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555-605 (Medium)|   Algebra|      
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Bunuel
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If x and y are positive integers such that x^2 - y^2 = 23. What is the 20 Dec 2019, 03:36
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B
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Difficulty:

25% (medium)

Question Stats:

71% (01:14) correct 29% (01:12) wrong based on 174 sessions

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If x and y are positive integers such that x^2 - y^2 = 23. What is the value of y?

A. 10
B. 11
C. 12
D. 23
E. Cannot be determined
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B
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rajatchopra1994
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If x and y are positive integers such that x^2 - y^2 = 23. What is the 20 Dec 2019, 03:47
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Bunuel
If x and y are positive integers such that x^2 - y^2 = 23. What is the value of y?

A. 10
B. 11
C. 12
D. 23
E. Cannot be determined
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Explanation:

x^2 - y^2 = 23
(x+y)(x-y) = 23 X 1 (only possible value, as 23 is prime)
so, only value for x & y is 12 & 11
so, therefore, y=11
IMO-B
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Schools: XLRI (A)
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If x and y are positive integers such that x^2 - y^2 = 23. What is the 11 Oct 2024, 09:12
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Given that x and y are positive integers and x^2 - y^2 = 23. And we need to find y

x^2 - y^2 = 23
=> (x-y) * (x+y) = 23 = 1*23
=> x - y = 1 ...(1)
=> x+y = 23 ...(2)

(2) - (1) we get
2y = 23-1 = 22
=> y = 11

So, Answer will be B
Hope it helps!
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If x and y are positive integers such that x^2 - y^2 = 23. What is the 07 Sep 2026, 11:57
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Bunuel
If x and y are positive integers such that x^2 - y^2 = 23. What is the value of y?

A. 10
B. 11
C. 12
D. 23
E. Cannot be determined
Show more
Alternative Approach: Backsolving

The answer choices can be easily substituted into the equation, so let's backsolve.

Start with choice A (10) because it's the easiest. If y = 10, then \(y^2 = 100\).

If x is then 11, \(x^2 = 121\), and \(121 - 100 = 21\). Not big enough.

But if x is 12, then \(x^2 = 144\), and \(144 - 100 = 44\). Too big.

However, if y = 11 and x = 12, then \(x^2 - y^2 = 144 - 121 = 23\). The answer is B.

Note that the gap between \(10^2\) and \(11^2\) is 21, and the gap between \(11^2\) and \(12^2\) is 23, an increase of 2. This pattern holds true for all squares of positive integers. The gap between squares of consecutive integers increases by 2 every time:
\(1^2 = 1\)
\(2^2 = 4\) (gap of 3)
\(3^2 = 9\) (gap of 5)
\(4^2 = 16\) (gap of 7)
etc.

Takeaways:
1. Backsolving is an effective approach on a problem like this because the answer choices can be easily substituted into the given equation.
2. Knowing the pattern for squares of consecutive integers can help you quickly zero in on the correct answer.
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Bunuel
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