Problem:
How many positive integer multiples of 1001 can be expressed in the form 10jβ10i, where i and j are integers and 0β€i<jβ€99?
Solution:
Because
10jβ10i=10i(10jβiβ1)
and 1001=7β
11β
13 is relatively prime to 10i, it is necessary to find i and j so that 10jβiβ1 is divisible by the primes 7,11, and 13. Notice that 106 is the smallest power of 10 that leaves a remainder of 1 when divided by 7 or 13, and that 102 is the smallest power of 10 that leaves a remainder of 1 when divided by 11. Hence 10i(10jβiβ1) is divisible by 1001 if and only if jβi=6n for some positive integer n. Thus it is necessary to count the number of integer solutions to
i+6n=j, where jβ€99,iβ₯0,n>0
For each n=1,2,3,β¦,16, there are 100β6n suitable values of i (and j), so the number of solutions is
94+88+82+β―+4=784β
The problems on this page are the property of the MAA's American Mathematics Competitions