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10 Feb 2022, 09:40
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
62% (01:50) correct
38%
(01:55)
wrong
based on 217
sessions
History
Date
Time
Result
Not Attempted Yet
Find the value of N which satisfies the equation
\(\frac{1}{100}\) + \(\frac{2}{100}\) + ... + \(\frac{N}{100}\) = 100 * N
A. 19999
B. 9999
C. 21999
D. 999
E. 1999
\(\frac{1}{100}\) + \(\frac{2}{100}\) + ... + \(\frac{N}{100}\) = 100 * N
A. 19999
B. 9999
C. 21999
D. 999
E. 1999
10 Feb 2022, 22:04
Kudos
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gmatcafe
\(\frac{1}{100}\) + \(\frac{2}{100}\) + ... + \(\frac{N}{100}\) = 100 * N
\(\frac{1+2+....+N}{100}=100*N\)
\(1+2+3+......N=100*100N\)
\(\frac{N(N+1)}{2}=10000N\)
\(N(N+1)=20000N\)
Since N is positive, we can divide both sides by N.
\(N+1=20000\)
\(N=19999\)
A
11 Feb 2022, 00:12
Kudos
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gmatcafe
N*(N+1) /2 = 100*100*N
=> N(N-19999) =0
Therefore IMO A
