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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Manhattan GMAT Adapative CAT (Taken Jan. 19, 09): 600 (Q-42, V-32)
Manhattan GMAT Adapative CAT (Taken Feb. 14, 09): 620 (Q-43, V-32)
GMATPrep CAT (Taken Feb 28, 09): 590 (Q-39, V-34)
Manhattan GMAT Adaptive CAT (Taken March 6, 09): 640 (Q-44, V-34)
800Score CAT (Taken March 14, 09): 630 (Q-41, V-36)
800score CAT (Taken March 20, 09): 600 (Q-41, V-31)
Manhattan GMAT Adaptive CAT (Taken March 22, 09): 600 (Q-39, V-34)
GMATPrep CAT (Taken March 28, 09): 570 (Q-37, V-31)
Knewton Diagnostic CAT (Taken April 3, 09): 640 (Q-45, V-33)
Manhattan GMAT Adapative CAT (Taken Feb. 14, 09): 620 (Q-43, V-32)
GMATPrep CAT (Taken Feb 28, 09): 590 (Q-39, V-34)
Manhattan GMAT Adaptive CAT (Taken March 6, 09): 640 (Q-44, V-34)
800Score CAT (Taken March 14, 09): 630 (Q-41, V-36)
800score CAT (Taken March 20, 09): 600 (Q-41, V-31)
Manhattan GMAT Adaptive CAT (Taken March 22, 09): 600 (Q-39, V-34)
GMATPrep CAT (Taken March 28, 09): 570 (Q-37, V-31)
Knewton Diagnostic CAT (Taken April 3, 09): 640 (Q-45, V-33)
10^50 -74=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
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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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).
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).
