TCS
Company
Numerical Ability
Number System
number of zeros in 1*2^2*3^3*4^4*....49^49
Read Solution (Total 6)
-
- In this problem, there are more 2's than 5's. so let us count the number of 5's.
5^5 contributes 5 5's
10^10 contributes 10 5's
15^15 -> 15 5's
20^20 -> 20 5's
25^25 -> 5^50 that is 50 5's
30^30-> 30 5's
35^ 35-> 35 5's
40^40-> 40 5's
45^45 -> 45 5's
so number of 5's = 5 + 10 + 15 + 20 + 50 + 30+ 35+40+45 = 250
so there are 250 0's. - 10 years agoHelpfull: Yes(14) No(0)
- total no. of 5's = 5 + 10 + 15 + 20 + 50 + 30 + 35 + 40 + 45 = 250
- 10 years agoHelpfull: Yes(2) No(0)
- @kumari manisha:how do u calculate the contribution of 5???pls explain
- 10 years agoHelpfull: Yes(1) No(0)
- no of zeros=(5+10+15+20+25+30+35+40+45)=225
- 10 years agoHelpfull: Yes(1) No(2)
- when u add 5+10+15....45 u will get 225 u will not get 250 can u explain how u got 250
- 10 years agoHelpfull: Yes(1) No(1)
- Ans is 250
- 10 years agoHelpfull: Yes(1) No(0)
TCS Other Question