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26 Sep 2009, 05:11
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If m is an integer, and m=10^32-32 what is the sum of it's digits?
all the answer choices are 2 digit numbers
267,275 and other..
I can't get the idea, or the problem is a test one?
Artuurs
all the answer choices are 2 digit numbers
267,275 and other..
I can't get the idea, or the problem is a test one?
Artuurs
Archived Topic
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26 Sep 2009, 06:29
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Answer 284
Sum of digits of 10^32 - 32 = 9x30 + 6 + 8 = 284
Explanation:
10^2 - 32 = 68 (sum of digits = 6 + 8)
10^3 - 32 = 968 (sum of digits = 9 + 6 + 8)
10^4 - 32 = 9968 (sum of digits = 9+9 + 6 + 8 = 9x2 + 6 + 8)
10^5 - 32 = 99968 (sum of digits = 9+9+9 + 6 + 8 = 9x3 + 6 + 8)
....
....
....
....
10^32 - 32 = A68 (sum of digits = 9x30 + 6 + 8)
(A is "9" digit repeated thirty times)
As a trend above, number of "9" digits is two less than the "power". So, in 10^32 - 32, there will be 32 - 2 = 30 times "9" digit.
Sum of digits of 10^32 - 32 = 9x30 + 6 + 8 = 284
Explanation:
10^2 - 32 = 68 (sum of digits = 6 + 8)
10^3 - 32 = 968 (sum of digits = 9 + 6 + 8)
10^4 - 32 = 9968 (sum of digits = 9+9 + 6 + 8 = 9x2 + 6 + 8)
10^5 - 32 = 99968 (sum of digits = 9+9+9 + 6 + 8 = 9x3 + 6 + 8)
....
....
....
....
10^32 - 32 = A68 (sum of digits = 9x30 + 6 + 8)
(A is "9" digit repeated thirty times)
As a trend above, number of "9" digits is two less than the "power". So, in 10^32 - 32, there will be 32 - 2 = 30 times "9" digit.
26 Sep 2009, 06:58
Kudos
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OK, my bad, this was really simple!!!
Thank you!!!
Thank you!!!
