Problem:
Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is 1/29 of the original integer.
Solution:
The desired integer has at least two digits. Let d be its leftmost digit, and let n be the integer that results when d is deleted. Then for some positive integer p,10p⋅d+n=29n, and so 10p⋅d=28n. Therefore 7 is a divisor of d, and because 1≤d≤9, it follows that d=7. Hence 10p=4n, so n=410p=4100⋅10p−2=25⋅10p−2. Thus every positive integer with the desired property must be of the form 7⋅10p+25⋅10p−2=10p−2(7⋅102+25)=725⋅10p−2 for some p≥2. The smallest such integer is 725.
OR
The directions for the AIME imply that the desired integer has at most three digits. Because it also has at least two digits, it is of the form abd or cd, where a,b,c, and d are digits, and a and c are positive. Thus bd⋅29=abd or d⋅29=cd. Note that no values of c and d satisfy d⋅29=cd, and that d must be 0 or 5. Thus b0⋅29=ab0 or b5⋅29=ab5. But b0⋅29=ab0 implies b⋅29=ab, which is not satisfied by any values of a and b. Now b5⋅29=ab5 implies that b5<1000/29<35, and so b=1 or b=2. Because 15⋅29=435 and 25⋅29=725, conclude that the desired integer is 725.
The problems on this page are the property of the MAA's American Mathematics Competitions