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by gmat740 » Fri Jul 24, 2009 9:47 am
If 10^50 - 74 is written as an integer in a 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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by yogami » Fri Jul 24, 2009 9:58 am
I dont remember my logs and bases and I highly doubt if GMAC will test you on these principles. Whats the source of this question?
200 or 800. It don't matter no more.
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by prindaroy » Fri Jul 24, 2009 11:24 am
The answer is C.

10 ^ 50 = 51 digits, i.e 10^2 = 100 = 3 digits, 10^3 = 1000 = 4 digits, etc...

so 10.........-74 = 999999........26 with 50 digits in total. So now, 2+6+(9*48)
= 440
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by gmat740 » Fri Jul 24, 2009 6:00 pm
prindaroy wrote:The answer is C.

10 ^ 50 = 51 digits, i.e 10^2 = 100 = 3 digits, 10^3 = 1000 = 4 digits, etc...

so 10.........-74 = 999999........26 with 50 digits in total. So now, 2+6+(9*48)
= 440
Can you explain the step marked with Bold.
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by Stuart@KaplanGMAT » Fri Jul 24, 2009 8:33 pm
gmat740 wrote:
prindaroy wrote:The answer is C.

10 ^ 50 = 51 digits, i.e 10^2 = 100 = 3 digits, 10^3 = 1000 = 4 digits, etc...

so 10.........-74 = 999999........26 with 50 digits in total. So now, 2+6+(9*48)
= 440
Can you explain the step marked with Bold.
The last two digits are 2 and 6.

The first 48 digits are 9.

So when we sum all the digits, we get 48 9s, one 2 and one 6 to gives us 48*9 + 2 + 6
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by gmat740 » Fri Jul 24, 2009 9:56 pm
Ahh Thanks a lot!

Number property questions are tricky :shock:
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