step - 2 : Add 7 to the Lcm

3640 + 7 = 3647

Hence; the smallest number which will be divided by 35 , 56 and 91.

Verification :

3647 ÷ 35 = 104 (Remainder = 7)

3647 ÷ 56 = 65 (Remainder = 7)

3647 ÷ 91 = 40 (Remainder = 7)