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† and ¥ represent nonzero digits, and (†¥)²  (¥†)²
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21 Jul 2014, 10:53
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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
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21 Jul 2014, 11:54
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. Answer: E.
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Re: † and ¥ represent nonzero digits, and (†¥)²  (¥†)²
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21 Jul 2014, 12:12
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
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14 Oct 2015, 22:13
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
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19 Feb 2016, 06:29
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



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Re: † and ¥ represent nonzero digits, and (†¥)²  (¥†)²
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28 Sep 2015, 00:45
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
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19 Feb 2016, 03:47
† 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
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21 May 2016, 12:48
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.
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
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05 Jun 2017, 00:35
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  10yx = 9x  9y = 9(xy) 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)(ab))
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
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05 Feb 2019, 20:48
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 9What 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
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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
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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
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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)(6556) = 121*9 = 1089. Success!



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Re: † and ¥ represent nonzero digits, and (†¥)²  (¥†)²
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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. Hence, Answer is E
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Re: † and ¥ represent nonzero digits, and (†¥)^2  (¥†)^2 is a perfect squ
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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 So. Answer E



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Re: † and ¥ represent nonzero digits, and (†¥)^2  (¥†)^2 is a perfect squ
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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.



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† and ¥ represent nonzero digits, and (†¥)²  (¥†)²
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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)(6556) = 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
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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
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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.. Answer E



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Re: † and ¥ represent nonzero digits, and (†¥)²  (¥†)²
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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^2y^2) Only number divisible by 99 in the option is E ANS:E




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