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† and ¥ represent nonzero digits, and (†¥)² - (¥†)² is a perfect square. What is that perfect square?

A. 121 B. 361 C. 576 D. 961 E. 1089

Lets' replace these weird symbols with a and b.

So, we have that a and b are nonzero digits, and (ab)² - (ba)² is a perfect square.

Two digit number ab can be represented as 10a + b (for example, 45 = 10*4 + 5). Two digit number ba can be represented as 10b + a (for example, 45 = 10*4 + 5).

\((10a + b)^2 - (10b + a)^2 = (10a + b - 10b - a)(10a + b + 10b + a) = 9(a - b)*11(a + b) = 11*9*(a^2 - b^2)\).

The correct answer must be divisible by both 9 and 11. Only C and E are divisible by 9 (chek whether the sm of the digits is divisible by 9). Check whether C is divisible by 11: NO. So, the answer is E.

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

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21 Jul 2014, 12:12

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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
_________________

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).

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

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14 Oct 2015, 09:41

I don't understand why we had to place a 10 in front of the two variables. Was there a rule or number property I should know for this type of question? Thanks.

I don't understand why we had to place a 10 in front of the two variables. Was there a rule or number property I should know for this type of question? Thanks.

If AB is a 2 digit number such that A is the tens digit and B is the units digit, the place value of A is 10 and place value of B is 1. So AB = 10A + B Say, AB = 15 A = 1 B = 5

Is AB = A + B? No. 15 is not equal to 1 + 5 = 6 AB = 10A + B = 10*1 + 5 = 15

When you square AB, what will you get?

Is it (A + B)^2 = (1 + 5)^2 = 36? No. 15^2 = 225

Then what is missing? A's place value of 10. So it will be (10A + B) = (10*1 + 5)^2 = 225

So, in terms of AB, AB^2 = (10A + B)^2.
_________________

Re: † and ¥ represent nonzero digits, and (†¥)^2 - (¥†)^2 is a perfect squ [#permalink]

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19 Feb 2016, 06:29

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Let the digits be a and b

(ab) can be written as (10a + b) (ba) can be written as (10b + a)

(ab)^2 - (ba)^2 = (ab + ba)(ab - ba) = (10a + b + 10b + a)(10a + b - 10b - a) = 11(a + b)*9(a - b) = 99(a^2 - b^2) 99(a^2 - b^2) is a perfect square. So the perfect square must be divisible by 99. Only option E is divisible by 99.

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

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21 May 2016, 12:48

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These types of questions can be hard until you get a set method for doing them. After that, they often are fairly similar.

Anytime I see questions where the ten and units digits switch places this is the method I use:

† changes to 10t, which is 10 times the tens digit ¥ changes to u, which is the units digit of the number So, †¥ is equal to 10t + u

E.g. †¥ = 81, 10(8) + 1 = 81

And for ¥† since the tens digit and units digit switch places: ¥ changes to 10u, which is 10 times the tens digit † changes to t, which is the units digit of the number So, ¥† is equal to 10u + t

E.g ¥† = 18, 10(1) + 8 = 18

Now you have two equations to work with! Anytime you see an equation in the format of (x)² - (y)² think the difference of two squares, which is (x + y)(x - y).

So when we plug our equations into the (x + y )(x - y) we come up with (10t + u + 10u + t)(10t + u - 10u - t)

(11t + 11u)(9t - 9u)

Here you can factor out 11 and 9.

11(t + u)9(t - u)

So now we can see that the answer choice must be a factor of 11 & 9.

a) Factor of 11 but not 9 b) Isn't a factor of 11 or 9 c) Factor of 9 but not 11 d) Isn't a factor of either e) Correct answer.

Side note on checking to see if numbers are a factor 9. When you add up all of the digits in the number the result must be a factor of 9.

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

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08 Oct 2016, 05:22

OE:

Let's begin by representing the two digit number †¥ as 10† + ¥ and the two digit number ¥† as 10¥ + †. We have (10† + ¥)² - (10¥ + †)², which simplifies as 99 * (†² - ¥²). 99 = 3 * 3 * 11, so our answer must divide by 9 and 11, and 1089 is the only answer choice meeting such a condition.

Just for fun, if we wanted to go further and actually find † and ¥, we could notice that (†² - ¥²) must divide by 11, which means that either († + ¥) or († - ¥) must = 11. Since † and ¥ are single digits, († + ¥) must = 11. († - ¥) must also equal a perfect square, so † - ¥ must equal 4 or 1. († + ¥) =11 and († - ¥) = 4 is a system without integer solutions, however, so we must have († + ¥) = 11 and († - ¥) = 1, or † = 6 and ¥ = 5. Notice that 65² - 56² = (65+56)(65-56) = 121*9 = 1089. Success!
_________________

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Re: † and ¥ represent nonzero digits, and (†¥)^2 - (¥†)^2 is a perfect squ [#permalink]

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05 Jun 2017, 00:35

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A 2 digit number xy is formed as: 10x + y Its reverse number yx is formed as: 10y + x

When we add these two numbers (original with reverse), we get: 10x+y + 10y+x = 11x + 11y = 11(x+y) We can see that this sum is Always a multiple of 11.

When we subtract these two numbers (reverse from original) we get: 10x+y - 10y-x = 9x - 9y = 9(x-y) We can see that this difference is Always a multiple of 9.

Thus we can subtract their squares, as given in the question: Subtraction of squares means addition of numbers multiplied by their difference (as in a^2 - b^2 = (a+b)(a-b))

Thus the result has to be a multiple of 11 as well as 9.. We can check the options to see which number is both a multiple of 11 as well as 9. Only option E - hence E answer

As we can see from above the answer choice should be divisible by \(11\) & \(9\)both and only option \(E - 1089\) can be divided.

Hence, Answer is E _________________

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Lets check the options . All options are perfect square. Now lets dividing each option by 99 to get a remainder 0. 121 =11^2 is not divisible by 99 361 = 19^2 is not divisible by 99 576 = 24^2 is not divisible by 99 961 = 31^2 is not divisible by 99 1089 = 33^2 is divisible by 99

Re: † and ¥ represent nonzero digits, and (†¥)^2 - (¥†)^2 is a perfect squ [#permalink]

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10 Sep 2017, 11:43

Vyshak wrote:

Let the digits be a and b

(ab) can be written as (10a + b) (ba) can be written as (10b + a)

(ab)^2 - (ba)^2 = (ab + ba)(ab - ba) = (10a + b + 10b + a)(10a + b - 10b - a) = 11(a + b)*9(a - b) = 99(a^2 - b^2) 99(a^2 - b^2) is a perfect square. So the perfect square must be divisible by 99. Only option E is divisible by 99.

Answer: E

Did it in a slightly different way. Always forget to check the options.

So 99(a^2 - b^2) is a perfect square. What does it mean? 99(a^2 - b^2) = 3^2*11*(a^2 - b^2)) It means that a^2 - b^2 has to be equal to 11. It happens when a = 6 and b = 5. Then we find the perfect suare: 99*11.

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

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20 Sep 2017, 08:36

Nevernevergiveup wrote:

OE:

Let's begin by representing the two digit number †¥ as 10† + ¥ and the two digit number ¥† as 10¥ + †. We have (10† + ¥)² - (10¥ + †)², which simplifies as 99 * (†² - ¥²). 99 = 3 * 3 * 11, so our answer must divide by 9 and 11, and 1089 is the only answer choice meeting such a condition.

Just for fun, if we wanted to go further and actually find † and ¥, we could notice that (†² - ¥²) must divide by 11, which means that either († + ¥) or († - ¥) must = 11. Since † and ¥ are single digits, († + ¥) must = 11. († - ¥) must also equal a perfect square, so † - ¥ must equal 4 or 1. († + ¥) =11 and († - ¥) = 4 is a system without integer solutions, however, so we must have († + ¥) = 11 and († - ¥) = 1, or † = 6 and ¥ = 5. Notice that 65² - 56² = (65+56)(65-56) = 121*9 = 1089. Success!

Nevernevergiveup wrote:

OE:

(† - ¥) must also equal a perfect square, so † - ¥ must equal 4 or 1.

I am having difficulty to understand why † - ¥ must be a perfect square.

Can you explain?

I find the solution Since I already have 3*3*11*11*(some variable) the only option is this last part to be a perfect square. ( We already have the pairs two 3 and two 11)