$33^{33}$ is the sum of $33$ consecutive odd numbers. Which one is the largest? (Q25 from AMC 2012) 
The number $33^{33}$ can be expressed as the sum of $33$ consecutive odd numbers. The largest of these odd numbers is
$\mathrm{A.}\ 33^{32} +32$
$\mathrm{B.}\ 33^{31} +32$
$\mathrm{C.}\ 33^{32} -32$
$\mathrm{D.}\ 33^{31} -32$
$\mathrm{E.}\ 33^{32}$

So no calculator is allowed, when I tried using the calculator,the answer is A.Why is it so and is there another way to do this question without using the calculator?
 A: $33^{33}=33\times 33^{32}+(-32-30-\cdots -2-0+2+\cdots +30+32)=$
$(33^{32}-32)+(33^{32}-30)+\cdots + (33^{32})+\cdots +(33^{32}+30)+(33^{32}+32)$
As $33^{32}$ is an odd number, then $(33^{32}\pm 2k)$ is odd as well. So, the largest one is $(33^{32}+32)$.
A: First we prove that $1+3+5+\ldots +(2n-1) = n^2$ by induction:
The base case $2\cdot 1-1=1=1^1$ holds. Suppose the claim is true for $n$. Then $1+3+\ldots + 2(n+1)-1 = (1+\ldots +2n-1)+(2n+1) = n^2+2n+1 = (n+1)^2$, as desired.
Now, starting at number $m$, we have that $(m+1)+(m+3)+\ldots+(m+2\cdot 33-1)=33^{33}$. Hence $33m+33^2=33^{33}$, and therefore $m=33^{32}-33$ and so the last number is $m+2\cdot 33-1 = 33^{32}+33-1 = 33^{32}+32$.
A: The sum of an arithmetic series is $\frac n2(a+l)$, where $n$ is the number of terms and $a$ and $l$ are the first and last terms respectively. You can think of it as the number of terms multiplied by the arithmetic means (simple average) of the bounding terms.
Here, $n=33$ and $l = a+(33-1)(2) = a + 64$ so the sum is $\frac{33}{2}(a + a + 64) = 33(a+32)$. Equating that to $33^{33}$, we find that $a + 32 = 33^{32}$, giving $l = a + 32 + 32 = 33^{32} + 32$.
A: The average of those $33$ consecutive odd numbers is $33^{33}/33=33^{32}$. And $16$ of them are larger than the average, so the largest is...
A: We have $x-32,x-30,\ldots,x,\ldots,x+30,x+32$ and thus $33x = 33^{33}$. Thus, the largest number is $33^{32}+32$.
