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

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

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.

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

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

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!

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

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

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)

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

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

GMAT assassins aren't born, they're made,
Rich

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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