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26 Jun 2023, 07:00
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B
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Difficulty:
45%
(medium)
Question Stats:
72% (02:29) correct
28%
(02:42)
wrong
based on 126
sessions
History
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Given that \(1+2+.......+n=\frac{n(n+1)}{2}\) and \(1^2+2^2+…..+n^2=\frac{n(n+1)(2n+1)}{6}\), what is the sum of the integers between \(1\) and \(100\) (inclusive) that are not squares of integers?
A. \(4050\)
B. \(4665\)
C. \(4775\)
D. \(5000\)
E. \(5050\)
A. \(4050\)
B. \(4665\)
C. \(4775\)
D. \(5000\)
E. \(5050\)
26 Jun 2023, 07:10
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Bunuel
Sum of all numbers from 1 to 100 = \(1+2+.......+100=\frac{100(100+1)}{2} = 5050\)
Sum of Squares between 1 and 100 = \(1^2+2^2+…..+10^2=\frac{10(10+1)(2*10+1)}{6} = 385\)
Sum of numbers from 1 to 100 that are NOT squares = 5050-385 = 4665
Answer: Option B
26 Jun 2023, 09:23
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Given that \(1+2+.......+n=\frac{n(n+1)}{2}\) and \(1^2+2^2+…..+n^2=\frac{n(n+1)(2n+1)}{6}\), what is the sum of the integers between \(1\) and \(100\) (inclusive) that are not squares of integers?
A. \(4050\)
B. \(4665\)
C. \(4775\)
D. \(5000\)
E. \(5050\)
Given that we have 100 numbers the sum of which would be
\(\frac{n(n+1)}{2}\) where n = 100
So,
\(\frac{n(n+1)}{2}\) = \frac{100*101}{2}[/m] = 5050
E is out.
Only people in hurry might chose this one if these somehow forgot to read question stem correctly.
Now, Squares are 1, 4, 9, 16... 100 totalling 10 such unique numbers i.e. 1^2, 2^2, 3^2 ... 10^2
Sum of these would be more than 100
Hence subtracting 100 from 5050 given a number less than 5000.
So, D is out
If highest square is 100 then sum of these squares should be less than 1000
therefore, A is put.
We are left with B and C.
Note: After that we calculated 5050, we can be sure that answer might between B and C(intelligent guess) since both are close figure.
So, we are better of calculating using the formula given
\(1^2+2^2+…..+n^2=\frac{n(n+1)(2n+1)}{6}\) where n = 10
Therefore \(1^2+2^2+…..+10^2=\frac{n(n+1)(2n+1)}{6} = \frac{10*11*21}{6} = 385\)
Now, answer = 5050-385 = 4775
Answer D.
A. \(4050\)
B. \(4665\)
C. \(4775\)
D. \(5000\)
E. \(5050\)
Given that we have 100 numbers the sum of which would be
\(\frac{n(n+1)}{2}\) where n = 100
So,
\(\frac{n(n+1)}{2}\) = \frac{100*101}{2}[/m] = 5050
E is out.
Only people in hurry might chose this one if these somehow forgot to read question stem correctly.
Now, Squares are 1, 4, 9, 16... 100 totalling 10 such unique numbers i.e. 1^2, 2^2, 3^2 ... 10^2
Sum of these would be more than 100
Hence subtracting 100 from 5050 given a number less than 5000.
So, D is out
If highest square is 100 then sum of these squares should be less than 1000
therefore, A is put.
We are left with B and C.
Note: After that we calculated 5050, we can be sure that answer might between B and C(intelligent guess) since both are close figure.
So, we are better of calculating using the formula given
\(1^2+2^2+…..+n^2=\frac{n(n+1)(2n+1)}{6}\) where n = 10
Therefore \(1^2+2^2+…..+10^2=\frac{n(n+1)(2n+1)}{6} = \frac{10*11*21}{6} = 385\)
Now, answer = 5050-385 = 4775
Answer D.
