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

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

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21 Jul 2014, 10:53
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75% (hard)

Question Stats:

58% (02:23) correct 42% (02:25) wrong based on 338 sessions

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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

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

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21 Jul 2014, 11:54
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WoundedTiger wrote:
† 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.

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

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14 Oct 2015, 22:13
5
1
Gonga wrote:
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.
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Re: † and ¥ represent nonzero digits, and (†¥)^2 - (¥†)^2 is a perfect squ  [#permalink]

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19 Feb 2016, 06:29
5
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.

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

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28 Sep 2015, 00:45
3
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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 $$(AB)^2 - (BA)^2 = (10A + B)^2 - (10B + A)^2 = (10A)^2 + B^2 + 2*10A*B - ((10B)^2 + A^2 + 2*10B*A)$$

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

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

So now you know that the required expression will be divisible by 9 as well as 11.
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† and ¥ represent nonzero digits, and (†¥)^2 - (¥†)^2 is a perfect squ  [#permalink]

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19 Feb 2016, 03:47
3
11
† 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

Kudos for correct solution.
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† and ¥ represent nonzero digits, and (†¥)² - (¥†)²  [#permalink]

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21 May 2016, 12:48
2
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

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.

E.g.

Answer a) 1+2+1 = 4, 4 is not a factor or 9

c) 5 + 7 + 6 = 18, 18 is a factor of 9
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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
1
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
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Re: † and ¥ represent nonzero digits, and (†¥)² - (¥†)²  [#permalink]

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05 Feb 2019, 20:48
1
jfranciscocuencag wrote:
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

Hello EMPOWERgmatRichC !

Could someone please explain to me this rule?

(x + y) (x - y)

an addition and a subtraction of two mirrored numbers are always a multiple of 11 and 9

What I know is that any number can be written as a difference of squares just if the number is either Odd or has a 4 as a factor.

Am I correct?

I am confused.

Kind regards!

This is not a rule/property. You can arrive at it 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 $$(AB)^2 - (BA)^2 = (10A + B)^2 - (10B + A)^2 = (10A)^2 + B^2 + 2*10A*B - ((10B)^2 + A^2 + 2*10B*A)$$

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

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

So now you know that the required expression will be divisible by 9 as well as 11.
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Re: † and ¥ represent nonzero digits, and (†¥)² - (¥†)²  [#permalink]

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21 Jul 2014, 18:02
I picked numbers in this one since I was able to ballpark the square root of the numbers shown:

A. 121 = 11
B. 361 = 19
C. 576 = 24
D. 961 = 31
E. 1089 = 33

Biggest mistake was not considering that 0 is also a perfect square. Immediately eliminated the OA because of this.
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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.
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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  [#permalink]

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08 Jul 2017, 12:33
WoundedTiger wrote:
† 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 assume the above symbols as $$x$$ and $$y$$ respectively for ease of understanding.

$$(xy)^2 - (yx)^2$$

as you know $$xy$$is a two digit number and can be written as $$10x + y$$

Similarly, $$yx$$ is a two digit number and can be written as $$10y + x$$

$$(10x + y)^2 - (10y + x)^2$$

$$(10x)^2 + (y)^2 + 2 * 10x * y - ((10y)^2 + (x)^2 + 2 * 10y * x)$$

$$(10x)^2 + (y)^2 + 20xy - ((10y)^2 + (x)^2 + 20xy)$$

$$(10x)^2 + (y)^2 + 20xy - (10y)^2 - (x)^2 - 20xy)$$

$$99x^2 - 99y^2$$

$$99 * (x^2 - y^2)$$

$$11 * 9 * (x^2 - y^2)$$

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

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

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21 Jul 2017, 06:43
Bunuel wrote:
† 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

Kudos for correct solution.

(†¥)^2 - (¥†)^2
= (10† + ¥)^2 - (10¥ + †)^2
= (100*†^2 + ¥^2 + 20†¥) - (100*¥^2 + †^2 + 20†¥)
= 99(†^2 - ¥^2)

So, the term is a multiple of 99.

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

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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.

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

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21 Sep 2017, 00:30
Assume the two symbols are T and V.

99 * (T^2 - V^2) is a perfect square.

So 9*11*(T + V)*(T - V) is a perfect square.

9 is already a perfect square.
If (T + V) is 11, we have 11^2.
So the leftover (T - V) should be a perfect square too.

Posted from my mobile device
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Re: † and ¥ represent nonzero digits, and (†¥)² - (¥†)²  [#permalink]

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21 Sep 2017, 00:47
WoundedTiger wrote:
† 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

(†¥)² - (¥†)² = (10†+¥)² - (10¥+†)² = (100†²+¥²+20†¥)² - (†²+100¥²+20†¥) = 99(†²-¥²)

So, the perfect square must be a multiple of 99

1089 is only multiple of 99 and a perfect square..

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

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21 Sep 2017, 08:02
WoundedTiger wrote:
† 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

First number is 10*x+y --> Its square is 100*x^2+y^2+20xy
2nd number is 10*y+x -->its square is 100*y^2+x^2+20xy

100*x^2+y^2+20xy - 100*y^2+x^2+20xy =99(x^2-y^2)

Only number divisible by 99 in the option is E
ANS:E
Re: † and ¥ represent nonzero digits, and (†¥)² - (¥†)²   [#permalink] 21 Sep 2017, 08:02

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