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by shashank.ism » Mon Feb 08, 2010 7:38 am
If N = 1000^8 - 8, what is the sum of its digits?

a) 200
b) 207
c) 208
d) 209
e) 175
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by Brent@GMATPrepNow » Mon Feb 08, 2010 7:47 am
shashank.ism wrote:If N = 1000^8 - 8, what is the sum of its digits?

a) 200
b) 207
c) 208
d) 209
e) 175
1000^8 = (10^3)^8 = 10^24
So, 1000^8 is a number with 1 followed by 24 zeroes.
This means that 1000^8 - 8 is a 24-digit number: 9999999.....9992
This number consists of 23 nines and 1 two.
So, the sum is 23x9 + 2 = 209 (D)
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by ajith » Mon Feb 08, 2010 10:37 am
shashank.ism wrote:If N = 1000^8 - 8, what is the sum of its digits?

a) 200
b) 207
c) 208
d) 209
e) 175
1000^8 = (10^3)^8 = 10^(8*3 ) = 10^24

10^24-8 = 99...(23times)2

Sum of digits= 9*23+2 = 209
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