We increase the fifth power of an odd natural number by 3. This number is divisible by:
(A) 4
(B) 5
(C) 4 and 5
(D) only 1
(E) there is no such number
A5. Let the odd number be $2 n-1$. The square of this number is $4 n^{2}-4 n+1$. We add 3 to this number and get $4 n^{2}-4 n+1+3$, factor out 4 and we get $4\left(n^{2}-n+1\right)$. The obtained number is a multiple of 4. Therefore, the number is divisible by 4.