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Problem Solving - Perfect squares

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Problem Solving - Perfect squares

by lucas211 » Sat Feb 20, 2016 3:27 am

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Hello everybody
Would appreciate a little help on the following question:

When a certain perfect square is increased by 148, the result is another perfect square. What is the value of the original perfect square?

a) 1296
b) 1369
c) 1681
d) 1764
e) 2500

Thanks in advance :-)
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by GMATGuruNY » Sat Feb 20, 2016 3:48 am
lucas211 wrote:Hello everybody
Would appreciate a little help on the following question:

When a certain perfect square is increased by 148, the result is another perfect square. What is the value of the original perfect square?

a) 1296
b) 1369
c) 1681
d) 1764
e) 2500
Let x² = the smaller perfect square and y² = the greater perfect square.

A certain perfect square is increased by 148.
The result is another perfect square.
Translated into math:
x² + 148 = y².

Thus:
x² - y² = -148

(x+y)(x-y) = (74)(-2)

(x+y)(x-y) = (36+38)(36-38).

Resulting values:
x=36, y=38.
x² = 36² = integer with a units digit of 6.

The correct answer is A.
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by [email protected] » Sat Feb 20, 2016 9:23 am
Hi lucas211,

If you don't see the 'quadratic approach' to this question, you can still solve it with a bit of 'brute force' and some Number Property knowledge. This approach won't be elegant, but if you're comfortable doing math by hand, it will get you the correct answer.

To start, we're dealing with two PERFECT SQUARES - and perfect squares can only end in certain digits. Try listing out the numbers 1-10, inclusive AND the results when you square those numbers. You'll find that the perfect squares can end in only the following digits: 0, 1, 4, 5, 6 and 9. All of the answer choices fit that limitation, BUT we're also told that adding 148 will create another perfect square.

As an example, adding 148 to a number that ends in 0 will create a new number that ends in 8.... but THAT new number won't be a perfect square (since it ends in 8). You can eliminate Answers B, D and E for that reason.

Between A and C, you just have to do a little more work to prove which is the answer:

Answer A: 1296

The square root of 1296 has to be between 30 and 40 (since 30^2 = 900 and 40^2 = 1600). Since it ends in a 6, the units digit has to be either a 4 or a 6. With a little multiplication, we can figure it out...

34^2 = 1156 which is too small
36^2 = 1296 which is a MATCH

Now we just have to see what happens when we add 148 to it...

1296+148 = 1444

From our prior work, we know that the square root of 1444 has to have a units digit of either 2 or 8. 1444 is greater than 1296, so the square root of 1444 has to be greater than 36 but less than 40. Let's try 38....

38^2 = 1444 which matches perfectly with everything else that we were told, so this must be the answer.

Final Answer: A

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by lucas211 » Sun Feb 21, 2016 10:17 am
Thanks to both of you.

Appreciate it! :-)
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by lucas211 » Tue Feb 23, 2016 4:34 am
Hi Mitch

Thank you for your answer.
Would you be kind to elaborate on the following steps:

Thus:
x² - y² = -148

(x+y)(x-y) = (74)(-2)

(x+y)(x-y) = (36+38)(36-38)

I am having troubles with understanding how we get to find the (74)(-2) and afterwards the (36+38)(36-38).

Thanks in advance
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by GMATGuruNY » Tue Feb 23, 2016 5:13 am
lucas211 wrote:Hi Mitch

Thank you for your answer.
Would you be kind to elaborate on the following steps:

Thus:
x² - y² = -148

(x+y)(x-y) = (74)(-2)

(x+y)(x-y) = (36+38)(36-38)

I am having troubles with understanding how we get to find the (74)(-2) and afterwards the (36+38)(36-38).

Thanks in advance
Since x² and y² are perfect squares, x and y are positive integers.

(x+y)(x-y) = (74)(-2).

This equation implies that x+y = 74 and x-y = -2.
In other words, x and y are two positive integers with a sum of 74 and difference of -2.
Some students will be able to deduce x=36 and y=38 by trial and error.

We could also solve for x and y algebraically.
Adding together x+y = 74 and x-y = -2, we get:
(x+y) + (x-y) = 74 + (-2)
2x = 72
x = 36.

Substituting x=36 into x+y = 74, we get:
36+y = 74
y = 38.
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by lucas211 » Tue Feb 23, 2016 2:47 pm
GMATGuruNY wrote:
lucas211 wrote:Hi Mitch

Thank you for your answer.
Would you be kind to elaborate on the following steps:

Thus:
x² - y² = -148

(x+y)(x-y) = (74)(-2)

(x+y)(x-y) = (36+38)(36-38)

I am having troubles with understanding how we get to find the (74)(-2) and afterwards the (36+38)(36-38).

Thanks in advance
Since x² and y² are perfect squares, x and y are positive integers.

(x+y)(x-y) = (74)(-2).

This equation implies that x+y = 74 and x-y = -2.
In other words, x and y are two positive integers with a sum of 74 and difference of -2.
Some students will be able to deduce x=36 and y=38 by trial and error.

We could also solve for x and y algebraically.
Adding together x+y = 74 and x-y = -2, we get:
(x+y) + (x-y) = 74 + (-2)
2x = 72
x = 36.

Substituting x=36 into x+y = 74, we get:
36+y = 74
y = 38.
Hi Mitch

And we know that (x+y) is 74 and (x-y) is -2 (and not for example 148 and -1 or 37 and -4 respectively) because:
the prime factors of 148 is 2,2,37, and
148, 1 and 37, 4 will not be able to give us the desired outcome with integers.
for example lets try with 148 and 1.

(x+y)(x-y) = (148)(-1)
(x+y)(x-y) => here we can´t make it fit while x and y have to be integers.

the same with 37 and -4

is this correct understood?

Thanks
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by GMATGuruNY » Wed Feb 24, 2016 4:37 am
lucas211 wrote:
GMATGuruNY wrote:
lucas211 wrote:Hi Mitch

Thank you for your answer.
Would you be kind to elaborate on the following steps:

Thus:
x² - y² = -148

(x+y)(x-y) = (74)(-2)

(x+y)(x-y) = (36+38)(36-38)

I am having troubles with understanding how we get to find the (74)(-2) and afterwards the (36+38)(36-38).

Thanks in advance
Since x² and y² are perfect squares, x and y are positive integers.

(x+y)(x-y) = (74)(-2).

This equation implies that x+y = 74 and x-y = -2.
In other words, x and y are two positive integers with a sum of 74 and difference of -2.
Some students will be able to deduce x=36 and y=38 by trial and error.

We could also solve for x and y algebraically.
Adding together x+y = 74 and x-y = -2, we get:
(x+y) + (x-y) = 74 + (-2)
2x = 72
x = 36.

Substituting x=36 into x+y = 74, we get:
36+y = 74
y = 38.
Hi Mitch

And we know that (x+y) is 74 and (x-y) is -2 (and not for example 148 and -1 or 37 and -4 respectively) because:
the prime factors of 148 is 2,2,37, and
148, 1 and 37, 4 will not be able to give us the desired outcome with integers.
for example lets try with 148 and 1.

(x+y)(x-y) = (148)(-1)
(x+y)(x-y) => here we can´t make it fit while x and y have to be integers.

the same with 37 and -4

is this correct understood?

Thanks
Your understanding is correct.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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by lucas211 » Wed Feb 24, 2016 5:25 am
Thanks a lot :-)
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by Matt@VeritasPrep » Tue Mar 01, 2016 11:36 pm
How about

x² + 148 = y²

148 = y² - x²

148 = (y + x)(y - x)

Since (y - x) and (y + x) are both positive integers, we must have one of these three pairs:

1 * 148
2 * 74
4 * 37

But we can't have one number be odd and the other be even, or we'll have to have either x or y be a decimal! So our only solution is 2 * 74, which leaves

y + x = 74
y - x = 2
y = 38, x = 36

From there, our number is x², or 36², which must end in 6 and hence be A.
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by Scott@TargetTestPrep » Tue Apr 24, 2018 10:40 am
lucas211 wrote: When a certain perfect square is increased by 148, the result is another perfect square. What is the value of the original perfect square?

a) 1296
b) 1369
c) 1681
d) 1764
e) 2500
We can let the original value (before it's squared) = x for some positive integer x, and the new value (before it's squared) = x + k for some positive integer k. We can create the equation:

x^2 + 148 = (x + k)^2

x^2 + 148 = x^2 + 2kx + k^2

148 = 2kx + k^2

Since both k and x are positive, we see that k^2 < 148. Thus k ≤ 12. Also, since 148 and 2kx are even, k must be even also. Thus k can only be 2, 4, 6, 8, 10 or 12. Let's analyze each of these values until we find a suitable value for x.

If k = 2, then

148 = 2(2)x + 2^2

144 = 4x

36 = x

We see that x can be 36 and 36^2 = 1296, which is choice A.

Answer: A

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