Latest TCS Aptitude Question SOLUTION: find the no. of zeros in the product of 1^1*2^2*3^3....*49^49?

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(#M40011736) TCS Logical Reasoning question Keep an EYE

find the no. of zeros in the product of 1^1*2^2*3^3....*49^49?

Asked In TCS Arit (11 years ago)

Unsolved Read Solution (14)

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BRIJESH KUMAR GUPTA  (11 years ago)

ans =250
0 will be produced when an even number be multiplied by 5
or a num multiplied by 10
now 10^10=10 0's
20^20= 20 0's
30^30=30 0's
40 ^40=40 0's
5^5=5 0's
15^15=15 0's
35^35=35 0's
45^45=45 0's
but 25^25 can be written as
5^25*5^25=50 0's
summing up 250 0's

Is this solution Helpfull? Yes (38) | No (4)  

udit narayana  (11 years ago)

ans is 250

Is this solution Helpfull? Yes (17) | No (6)  

Neeraj Kumar  (10 years ago)

I apologize for my incorrect solution as I read the question incorrectly.
the number of zero's is 225

the number of zero's can be found by finding number of 2's and 5's.
since in this number, number of occurrences 2's will always be greater than number of occurrences of 5's.
therefore we need to find number of 5's only.
number of occurrences of 5's
in 5^5 = 5
10^10=10
15^15=15
20^20=20
25^25=25
30^30=30
35^35=35
40^40=40
45^45=45
sum=225.
therefore 225 5's can be paired with 225 2's.
total number of zeros = 225

Is this solution Helpfull? Yes (8) | No (7)  

ankit shukla  (11 years ago)

when 25^ 25 then no of 5 is 50 not 25....
bcoz ((5^2)25)....5^50 so total is 250 exact...

Is this solution Helpfull? Yes (6) | No (4)  

khushboo singh  (11 years ago)

total zeros =5+10+15+20+25+30+35+40+45
= 225

Is this solution Helpfull? Yes (4) | No (14)  

shubham kumar Gupta  (9 years ago)

we know that 0 will occur when 2 will be multiplied by 5 ,here 5 will be less than 2 so we will calculate power of 5
now 10^10=10 0's
20^20= 20 0's
30^30=30 0's
40 ^40=40 0's
5^5=5 0's
15^15=15 0's
35^35=35 0's
45^45=45 0's
but 25^25 can be written as
5^25*5^25=50 0's

total ans =250

Is this solution Helpfull? Yes (4) | No   

Sonal   (11 years ago)

we calculate no of 5 on once place 5^5+15^15+25^25+35^35+45^45+(2*5)^10+(2*5)^20+(2*5)^30+(2*5)^40=225

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Neeraj Kumar  (10 years ago)

20 zeros.


this number can be viewed as (49!)^2
now to find the number of zeros we need to find the occurrences of 2's and 5's only.
[] denotes the floor value.
occurrences of 2's =[49/2]+[49/(2^2)]+[49/(2^3)]+[49/(2^4)]+[49/(2^5)]
=24+12+6+3+1
=46
occurrences of 5's =[49/5]+[49/(5^2)]
=9+1
=10
10 5's can be paired with 10 2's out of 46.
therefore number of 10's in 49! = 10.
therefore number of 10's in (49!)^2=20.

total number of 0's = 20.

Is this solution Helpfull? Yes (1) | No (6)  

Ishwarya  (9 years ago)

neeraj kumar note this.. 25^25 = 50 zeros...therefore 250...

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Saquib Akhter  (9 years ago)

the fives(5) will be less than twos(2).Hence we need to count only the fives.
thus:5^5*10^10*15^15*20^20*25^25*30^30*35^35*40^40*45^45
gives us:5+10+15+20+25+30+35+40+45 fives.Thus the product has 250 zeros

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deepak gupta  (11 years ago)

how plz explain it clearly........

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Swati Gupta  (9 years ago)

it is 49!*49!
(49/5 + 49/(5^2) )*2 = 20 zeros

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Amarpal Singh  (9 years ago)

18 we have 10*10,20*20,30*30,40*40;
and 5*5,15*15,25*25,35*35,45*45 each 5 will result a 0(5*2=10)
hence total zero=8+10=18

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Ankit Kumar  (7 years ago)

(5^5)*(2*5)^10*(3*5)^15*(2*2*5)^20*(5*5)^25*(2*3*5)^30*(5*7)^35*(2*2*2*5)^40*(3*3*5)^45
here we'll take only 5
Total number of five
(5^5) = 5
(2*5)^10 =(2^10*5^10) 10
(3*5)^15 =(3^15*5^15) 15
(2*2*5)^20 =(2^20*2^20*5^20) 20
(5*5)^25 =(5^25*5^25) 25+25
(2*3*5)^30 =(2^30*3^30*5^30) 30
(5*7)^35 =(5^35*7^35) 35
(2*2*2*5)^40 =(2^40*2^40*2^40*5^40) 40
(3*3*5)^45 =(3^45*3^45*5^45) 45

total = 250

Note :: most of person confused ..why (25+25) ...because when we will (5*5)^25=(5)^50

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