Last two digits of odd products Is there any proof where we can find the last two digits of odd number product.
$$1\cdot3\cdot5\cdot7\cdot\ldots\cdot99 = a_i$$ Find the last two digits of $a_i$. The answer would be $75$ as any multiples of $5$ will always end with $5$.
Are there any solid proof or trick to find the last two digits of the above product?

$1,4 + 1=5, 2*8+1=9,....,4*24+1=97$ are are $1 \pmod 4$
$3,7,......, 4*24+3=99$ are all $-1\pmod 4$.
So $N = 1\cdot 3.... \cdot 97\cdot 99 \equiv 1^{25}(-1)^{25}\equiv -1\equiv 3\pmod 4$.
And obviously $N \equiv 0 \pmod {25}$.
So by Chinese Remainder theorem we can always solve $N\pmod{100 = 4*25}\equiv 75$. 
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Most statements of CRT give the formula for solving but I can never remember the proper variables so I always do it by hand each time.
In this case $3 + 4k = 25j;0\le k < 24; 0\le j < 4;$.  So $  3 + 4k= j + 4(6j)$.  $j = 3; k=6j=18$ 
and $N \equiv 3+4k = 25j = 75 \pmod {100}$
