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GRE Prep Club Team Member
Joined: 20 Feb 2017
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The sum of all the digits of the positive integer q is equal to the
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07 Apr 2021, 00:13
07 Apr 2021, 00:13
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Question Stats:
45% (02:27) correct
54% (03:06) wrong based on 11 sessions
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The sum of all the digits of the positive integer q is equal to the three-digit number x13. If \(q = 10^n – 49\), what is the value of \(n\)?
(A) 24
(B) 25
(C) 26
(D) 27
(E) 28
(A) 24
(B) 25
(C) 26
(D) 27
(E) 28
Re: The sum of all the digits of the positive integer q is equal to the
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08 Apr 2021, 16:57
08 Apr 2021, 16:57
solution plz
The sum of all the digits of the positive integer q is equal to the
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08 Apr 2021, 19:33
08 Apr 2021, 19:33
2
\(10^n\) - 49 starts producing a series of 9s (as n increases) with a 5 in the tens position and a 1 in the units position.
At 10^3 - 49, you end up with 951
At 10^4 - 49, you end up with 9951
1000 - 49 = 951, sum of digits = 15
10,000 - 49 = 9951, sum of digits = 24
on and on
Hence formula is: 9*(n-2) + 6 is the sum of the digits of \(10^n\) - 49 .
9*(25 - 2) + 6 = 213
Answer is B
At 10^3 - 49, you end up with 951
At 10^4 - 49, you end up with 9951
1000 - 49 = 951, sum of digits = 15
10,000 - 49 = 9951, sum of digits = 24
on and on
Hence formula is: 9*(n-2) + 6 is the sum of the digits of \(10^n\) - 49 .
9*(25 - 2) + 6 = 213
Answer is B
