Upto how many integers we should take the product 1*2*3*4*5.......... to make 31 zeros at last of the number.

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08 Apr 2014 Solve Upto how many integers we should take the product 1*2*3*4*5.......... to make 31 zeros at last of the number.: OPtion
1) 140
2) 130
3) 125
4) 110
5) 100
6) 105
7) 135
8) 115
9) 145
10)None of these; Correct Option: 3 Best Solution (24)

Dr.Dipin Singh 10 years AGO

No of zeros at last of product 1*2*3*4*5.......n = n/5 +n/25+n/125 =31
n(25+5+1)=31*125
31*n=31*125
n=125 .... option 3

Like? Yes (10) | No (3)

Aman 10 years AGO

As
125/5 +125/25 +125/125= 25+5+1=31
hence we should take 125 integers for the product 1*2*3*4*5.......... to make 31 zeros at last of the number.

Like? Yes (9) | No (3)

Saranjit Singh 10 years AGO

zeros at last of number in n!= n/5 +n/25+n/125+n/625
when n=125,
no. of zeros= 125/5+125/25+125/125=31

Like? Yes (7) | No (2)

Pargat Pal Singh 10 years AGO

1*2*3*4*5.....n = n!
no of zeros at end of n! is given by
n/5 +n/25+n/125.
so n/5 +n/25+n/125 = 31
n=125

Like? Yes (7) | No (3)

llamar 10 years AGO

1-10 you get 2 zero
10-20 u get 2 as well
...............upto 100 u get 21 zeros then after
100-110=>4 zero
110-120=>4 zeros
125 u get 2 zeros

Like? Yes (1) | No

vishnuvardhan 10 years AGO

125!=
[125/5]+[125/25]+[125/125]=25+5+1=31

Like? Yes (1) | No

Vaishali Dhiman 10 years AGO

ans is 125

Like? Yes (1) | No

AMAN 10 years AGO

125/5=25,125/25=5,125/125=1,add them 31

Like? Yes (1) | No (2)

ashish kumar 10 years AGO

for this the rule is
(N/5)+(N/5^2)+(N/5^3)+(N/5^4)+.......
take N =125 from option u will dirtectly able to tel l that this is answer.
(125/5)+(125/5^2)+(125/5^3)+.....
25+5+1=31
take only integers

Like? Yes (3) | No (4)

MD SOHAIL AKHTER 10 years AGO

To calculate no. of zeroes let the no. be 125 then
125/5=25
25/5=5
5/5=1
now, adding all the quotient-25+5+1=31
total no. of zeroes =31
ans- 125

Like? Yes (3) | No (6)

Md Shahid Akhter 10 years AGO

To calculate no. of zeroes let the no. be 125 then
125/5=25
25/5=5
5/5=1
now, adding all the quotient-25+5+1=31
total no. no. of zeroes =31
the no. will be 125

Like? Yes (2) | No (5)

ASHISH MITTAL 10 years AGO

125/5=25
25/5=5
5/5=1
25+5+1=31

Like? Yes (1) | No (4)

Ann Theressa 10 years AGO

For finding the number of trailing zeros at the end of a factorial(let it be n!), the equation is,

number of trailing zeros
= n/5 + n/5^2 + n/5^3+....+n/5^k
where 5^(k+1) > n

Here number of trailing zeros = 31

n/5 + n/25 + n/125+ ... = 31

Solving n = 125

Like? Yes (1) | No (4)

AVIJIT 10 years AGO

125/5=25
125/25=5
125/125=1
25+5+1=31

Like? Yes (1) | No (4)

ABY 10 years AGO

125/5=25
125/25=5
125/125=1
25+5+1=31

Like? Yes (1) | No (4)

ABYSUP 10 years AGO

125/5=25
125/25=5
125/125=1
25+5+1=31

Like? Yes (1) | No (4)

mohit gupta 10 years AGO

125!= to calculate zero in product we formula divide 125/5=25
125/25=5
125/125=1
total 25+5+1=31
hence c option

Like? Yes (1) | No (4)

Mukesh Nayal 10 years AGO

the method is like this
125/5+125/5*5+125/5*5*5=31
no of zeroes are found by calculating the number of 5 for the factorial of a number.

Like? Yes (3) | No (6)

VIKASH RAY 10 years AGO

We know to find out the zeros last in the number we should divide the number by 5
here the number of zeroes is 31 ....
so the last in the required sequence is---
31*5=125
so option 3 is correct.

Like? Yes (1) | No (4)

Surya Narayana K 10 years AGO

Ans is 125 (option 3).
The formula for number of zeros in n! is n/5+(n/5^2)+(n/5^3)+....

Consider 140 = 140/5+140/25+140/125=28+5+1=34
130=130/5+130/25+130/125=26+5+1=32
125=125/5+125/25+125/125=25+5+1=31.

So answer is 125.

Like? Yes (1) | No (4)

Sivanjali 10 years AGO