Question

On dividing a number by 3, 4, 5, 6 as well as 7, we get 1 as remainder in each
case. What can be the smallest such number?
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Answer 1

Answer:

146461

Step-by-step explanation:

Step 1- Find out the LCM of 2, 3, 4, 5, 6 .

=2 x 3 x 2 x 2 x 5 x 2 x 3

=2 x 2 x 3 x 5

= 60

which is 60

Step 2- Add 1 to 60

60+1 = 61

which is 61.

Step 3- Multiple 61 by 7 repeatedly till it fulfills the condition that remainder should be 1.

Step 4- I got the answer 146461 which seems to correct.

Answer 2

421 is the smallest number that leaves a remainder 1 on dividing by 3, 4, 5, 6 as well as 7.

Given:

On dividing a number by 3, 4, 5, 6, or 7, we get 1 as the remainder.

To Find:

The smallest such number.

Solution:

Having a remainder 1 means that our final answer - 1 gives a remainder of zero and is a multiple of 3, 4, 5, 6, and 7, which is essentially the L.C.M. of these numbers.

L.C.M of ( 3, 4, 5, 6, 7) = 420

On dividing 420 by 3, 4, 5, 6 or 7 we get a remainder 0.

So to get a remainder, 1 add the number by 1.

So 420 + 1 = 421 is the smallest number that leaves a remainder of 1 on dividing by 3, 4, 5, 6 as well as 7.

∴ 421 is the smallest number that leaves a remainder of 1 on dividing by 3, 4, 5, 6 as well as 7.

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