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9 Apr 2010 21:49:15 IST

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how to find no of zeros at the end of 100!

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how to find no of zeros at the end of 100!

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ROHIT SHARMA

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9 Apr 2010 22:38:24 IST

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its simple...........the exponent of prime p in n ! is [n/p]+[n/psquare....so on..............so find the exponents of all such primes in 100! and then multiply those of them that give 0 at the end

Harshul bhutani

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9 Apr 2010 22:41:53 IST

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acc. to me ans would be 24

i will show u

no like 50,25,75 when multiplied with some small no (2,4,6) then extra 0 is generated

no ending with 5 will also generate 0

NUMBER ZEROES NUMBER ZEROES

100 2 95 1

90 1 85 1

80 1 75 2

70 1 65 1

60 1 55 1

50 2 45 1

40 1 35 1

30 1 25 2

20 1 15 1

10 1 5 1

TOTAL = 12 TOTAL = 12 TOTAL NO OF 0=24

Himaya Iyer

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9 Apr 2010 22:53:41 IST

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see..the number of zeroes in 100! is equal to no of zeroes in 5! no of zeroes in 5!=(100\5)+(100\25)so the no of zeroes is 24.

Genius Dude

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10 Apr 2010 20:05:34 IST

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the no. of zeroes is equal no of times 2x5 appears in the expansion

as there are sure plenty of 2s in there so ur worry is find no of 5

apply the prime exponent rule to find it

no of occurences of p in n! is [n/p] + [n/p^2] +[n/p^3]+....

so no of occurences of 5=[100/5] + [100/25] =20+4=24

this is the required no of zeroes

NOTE: [x] is the greatest integer function and yields 0 for decimal only so no need to continue division after 0.something

edison

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7 May 2010 15:32:48 IST

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We factorize 1.2.3.4......100 in which we find how many times 5 occurs which will multiply with even numbers to produce zeros at the end of the product.

5,10,15,20,30,35,40,45,55,60,65,70,80,85,90,95 each contain one 5 as their factors.
Number of 5's due to these in the product = 16

25 = 5x5 has two 5's as factors
similarly 50 = 5x5x2 and 75 = 5x5xx3 and 100 = 5x5x4 each have two 5's.
so no. of 5's due to these 4 numbers = 8

hence if we factorize 1.2.3.4.....100 total no. of 5's which came as factors = 16+8 = 24

all these 24 fives would multiply with even numbers to produce 22 zeros.

Alternatively : If p is a prime no. then its power in n! i.e. 1.2.3.....n is given by :

E = [n/p] + [n/p²] + [n/p³] + ......
where [.] is the greatest integer function.

here n=100 and p=5

E = [100/5] + [100/5²] + [100/5³] + ....

E = [20] + [4] + [0.8] + .....

except 1st two numbers all the numbers in brackets are less than and 1 and hence their greatest integers are 0.

so, E = 20 + 4 = 24

so 5 would occur 24 times in the expression and they would multiply with even numbers to produce 24 zeros.

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