Problem:
Suppose a,b, and c are positive integers with a+b+c=2006, and a!b!c!=m⋅10n, where m and n are integers and m is not divisible by 10 . What is the smallest possible value of n ?
Answer Choices:
A. 489
B. 492
C. 495
D. 498
E. 501
Solution:
Note that n is the number of factors of 5 in the product a!b!c!, and 2006<55. Thus
n=k=1∑4(⌊a/5k⌋+⌊b/5k⌋+⌊c/5k⌋)
Because ⌊x⌋+⌊y⌋+⌊z⌋≥⌊x+y+z⌋−2 for all real numbers x,y, and z, it follows that
n≥k=1∑4(⌊(a+b+c)/5k⌋−2)=k=1∑4(⌊2006/5k⌋−2)=401+80+16+3−4⋅2=492.
The minimum value of 492 is achieved, for example, when a=b=624 and c=758.
The problems on this page are the property of the MAA's American Mathematics Competitions