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# † and ¥ represent nonzero digits, and (†¥)² - (¥†)²

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Math Expert
Joined: 02 Sep 2009
Posts: 52438
Re: † and ¥ represent nonzero digits, and (†¥)² - (¥†)²  [#permalink]

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16 Nov 2017, 19:36
In this question, it needs to say one of the following:

1) The digits are distinct
2) The perfect square is positive
3) The question would be "which *of the following*..."

If you make the digits equal, you get 0 which is clearly a perfect square.

We are told that † and ¥ represent nonzero digits.
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Re: † and ¥ represent nonzero digits, and (†¥)² - (¥†)²  [#permalink]

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16 Nov 2017, 19:54
Bunuel wrote:
We are told that † and ¥ represent nonzero digits.

Yes, but if they both equal 2 (or any other nonzero digit) the result would also be a perfect square: 0. It doesn't say the result is nonzero, just the digits.
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Re: † and ¥ represent nonzero digits, and (†¥)² - (¥†)²  [#permalink]

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13 Mar 2018, 11:51
Hi All,

This question is tougher than a typical GMAT "symbolism" question (most symbolism question are based around basic arithmetic or algebra) and whoever wrote it didn't use proper phrasing (the question should ask "Which of the following COULD be that perfect square?"

The logic behind this prompt is built around some rarer arithmetic Number Property rules….

First off, the prompt can be re-written as X^2 - Y^2 = a perfect square (note that X and Y are both 2-digit numbers with none of the digits as 0 and the two numbers are "mirrors" of one another e.g. 14 and 41).

X^2 - Y^2 = (X + Y)(X - Y)

Now, as to the Number Properties:

1) If you add two "mirrored" 2-digit numbers, then you ALWAYS get a multiple of 11.

eg
14 + 41 = 55…..a multiple of 11
27 and 72 = 99….a multiple of 11
87 and 78 = 165….a multiple of 11

2) If you subtract two "mirrored" 2-digit numbers, then you ALWAYS get a multiple of 9.

41 - 14 = 27…a multiple of 9
72 - 27 = 45…a multiple of 9
87 - 78 = 9…a multiple of 9

This ultimately means that the final answer MUST be a multiple of 11 (because X + Y is a multiple of 11) AND a multiple of 9 (because X - Y is a multiple of 9).

The only answer that fits these rules is

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Re: † and ¥ represent nonzero digits, and (†¥)² - (¥†)²  [#permalink]

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04 Jan 2019, 01:06
carcass wrote:
Not a standard gmat question in my humble opinion; it tastes a quite rare rule: (x + y) (x - y) an addition and a subtrction of two mirrored numbers are always a multiple of 11 and 9. So only E fits the bill.

Symbol question could be quite convoluted but never seen a question that tastes a so obscure number property like this one.

My opinion. Anyhow: kudo for the question

It doesn't expect you to know this property. It expects you to arrive at it - and that is done very simply using a very intuitive process.

Given (AB)^2 - where A and B are digits, how will you square it keeping the variables?
You will convert it to (10A + B)^2 as you have to if you want a two digit number with variables to undergo some algebraic manipulations. It is what you do when you have AB + BA. You write it as (10A + B) + (10B + A).

So similarly, you write $$(AB)^2 - (BA)^2 = (10A + B)^2 - (10B + A)^2 = (10A)^2 + B^2 + 2*10A*B - ((10B)^2 + A^2 + 2*10B*A)$$

H

$$= 99A^2 - 99B^2$$

$$= 9*11(A^2 - B^2)$$

So now you know that the required expression will be divisible by 9 as well as 11.

Hi, in the official solution to further find the number,
I couldn't understand this, can you please explain?

"(† - ¥) must also equal a perfect square, so † - ¥ must equal 4 or 1. († + ¥) =11 and († - ¥) = 4 is a system without integer solutions, "

Why should († - ¥) must also equal a perfect square?

Thanks.
Re: † and ¥ represent nonzero digits, and (†¥)² - (¥†)² &nbs [#permalink] 04 Jan 2019, 01:06

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# † and ¥ represent nonzero digits, and (†¥)² - (¥†)²

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