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If S = x + (x + 1) + ... + (x + 80), where x is a positive integer 03 May 2024, 01:30
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65% (hard)

Question Stats:

63% (01:57) correct 37% (02:32) wrong based on 131 sessions

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­If S = x + (x + 1) + ... + (x + 80), where x is a positive integer, what is the minimum value of x such that S is the square of an integer?

A. 4
B. 9
C. 24
D. 41
E. 81­
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If S = x + (x + 1) + ... + (x + 80), where x is a positive integer 03 May 2024, 01:39
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Bunuel
­If S = x + (x + 1) + ... + (x + 80), where x is a positive integer, what is the minimum value of x such that S is the square of an integer?

A. 4
B. 9
C. 24
D. 41
E. 81­


 
­
\(S = x + (x + 1) + ... + (x + 80)\)

\(S = (x + x + x .... \text{upto 81 times}) + (1 + 2 + 3 + ...80) \)

\(S = 81x + \frac{80*81}{2}\)

\(S = 81x +40*81\)

\(S = 81(x+40)\)

81 is already a square of an integer. Hence, we have to find the minimum value of x such that (x + 40) is a square.

Answer choice elimination

A. 4

40+ 4 = 44 → Not a square of an integer

B. 9

40+ 9 = 49 → Square of an integer

Option B
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If S = x + (x + 1) + ... + (x + 80), where x is a positive integer 03 May 2024, 03:45
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­A square number will, when primefactorised, have even exponent values for all of its prime factors.

\(S = x + (x + 1) + ... + (x + 80)\) 

This equation will have 81 \(x\)'s and the rest of the numbers will sum to \(\frac{80}{2}(1+80)\),  which simplifies to \(40*81\)


\(S = x + (x + 1) + ... + (x + 80)\) can therefore be written as: \(81x + 40*81\)

\(81(x+40)\)

\(3^4(x+40)\) 

The 3 has an even exponent so it is already a perfect square. We need to solely focus on making the bracket a perfect square. As we are adding 40 to x, the smallest perfect square most likely is in the 40's. Off the bat we know that \(7^2 = 49\). So for \(S = x + (x + 1) + ... + (x + 80)\) to be the smallest possible perfect square, we need \(x = 9\)

ANSWER B­
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If S = x + (x + 1) + ... + (x + 80), where x is a positive integer 06 May 2024, 00:15
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Official Solution:

­If \(S = x + (x + 1) + ... + (x + 80)\), where \(x\) is a positive integer, what is the minimum value of \(x\) such that S is the square of an integer?

A. 4
B. 9
C. 24
D. 41
E. 81


\(S = x + (x + 1) + ... + (x + 80)\), represents the sum of 81 consecutive integers from \(x\) to \(x+80\), inclusive. Thus, \(S = \frac{{\text{first} + \text{last}}}{2} * \text{the number of terms} = \frac{{x + (x + 80)}}{2} * 81 = 81(x+40)\). Since 81 is a perfect square, \(9^2\), for the entire expression to be a perfect square \((x+40)\) must also be a perfect square. Given that \(x\) is a positive integer, the minimum value of \(x\) is 9.


Answer: B­
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