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A and B are 2-digit numbers with non-zero digits. Both digits in A are

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A and B are 2-digit numbers with non-zero digits. Both digits in A are  [#permalink]

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New post 27 Mar 2019, 22:52
1
5
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

79% (02:01) correct 21% (01:58) wrong based on 94 sessions

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A and B are 2-digit numbers with non-zero digits. Both digits in A are distinct, and B is formed by reversing the digits of A. Which of the following is always a factor of \(A^2 – B^2\)?

    A. 5
    B. 15
    C. 45
    D. 75
    E. 99

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Re: A and B are 2-digit numbers with non-zero digits. Both digits in A are  [#permalink]

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New post 28 Mar 2019, 02:33
Let the two digits in A be x and y. Then A = 10x+y.
B = 10y+x.

\(A^2-B^2 = 100x^2+y^2+20xy-100y^2-x^2-20xy = 99x^2-99y^2 = 99(x^2-y^2)\)

So \(A^2-B^2\) will always be a factor of 99.

E is the answer.
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Re: A and B are 2-digit numbers with non-zero digits. Both digits in A are  [#permalink]

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New post 28 Mar 2019, 12:45
EgmatQuantExpert wrote:
A and B are 2-digit numbers with non-zero digits. Both digits in A are distinct, and B is formed by reversing the digits of A. Which of the following is always a factor of \(A^2 – B^2\)?

    A. 5
    B. 15
    C. 45
    D. 75
    E. 99


21^2-12^2=297
99
E
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Re: A and B are 2-digit numbers with non-zero digits. Both digits in A are  [#permalink]

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New post 31 Mar 2019, 22:05

Solution


Given:
In this question, we are given
    • A and B are 2-digit numbers, with non-zero digits.
    • Both digits in A are distinct.
    • B is formed by reversing the digits of A.

To find:
    • Among the given options, which number is always a factor of \(A^2 – B^2\).

Approach and Working:
Let us assume that A = pq and B = qp.
    • Therefore, the value of A = 10p + q
    • And, value of B = 10q + p

Now, \(A^2 – B^2 = (A + B) (A – B) = [(10p + q) + (10q + p)][ (10p + q) – (10q + p)]\)
    • Or, \(A^2 – B^2 = [11 (p + q)] [9 (p – q)] = 99 (p + q) (p – q)\)

Now, irrespective of the values of p and q, the expression 99 (p + q) (p – q) is always divisible by 99.

Hence, the correct answer is option E.

Answer: E
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A and B are 2-digit numbers with non-zero digits. Both digits in A are  [#permalink]

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New post 25 Jul 2019, 06:26
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EgmatQuantExpert wrote:
A and B are 2-digit numbers with non-zero digits. Both digits in A are distinct, and B is formed by reversing the digits of A. Which of the following is always a factor of \(A^2 – B^2\)?

    A. 5
    B. 15
    C. 45
    D. 75
    E. 99

Image


A fast approach is to test some values of A and B

For example, it COULD be the case that A = 21 and B =12
In this case, A² - B² = 21² - 12²
= (21 + 12)(21 - 12) [aside: since 21² - 12² is a DIFFERENCE OF SQUARES, we can first factor it before evaluating it]
= (33)(9)
= (3)(11)(3)(3)

Since 5 is not a factor of (3)(11)(3)(3), we can eliminate answer choice A
Since 15 is not a factor of (3)(11)(3)(3), we can eliminate answer choice B
Since 45 is not a factor of (3)(11)(3)(3), we can eliminate answer choice C
Since 75 is not a factor of (3)(11)(3)(3), we can eliminate answer choice D
HOWEVER, 99 IS a factor of (3)(11)(3)(3). So, KEEP answer choice E

Answer: E

Cheers,
Brent
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A and B are 2-digit numbers with non-zero digits. Both digits in A are   [#permalink] 25 Jul 2019, 06:26
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