Question

find the smallest number which when divided by 39,52and 65, leaves a remainder of 5 in each case

Answer 1

To find the smallest number which when divided by 161, 207 and 184 leaves remainder 21 in all case, we have to compute the L.C.M. of 161, 207 and 184 and then we will add 21 in the L.C.M. of 161, 207 and 184. The number we have after adding will be the required number.

2 | 161, 207, 184
|. ______________
2 | 161, 207, 92
|______________
2 | 161, 207, 46
|______________
3 | 161, 207, 23
|______________
3 | 161, 69, 23
|______________
7 | 161, 23, 23
|______________
7 | 23, 23, 23
|______________
23 | 1, 1, 1
|


L.C.M. of 161, 207 and 184 is 2*2*2*3*3*7*23

= 11592

So. L.C.M. of 161, 207 and 184 is 11592

Now, we will add 21 and 11592

11592 + 21 = 11613

The required smallest number is 11613, which when divided by 161, 207 and 184, leaves a remainder of 21 in each case.

Let us check our answer.

11613 ÷ 161
Quotient = 72 and Remainder = 21

11613 ÷ 207
Quotient = 56 and Remainder = 21

11613 ÷ 184
Quotient = 63 and Remainder = 21