**Problem 4 of the Canadian Mathematical Olympiad 1970**

a) Find all positive integers with initial digit 6 such that the integer formed by deleting this 6 is 1/25 of the original integer.

b) Show that there is no integer such that deletion of the first digit produces a result which is 1/35 of the original integer.

**Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to George C. Peregoy of Morgan Hill, CA)**

a) Let N = N_{0}N_{1}N_{2}...N_{n} ( n –> infinity ) be such integers.

N_{0} = 6 and N = 6N_{1}N_{2}...N_{n}

By deleting the initial digit 6, we have M = N_{1}N_{2}...N_{n} and N / M = 25

and N – M = 60...0 (with n 0’s) = 24M or (N - M) /24 = M =

60...0 / 24 = 6 × 10...0 / (6 × 4) = 10...0 / 4 = 250...0 (with n-2 0’s)

So N = 625, 6250, 62500, 625000, 6250000, 62500000, etc…

b) With the similar approach, assuming there are such integers N = N_{0}N_{1}N_{2}……..N_{n}
( n –> infinity ) where

N_{0} = 1-9 (integers) and N = N_{0}N_{1}N_{2}...N_{n}

By deleting the initial digit, we have M = N_{1}N_{2}...N_{n} and N / M = 35

and N – M = N_{0}0...0 (with n 0’s) = 34M or (N - M) /34 = M = N_{0}0...0 / 34 = N_{0} × 10...0 / 34

and since N_{0} takes on the integer values of 1-9, N_{0} × 10...0 is not divisible by 34. Therefore, there are no such integers N as we
had assumed.