Find all five digit numbers such that the number formed by deleting the middle digit divides the original number.

Source: Canadian math. contest

If abcde is the 5 digit number then abde is the 4 digit number.
Is it clear that abcde/abde must be roughly 10.
In fact the smallest quotient appears to be 10099/1099≈9.189 and the largest 10900/1000=10.9
So the quotient must be 10.
10*abde = abcde so e must be 0
10*abd0 = abcd0 so d must be 0
10*ab00 = abc00 so c must be 0
10*ab00 = ab000.
This works for any values of a and b.
So all 90 five digit multiple of 1000 work.

Incidentally if we allow deletions of the second digit there are additional solutions like 12100/1100 = 11
d=e=0, a+c=b

Posted by Jer on 2014-12-26 20:41:27