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# 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.

Note by Kishan K
4 years ago

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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 $$\boxed{75}$$. · 4 years ago

Argh, you beat me to it xD Well done. · 4 years ago

how these type of shortcuts can be found? I want to know .if in any website it is plz tell me. · 4 years 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. · 4 years ago

refer elemantary number theory by David M Burton · 4 years ago

I think you used Chinese remainder theorem.If not then please tell me the complete theorem. · 4 years ago

He did not use Chinese Remainder Theorem. Which part of his solution are you confused with? · 4 years ago

I am confused over the multiplication modulo 4... · 4 years ago

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. · 4 years ago

why only with 4, why not other? · 4 years ago

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. · 4 years ago

thank you · 4 years ago

Its 75... courtesy mathematica ... :-P...Bt how do we solve these sort of problems is still a mystery to me.. · 4 years ago