Question

10. Find the least number of seven digits which when
divided by 15, 20, 25, 30, 35 and 40 leaves remainders
10, 15, 20, 25, 30 and 35, respectively.​

Answer 1

Answer:

Let NN be the smallest number.

This is a question of multiple divisors with multiple remainders.

Although the remainders seem different, but they are the same when we take their negative counterparts.

R[N20]=14R[N20]=14 or (−6)(−6)

R[N25]=19R[N25]=19 or (−6)(−6)

R[N35]=29R[N35]=29 or (−6)(−6)

R[N40]=34R[N40]=34 or (−6)(−6)

So, we can see NN leaves the same remainder (-6) when divided by 20, 25, 35 or 40.

So, N=LCM(20,25,35,40)+(−6)N=LCM(20,25,35,40)+(−6)

Or, N=1394N=1394 (Answer)