jst chek out da ansr..
ques How many zeroes will be there at the end of 1^1 x 2^2 x 3^3x . . . . . . x 100^100 ?
um gettin 1220 zeroes.. wots ur ansr dudes...???
It’s like calculating how
It’s like calculating how many 5’s are there as the number of 2’s ll be much more than 5
We have 5^5 10^10 15^15 . . .
So total fives = 5 + 10 + 15 . . . . + 100
= 5 ( 1 + 2 + 3 . . . + 20) = 10 x 21 x 5 = 1050
Now 25 has two five – extra 25 fives
Now 50 has two five – extra 50 fives
Now 75 has two five – extra 75 fives
Now 100 has two five – extra 100 fives
So extra fives are 250
In total we have 1050 + 250 = 1300 fives
i hope this is correct
Correct me if I am wrong
We know that first number having zero is 10.
Also 2x5=10
So 2x5 will give us one zero.
In 1^1 x 2^2 x 3^3 x 4^4………..100^100 it is evident that number of 2`s is more that number of 5`s. So number of 5`s will decide the number of zero’s.
So we will see only multiples of five.
5^5 means 5 times 5
10^10 means 10 times 5
15^15 means 15 times 5
20^20 means 20 times 5
………and so on till 100^100 means 100 times 5
So we have a series
5+10+15+20+………..+100
Sum of this series that is 1050 zero`s.
Now 25 has two five so extra 25 fives
Now 50 has two five so extra 50 fives
Now 75 has two five so extra 75 fives
Now 100 has two five so extra 100 fives
So extra fives are 250
In total we have 1050 + 250 = 1300 fives
so the number of zeros is 1300
hii
ma approach wz crrect..may b sum calculation problem...i guess
hii
hiii.....1941...
ur approach is correct...
thanks for correcting.
i saw the solution and changed mine to avoid the confusion to others.
Solution:
The number of zeroes will depend on the number of 5s present.
In all the multiples of 5 (5,10,15,......,100) 5 is present atleast once and in 4 of these multiples 5 is present twice (25,50,75,100).
Thus number of zeroes = (5+10+15+......+100) + (25+50+75+100) = 1050 + 250 = 1300.