Maybe you would have already noticed, but see there are 2 pairs of 99!,98!2!,97!3!.. 100!.
also there are 2 pairs of 50! factorial, in S(n).
Maybe you could use it, or find a way to use it.

yes,you are right to "guess"....but would you kindly describe to calculate such type of problem at exam hall without spending much of time...thanking you for any help if you yield.......

actually i got the answer what the integer S(n) is...but you see i have to calculate it in short time within hardly 3 minute in hand....so i want to learn the technique behind it.....

@Raja Metronetizen
–
can you give me a simple and neat solution for...this...or few examples...to solve these type of mathematical...operations..using....bases.i still dont...know how to...calculate these...using bases

there, ll be 144 zeroes in this expression, because the largest term in this series is 50!50! and all other terms are smaller and have very less zeroes in them as compared to 144 zeroes in 50!50! therfore, their additon will not make much of a difference in the no of zeroes of whole expression.

so far to my calculation,.....you are absurdly wrong.....there can never be 144 zeroes in the term 50!50! because the number 50!50! only contains 129 decimal digits in base 10....your reason that being other terms so smaller they would not make much of a difference in the number of zeroes also sound to be halting and proofless...actually there is only 22 zeroes at last of the number S(n)....only find the technique that could solve it in time saving process.....i only took help of calculator to do this but you see that's not a process,not yours too....better try...enjoy it...:)

There are 24 zeroes at the end of 50!50!. There are at least 22 zeroes at the end of the sum.
To have a zero at the end, we need a 2 and 5 in the factorization (2*5=10). So to count the zeroes at the end of an expression, you count the number of pairings of 2 and 5. In factorials, we only need to count 5s because there are less 5s than 2s. Counting, we see that there are 10 multiples of 5 in 50!. However, two of these multiples are also multiples of (5^), or 25. We add another 5 for each of these, coming with a total of 12. %0!50! therefore has 24 zeroes or pairs of (25) in its decimal representation in base 10.
After a little more work, it seems as like every number (from 1!99! to 49!51!) except 50!50! could have 22 zeroes at the end. However, 25!75! and 75!25! also have 24 zeroes, as does 50!50 (do you see why?) More work yields that when r is congruent to 0(mod5), it has 23 zeroes. So the least zeroes that any component of the sum is 22; hence there are at least 22 zeroes at the end.
But when adding numbers, we can add more zeros to the end if in some 10^n th digit-place the digits add up to a number congruent to 0(mod10). So there might be more zeroes, but I am too lazy right now/under-educated to go any further.
Concerning the second part, we could divide the sum by 10^z (where z is the number of zeroes at the end) and do some modular arithmetic/Chinese Remainder Theorem or something.

thanx a lot for you gave perfect and neat technique for the solution of the problem.....however i got to the perception for solving such kind of problems...once again thanx....

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## Comments

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TopNewestMaybe you would have already noticed, but see there are 2 pairs of 99!,98!

2!,97!3!.. 100!. also there are 2 pairs of 50! factorial, in S(n). Maybe you could use it, or find a way to use it.Log in to reply

I think there are 22 zeroes at the end.

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yes,you are right to "guess"....but would you kindly describe to calculate such type of problem at exam hall without spending much of time...thanking you for any help if you yield.......

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Actually, nvm. I didn't account for the 23rd digits to add up to a multiple of 10... idk.

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actually i got the answer what the integer S(n) is...but you see i have to calculate it in short time within hardly 3 minute in hand....so i want to learn the technique behind it.....

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plz

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thank u soooo much :)

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isn't there anybody to encounter this tumbling question....?

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there, ll be 144 zeroes in this expression, because the largest term in this series is 50!

50! and all other terms are smaller and have very less zeroes in them as compared to 144 zeroes in 50!50! therfore, their additon will not make much of a difference in the no of zeroes of whole expression.Log in to reply

so far to my calculation,.....you are absurdly wrong.....there can never be 144 zeroes in the term 50!

50! because the number 50!50! only contains 129 decimal digits in base 10....your reason that being other terms so smaller they would not make much of a difference in the number of zeroes also sound to be halting and proofless...actually there is only 22 zeroes at last of the number S(n)....only find the technique that could solve it in time saving process.....i only took help of calculator to do this but you see that's not a process,not yours too....better try...enjoy it...:)Log in to reply

There are 24 zeroes at the end of 50!50!. There are at least 22 zeroes at the end of the sum. To have a zero at the end, we need a 2 and 5 in the factorization (2*5=10). So to count the zeroes at the end of an expression, you count the number of pairings of 2 and 5. In factorials, we only need to count 5s because there are less 5s than 2s. Counting, we see that there are 10 multiples of 5 in 50!. However, two of these multiples are also multiples of (5^), or 25. We add another 5 for each of these, coming with a total of 12. %0!50! therefore has 24 zeroes or pairs of (25) in its decimal representation in base 10. After a little more work, it seems as like every number (from 1!99! to 49!51!) except 50!50! could have 22 zeroes at the end. However, 25!75! and 75!25! also have 24 zeroes, as does 50!50 (do you see why?) More work yields that when r is congruent to 0(mod5), it has 23 zeroes. So the least zeroes that any component of the sum is 22; hence there are at least 22 zeroes at the end. But when adding numbers, we can add more zeros to the end if in some 10^n th digit-place the digits add up to a number congruent to 0(mod10). So there might be more zeroes, but I am too lazy right now/under-educated to go any further. Concerning the second part, we could divide the sum by 10^z (where z is the number of zeroes at the end) and do some modular arithmetic/Chinese Remainder Theorem or something.

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thanx a lot for you gave perfect and neat technique for the solution of the problem.....however i got to the perception for solving such kind of problems...once again thanx....

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