<s>[INST] Interview Question -- confused about the answer If $a$, $b$ and $c$ are positive integers then how many possible solutions exist to the equation $a! = 5b! + 27c!$
A. No Solution
B. $1$ Solution
C. $2$
D. $3$
E. Infinite Solution
 [/INST] The only solution is $32!=5.31!+27.31!$.
Otherwise if $b\ne c$, then either
(i) $b<c<a$, so $$a(a-1)\cdots(b+1)=5+27c(c-1)\cdots(b+1)$$ In this case $b+1=5$ since it is a common factor. There can be no other common factors, so $c=b+1=5$. But $a!=5.4!+27.5!$ has no solutions.
(ii) $c<b<a$, so $$a(a-1)\cdots(c+1)=5b\cdots(c+1)+27$$. In this case $c+1$, $c+2$,$\ldots$, must divide $27$. Since consecutive numbers cannot be multiples of $3$, $b=c+1=3,9,27$. But $a!=5.3!+27.2!=14.3!$, $a!=5.9!+27.8!=8.9!$, and $a!=5.27!+27.26!= 6.27!$ have no solutions.
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