Amazing problem!

Hello friends, We have to find the last two digits of 1×3×5×7×9....×991\times3\times5\times7\times9....\times99, I did and came out with 25.But I want to verify.

Note by Kishan K
6 years, 3 months ago

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11 votes

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From the fact that it is divisible by 2525 but not by 22, we can deduce that it ends with either 2525 or 7575. Now, consider the multiplications modulo 44: 1×3×5×7×9××991×3×1×3××3125×325125×(1)25175≢25 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 75\boxed{75}.

Tim Vermeulen - 6 years, 3 months ago

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Argh, you beat me to it xD Well done.

Sotiri Komissopoulos - 6 years, 3 months ago

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

Budha Chaitanya - 6 years, 2 months ago

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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 - 6 years, 2 months ago

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refer elemantary number theory by David M Burton

Naga Teja - 6 years, 2 months ago

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

Kishan k - 6 years, 3 months ago

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

Sotiri Komissopoulos - 6 years, 3 months ago

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

Krishna Jha - 6 years, 3 months ago

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@Krishna Jha That is simply the individual remainders left when 44 divides the given odd numbers which will obviously be 33 or 11 and you can easily find out the number of such terms which yield 11 or 33 as remainder.

Aditya Parson - 6 years, 3 months ago

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

Budha Chaitanya - 6 years, 2 months ago

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@Budha Chaitanya Because 44 divides 100100, and looking at the last two digits of a number is essentially considering de remainder when that number is divided by 100100. As 44 divides 100100, and we know that the number is one less than a multiple of 44, then the last two digits of the number are also one less than a multiple of 44. Hence, it's a nice trick to use here, as 7575 is one less than a multiple of 44, and 2525 is not.

Tim Vermeulen - 6 years, 2 months ago

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@Tim Vermeulen thank you

Budha Chaitanya - 6 years, 2 months ago

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

Krishna Jha - 6 years, 3 months ago

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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 - 6 years, 3 months ago

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