Problem:
For every integer n≥2, let pow (n) be the largest power of the largest prime that divides n. For example pow (144)=pow(24⋅32)=32. What is the largest integer m such that 2010m divides
n=2∏5300pow(n)?
Answer Choices:
A. 74
B. 75
C. 76
D. 77
E. 78
Solution:
Observe that 2010=2⋅3⋅5⋅67. Let P=∏n=25300pow(n)= 2a⋅3b⋅5c⋅67d⋅Q where Q is relatively prime with 2,3,5, and 67 . The largest power of 2010 that divides P is equal to 2010m where m=min(a,b,c,d).
By definition pow (n)=2k if and only if n=2k. Because 212=4096<5300< 8192=213, it follows that
a=1+2+⋯+12=212⋅13=78
Similarly, pow (n)=67 if and only if n=67N and the largest prime dividing N is smaller than 67 . Because 5300=79⋅67+7 and 71,73 , and 79 are the only primes p in the range 67<p≤79; it follows that for n≤5300, pow (n)=67 if and only if
n∈{67k:1≤k≤79}\{672,67⋅71,67⋅73,67⋅79}
Because 672<5300<2⋅672, the only n≤5300 for which pow (n)=67k with k≥2, is n=672. Therefore
d=79−4+2=77
If n=2j⋅3k for j≥0 and k≥1, then pow (n)=3k. Moreover, if 0≤j≤2 and 1≤k≤6, or if 0≤j≤1 and k=7; then n=2j⋅3k≤2⋅37=4374<5300. Thus
b≥3(1+2+⋯+6)+7+7=3⋅21+14=77
If n=2i⋅3j⋅5k for i,j≥0 and k≥1, then pow (n)=5k. Moreover, If 2i⋅3j∈ {1,2,3,22,2⋅3,23,32,22⋅3} and 1≤k≤3, or if 2i⋅3j∈{1,2,3,22,2⋅3,23} and k=4, or if 2i⋅3j=1 and k=5; then n=2i⋅3j⋅5k≤8⋅54=5000<5300. Thus
c≥8(1+2+3)+6⋅4+5=77
Therefore m=d=77.
The problems on this page are the property of the MAA's American Mathematics Competitions