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integers

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Junior | Next Rank: 30 Posts
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by jkwan » Thu Feb 05, 2009 10:10 pm
If 10^50 – 74 is written as an integer in base 10 notation, what is the sum of the digits in that integer?

A. 424
B. 433
C. 440
D. 449
E. 467
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Re: integers

by betamax » Thu Feb 05, 2009 11:06 pm
jkwan wrote:If 10^50 – 74 is written as an integer in base 10 notation, what is the sum of the digits in that integer?

A. 424
B. 433
C. 440
D. 449
E. 467
10^50 -74=
9*10^49+9*10^48+9*10^47....2*10^1+6*10^0...
basically
99999999999999999....99999926 <-------48 9's followed by a 2 and a 6
9*48+2+6=440

For this problem I think the important thing to realize is that 10^50 is 1 followed by 50 zeros. Subtracting 76 means the last 2 digits will be 26. Since you subtracted you effectively decrease the number of digits by 1, so from 51 to 50. You know the last 2 digits must be 26 so you must have 48 nines. But wow, this is pretty tricky.
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by DanaJ » Thu Feb 05, 2009 11:10 pm
Let's look at a few examples:
10^3 = 1000 and you have that 10^3 - 74 = 926.
10^4 = 10 000 and you get 10^4 - 74 = 9 926
10^5 = 100 000 and you have 10^5 - 74 = 99 926.
You notice that in all cases you have 2, 6 and (power of 10) - 2 nines. So you can devise a rule:
10^n - 74 has (n - 2) nines, a 2 and a 6.
This means that the sum of digits for 10^50 - 74 = (50 - 2)*9 + 2 + 6 = 432 + 2 + 6 = 440.

IMO answer is C

(I've been asked to use the spoiler function on my answers).
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