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by shashank.ism » Tue Feb 09, 2010 12:51 pm
If N = 1000^8 - 8, what is the sum of its digits?

200
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by sparky_paris » Thu Feb 11, 2010 6:39 pm
The result will have 22 9s and a 2 the sum of which is 200. So its option A
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by money9111 » Thu Feb 11, 2010 8:44 pm
what's the quickest way to realize how many zero's 1000^8 will produce?
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by ajith » Fri Feb 12, 2010 12:04 am
money9111 wrote:what's the quickest way to realize how many zero's 1000^8 will produce?
Make it a power of 10 and the power tells the number
1000^8 = (10^3)^8 = 10^3*8 = 10^24 and hence 1 followed by 24 zeroes
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by money9111 » Fri Feb 12, 2010 6:16 pm
perfect... because 8*3 is 24... therefore 24 zeros follow but then you must substract 8... etc etc got it.. thanks
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by harsh.champ » Thu Feb 18, 2010 7:10 am
ajith wrote:
money9111 wrote:what's the quickest way to realize how many zero's 1000^8 will produce?
Make it a power of 10 and the power tells the number
1000^8 = (10^3)^8 = 10^3*8 = 10^24 and hence 1 followed by 24 zeroes
Hey ajith,
It seems u are very gud with questions involving power.[Like you also devised the method of getting 0.9^10 quickly to 0.35]
Anyways,thanks for the soln. approach.
The result will have 22 9s and a 2 the sum of which is 200. So its option A
I was very confused with the sparky's answer and didn't know how he got those 22 9s.
Now,it seems very simple.
I would definitely mark this ques. for future reference.
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