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Find the highest power of 10 in 203!
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- 10=2*5 (prime factors of 10)
highest power of 10 in 203!=min(power of 2 in 203!,power of 5 in 203!)
power of 2 in 203! = |203/2| + |203/4| + |203/8| + |203/16| + |203/32| + |203/64| + |203/128| =101 + 50 + 25 + 12 + 6 + 3 + 1= 198
power of 5 in 203! = |203/5| + |203/25| + |203/125|= 40 + 8 + 1 = 49
=> min( 198,49)=49 - 11 years agoHelpfull: Yes(47) No(1)
- 203/2 = 101
101/2 = 50
50/2 = 25
25/2 = 12
12/2 = 6
6/2 = 3
3/2 = 1
Hence the power of 2 in 203! = 101 + 50 + 25 + 12 + 6 + 3 + 1 = 198
Now let’s divide 203 with 5
203/5 = 40
40/5 = 8
8/5 = 1
Hence the power of 5 in 203! = 40 + 8 + 1 = 49
Therefore, the power of 10 in 203! is 49. - 11 years agoHelpfull: Yes(31) No(1)
- simplest method to obtain solution of the highest power in factorial of any no. is to divide it by 5 and keep dividing :
203/5= 40 ignore the remainder part
now 40/5= 8
now 8/5=1 just write the quotients
now add the quotients: 40+8+1=49 - 11 years agoHelpfull: Yes(26) No(0)
- find no of zeros in 203!
that is49
203/5=40
40/5=8
8/5=1
40+8+1=49
48/2=24 and 1 10
10 power 25
- 10 years agoHelpfull: Yes(1) No(1)
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