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Trailing Zeros

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Post questions based on trailing zeros in this forum 

This topic was modified 1 month ago by TathaGat
9 questions & discussions are there under this sub-topic
0

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

TG Team 17/07/2018 11:32 am

 

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 . 

0

What power of 8 exactly divides 25! ?

TG Team 15/12/2018 2:26 pm
Capture

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 

0

Find the number of trailing zeroes in 142!

(a) 36    (b) 30     (c) 34    (d)  32

Neraj (Tg)Faculty 28/04/2024 5:52 pm
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image

(a) 2    (b) 192    (c) 199    (d) 7 

Neraj (Tg)Faculty 10/06/2024 6:41 pm

@neha-singh 

(800)!

number of trailing zeroes = 800/5 + 160/5 + 32/5 + 6/5

= 160 + 32 + 6 + 1 = 199

0

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

TathaGat 09/06/2024 1:21 am

@mayanksri 

Please find the solution below.

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.

0

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

100!*200!

TathaGat 09/06/2024 1:24 am

@mayanksri 

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

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

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

LydiaMadini 09/06/2024 1:24 am

@anna 

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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The number of zeroes at the end on N

N =  4 x 8 x 12 . . . .1000

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

TathaGat 09/06/2024 1:27 am

@mayanksri 

Please find the solution below,

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.

0

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

Neraj (Tg)Faculty 10/06/2024 6:31 pm