Latest TCS Aptitude Question SOLUTION: find the number of zeros in the product 1^1+2^2.......+49^49?

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find the number of zeros in the product 1^1+2^2.......+49^49?

Asked In TCS PRIYA NAMDEO (11 years ago)

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sirisha  (11 years ago)

we have to count no.of 5's and no.of 10's
5^5=5
10^10=10
15^15=15
20^20=20
25^25=(5*5)^2
30^30=30
35^35=35
40^40=40
45^45=45
i.e total=5+10+15+20+50+30+35+40+45=250 zeros

Is this solution Helpfull? Yes (17) | No (1)  

Abira Deb  (11 years ago)

10^10,20^20,30^30,40^40 are the numbers which contains zero.
so,the number of zeros are: 10+20+30+40=100.

Is this solution Helpfull? Yes (14) | No (9)  

RAKESH  (11 years ago)

qn should be find the number of zeros in the product 1^1*2^2*3^3....48^48*49^49
count the no. of 5's in 1^1*2^2*3^3....48^48*49^49
5^5*10^10*15^15*20^20*25^25*30^30*35^35*40^40*45^45(only these terms consists 5)
=5^5*5^10*5^15*..............*5^45
so no. of 5's in 1^1*2^2*3^3....48^48*49^49=5+10+15+20+25+25+30+35+40+45( as 25^25=5^50)
=5(1+2+3+........+9)+25
=5*45 +25=250
no. of zZEROES=250

Is this solution Helpfull? Yes (9) | No (1)  

ankush vaishya  (11 years ago)

we have to count no. of 5's so count powers of these as...
5^5..+10^10+15^15.....45^45
=>sum of powers 5+10+15+20+25+25+30+35+40+45=250
25 is two times cz 25^25=(5^2)^25=>5^50.

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ankush vaishya  (11 years ago)

Sorry guys the solution by ABIRA is correct.
250 is ans in case 1^1*2^2*3*3.....

Is this solution Helpfull? Yes (3) | No (1)  

Tarun Sharma  (11 years ago)

this is not the correct qsn. . it should be as
1^1*2^2*3^3........49^49.

then no of zeros are = 250.

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ankush vaishya  (11 years ago)

250 is the answer...

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

Prem kumar  (11 years ago)

formula used=n(n+1)(2n+1)/6
ans=40425

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avnish pandey  (11 years ago)

To each reducing power 2 zeros will be added.. so 49x2=98 zeros..

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