Latest TCS Aptitude Question SOLUTION: find the number of zeroes in 1^1*2^2*3^3*4^4 …. 48^48*49^49 ? a. 250 b. 225 c. 545 d. 135

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(#M40033480) TCS Numerical Ability question Keep an EYE

find the number of zeroes in 1^1*2^2*3^3*4^4 …. 48^48*49^49 ?
a. 250
b. 225
c. 545
d. 135

Asked In TCS Sruthi karuturi (10 years ago)

Unsolved Read Solution (2)

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Read Solution (2) :

RAKESH  (10 years ago)

only 2*5 gives 10

find total exponent of 5 in given product

5^5 =(1*5)^5 => 5^5
10^10=(2*5)^10 => 5^10
15^15=(3*5)^15 => 5^15
20^20=(4*5)^20 => 5^20
25^25=(5*5)^25 => (5^25)^2
30^30=(6*5)^30 => 5^30
35^35=(7*5)^35 => 5^35
40^40=(8*5)^40 => 5^40
45^45=(9*5)^45 => 5^45

exponent of 5 in product = 5+10+15+20+(2*25)+30+35+40+45
= 5*(1+2+...+9)+25
= 250

i.e product has a factor 5^250 & exponent of 2 is much more than 5

so, number of zeroes = 250

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srishti singh  (10 years ago)

consider only 5,10,15,20.....45
so add the number of 5 in these occurence
so 5+10+15+20+50+30+35+40+45=250
a is the answer

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