Trailing Zeros

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# Trailing Zeros

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Find the no. of consecutive zeros at the end of the following no.

100!*200!

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How to calculate the last rightmost non zero digit in 30!

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Hello Vidisha,

30! = 2^26 x 3^14 x 5^7 x 7^4 x 11^2 x 13^2 x 17^1 x 19^1 x 23^1 x 29^1

Now the for the first non zero digits,lets first remove the terms which are causing the trail of zeroes at the end...which is 5^7 x 2^7....so u r left with....

2^19 x 3^14 x 7^4 x 11^2 x 13^2 x 17^1 x 19^1 x 23^1 x 29^1

the last digit of this pdt will be 8 x 9 x 1 x 1 x 9 x 7 x 9 x 3 x9 = 8..
so the first non zero digit from the right will be 8.

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Find the no. of consecutive zeros at the end of the following no.

1!*2!*3!*4!*5*................................*50!

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Power of 5 in 1!*2!*3!*4!*5*................................*50! = 262. Thus number of zeros at the end of this num = 262.

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power of 5 in 100!*200!=73. Thus number of zeros at the end of this num = 73.

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The number of zeroes at the end on N

N =  4 x 8 x 12 . . . .1000

a. 11                 b. 6              c. 16                     d. 62

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Hello, Vidisha.nagpal

N = 4*8*12*....*1000 = (4*1)*(4*2)*(4*3)*.....*(4*250) = 4250*(1*2*3*...250) = 2500*250!
Now 250! = 2244*3123*562*740*1124*1320*1714*1913*2310*298*318*376*......
Hence the number of zeroes at the end of N is 62.

Nice instruction, thanks

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How many zeroes are present at the end of 25!+26!+27!+28!+30!?

Hi Samyak

25! ( 1 + 26 + 26 × 27 + 26 × 27 × 28 + 26 × 27 × 28 × 29 × 30 )

25! ( 1 + 26 + 26 × 27 + 26 × 27 × 28 + 26 × 27 × 28 × 29 × 30 )

25! has 6 trailing zeros and, the term inside the bracket is divisible by 5

Hence, 6 +1  , 7 trailing zeros .

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What power of 8 exactly divides 25! ?

Highest power of 2 in 25! = [25/2] + [25/2^2] + [25/2^3] +........

12 + 6 + 3 + 1 = 22

So , the highest power of 8 in 25! = [22/3] = 7

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For how many positive integers does n! End with exactly 100 zeroes?

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Good

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ok

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tuyet voi

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good

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