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 \(\boxed{75}\).

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.

@Krishna Jha
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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.

@Budha Chaitanya
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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.

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.

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## Comments

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TopNewestFrom 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 \(\boxed{75}\).

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Argh, you beat me to it xD Well done.

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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.

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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.

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refer elemantary number theory by David M Burton

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

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

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

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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.

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