Amazing problem!

Hello friends, We have to find the last two digits of $1\times3\times5\times7\times9....\times99$ , I did and came out with 25.But I want to verify.

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

From the fact that it is divisible by $25$ but not by $2$ , we can deduce that it ends with either $25$ or $75$ . Now, consider the multiplications modulo $4$ : $1 \times 3 \times 5 \times 7 \times 9 \times \dots \times 99 \equiv 1 \times 3 \times 1 \times 3 \times \dots \times 3 \equiv 1^{25} \times 3^{25} \equiv 1^{25} \times (-1)^{25} \equiv -1 \equiv 75 \not\equiv 25$ , so the answer is $\boxed{75}$ .

Tim Vermeulen - 7 years, 11 months ago

Argh, you beat me to it xD Well done.

Sotiri Komissopoulos - 7 years, 11 months ago

how these type of shortcuts can be found? I want to know .if in any website it is plz tell me.

Budha Chaitanya - 7 years, 11 months ago

I did not use any website. In fact, I never solved such a problem before. I guess you'll just need to be familiar with prime factors and the modulo operator.

Tim Vermeulen - 7 years, 11 months ago

refer elemantary number theory by David M Burton

Naga Teja - 7 years, 11 months ago

I think you used Chinese remainder theorem.If not then please tell me the complete theorem.

Kishan k - 7 years, 11 months ago

He did not use Chinese Remainder Theorem. Which part of his solution are you confused with?

Sotiri Komissopoulos - 7 years, 11 months ago

@Sotiri Komissopoulos – I am confused over the multiplication modulo 4...

Krishna Jha - 7 years, 11 months ago

@Krishna Jha – That is simply the individual remainders left when $4$ divides the given odd numbers which will obviously be $3$ or $1$ and you can easily find out the number of such terms which yield $1$ or $3$ as remainder.

Aditya Parson - 7 years, 11 months ago

@Aditya Parson – why only with 4, why not other?

Budha Chaitanya - 7 years, 11 months ago

@Budha Chaitanya – Because $4$ divides $100$ , and looking at the last two digits of a number is essentially considering de remainder when that number is divided by $100$ . As $4$ divides $100$ , and we know that the number is one less than a multiple of $4$ , then the last two digits of the number are also one less than a multiple of $4$ . Hence, it's a nice trick to use here, as $75$ is one less than a multiple of $4$ , and $25$ is not.

Tim Vermeulen - 7 years, 11 months ago

@Tim Vermeulen – thank you

Budha Chaitanya - 7 years, 11 months ago

Its 75... courtesy mathematica ... :-P...Bt how do we solve these sort of problems is still a mystery to me..

Krishna Jha - 7 years, 11 months ago

There is no theorem that says "oh, if you multiply the first 50 odd digits then the last 2 digits are 75." Rather, it takes intuition along with a knowledge of mathematical theorems. In this case, somebody figured "okay, well, there are more than 1 factors of "5" in this, but no factors of 2 so it must end in either 25 or 75. Then, to figure out if it was a 75 or 25 at the end, he counted the numbers of factors of 3, because any number ending in 25 when multiplied by 3 ends with 75, and any number ending in 75 multiplied by 3 ends with 25.

Michael Tong - 7 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$