Problem:
Find the smallest positive integer whose cube ends in 888.
Solution:
If the cube of an integer ends in 8, then the integer itself must end in 2; i.e., must be of the form 10k+2. Therefore,
n3=(10k+2)3=1000k3+600k2+120k+8(1)
where the penultimate term, 120k, determines the penultimate digit of n3, which must also be 8. In view of this, 12k must also end in 8; i.e., k must end in 4 or 9, and hence be of the form 5m+4. Thus
n3​=(10(5m+4)+2)3=125000m3+315000m2+264600m+74088​(2)
Since the first two terms on the right of (2) end in 000, while the last term ends in 088, it follows that 264600m must end in 800. The smallest m which will ensure this is m=3, implying that k=5⋅3+4=19, and n=10⋅19+2=192​. (Indeed, 1923=7,077,888.)
The problems on this page are the property of the MAA's American Mathematics Competitions