Problem:
There are integers a,b, and c, each greater than 1 , such that
aNbNcN=36N25
for all N>1. What is b ?
Answer Choices:
A. 2
B. 3
C. 4
D. 5
E. 6
Solution:
aNbNcN can be simplified to Na1+ab1+abc1.
The equation is then Na1+ab1+abc1=N3625 which implies that a1+ab1+abc1=3625.
a has to be 2 since 3625>127. 127 is the result when a,b, and c are 3,2 , and 2
b being 3 will make the fraction 32 which is close to 3625.
Finally, with c being 6 , the fraction becomes 3625. In this case a,b, and c work, which means that b must equal
(B)3.
~lopkiloinm
OR
As above, notice that you get a1+ab1+abc1=3625.
Now, combine the fractions to get abcbc+c+1=3625.
Let us assume bc+c+1=25 and abc=36 as this is the most convenient. (EDIT: This used to say WLOG but that is inaccurate)
From the first equation, we get c(b+1)=24. Note also that from the second equation, b and c must be factors of 36 .
After listing out the factors of 36 and utilising trial and error, we find that c=6 and b=3 works, with a=2. So our answer is (B)3 .
The problems on this page are the property of the MAA's American Mathematics Competitions