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If S = x + (x + 1) + ... + (x + 80), where x is a positive integer
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03 May 2024, 01:30
03 May 2024, 01:30
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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
A. 4
B. 9
C. 24
D. 41
E. 81
Quant Chat Moderator
Joined: 22 Dec 2016
Posts: 3133
Given Kudos: 1856
Location: India
Concentration: Strategy, Leadership
Re: If S = x + (x + 1) + ... + (x + 80), where x is a positive integer
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03 May 2024, 01:39
03 May 2024, 01:39
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Bunuel wrote:
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
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
Re: If S = x + (x + 1) + ... + (x + 80), where x is a positive integer
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03 May 2024, 03:45
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
\(S = x + (x + 1) + ... + (x + 80)\)
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
