Problem:
The number obtained from the last two nonzero digits of 90! is equal to n. What is n?
Answer Choices:
A. 12
B. 32
C. 48
D. 52
E. 68
Solution:
There are 18 factors of 90! that are multiples of 5,3 factors that are multiples of 25, and no factors that are multiples of higher powers of 5. Also, there are more than 45 factors of 2 in 90!. Thus 90!=1021N where N is an integer not divisible by 10, and if N≡n(mod100) with 0<n≤99, then n is a multiple of 4.
Let 90!=AB where A consists of the factors that are relatively prime to 5 and B consists of the factors that are divisible by 5. Note that ∏j=14(5k+j)≡ 5k(1+2+3+4)+1⋅2⋅3⋅4≡24(mod25), thus
A=(1⋅2⋅3⋅4)⋅(6⋅7⋅8⋅9)⋯⋯(86⋅87⋅88⋅89)≡2418≡(−1)18≡1(mod25)
Similarly,
B=(5⋅10⋅15⋅20)⋅(30⋅35⋅40⋅45)⋅(55⋅60⋅65⋅70)⋅(80⋅85⋅90)⋅(25⋅50⋅75), thus
521B=(1⋅2⋅3⋅4)⋅(6⋅7⋅8⋅9)⋅(11⋅12⋅13⋅14)⋅(16⋅17⋅18)⋅(1⋅2⋅3)≡243⋅(−9)⋅(−8)⋅(−7)⋅6≡(−1)3⋅1≡−1(mod25)
Finally, 221=2⋅(210)2=2⋅(1024)2≡2⋅(−1)2≡2(mod25), so 13⋅221≡ 13⋅2≡1(mod25). Therefore
N≡(13⋅221)N=13⋅52190!=13⋅A⋅521B≡13⋅1⋅(−1)(mod25)≡−13≡12(mod25)
Thus n is equal to 12,37,62, or 87, and because n is a multiple of 4, it follows that n=12.
The problems on this page are the property of the MAA's American Mathematics Competitions