Last two digits of odd products

Question

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?

Answer 1

One of the terms in $1\cdot3\cdot5\cdot...\cdot99$ will be $25$, so indeed the answer is either $25$ or $75$. Note that $25k$ ends in $25$ if $k\equiv1$ (mod $4$) and $25k$ ends in $75$ if $k\equiv3$ (mod $4$). We have $$1\cdot3\cdot5\cdot...\cdot99\equiv1\cdot3\cdot1\cdot...\cdot1\cdot3=3^{25}=(3^2)^{12}\cdot3\equiv1^{12}\cdot3=3\mbox{ (mod }4)$$ and $25\equiv1$ (mod $4$), so $k\equiv3$ (mod $4$). Conclusion: The last two digits of $1\cdot3\cdot5\cdot...\cdot99$ are $75$.

Indeed, as Bill Dubuque pointed out, our key observation can be written as $25k\mbox{ (mod }25\cdot4)=25\cdot(k\mbox{ (mod }4))$. In this particular case it can be easily seen by just a little bit of experimentation and a proof by induction is also not very difficult, however writing it in this form shows that this is a particular case of a much more general fact. See the mod distributive law for details.

Answer 2

Hint: The number $5^{10}$ ends with $25$. On the other hand, if $k$ is odd, then the number $25k$ ends with:

• $25$ if $k=4n+1$ for some non-negative integer $n$;
• $75$ if $4=4n+3$ for some non-negative integer $n$.

Answer 3

Hint $\ \ \ 25\mid n\:\Rightarrow\: n\bmod 100 = 25(n/25 \bmod 4) = 25(\color{#c00}{n\bmod 4}) = 25\cdot\color{#c00}3$

since here $\!\color{#c00}{\bmod 4\!:\ n }\equiv (1\cdot 3)(5\cdot 7)\cdots (97\cdot 99)\equiv (1\cdot(-1))^{\large25}\!\equiv -1\equiv\color{#c00} 3$

Answer 4

$1,4 + 1=5, 2*8+1=9,....,4*24+1=97$ 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 $N\pmod{100 = 4*25}\equiv 75$.

...

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 $N \equiv 3+4k = 25j = 75 \pmod {100}$