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When a certain perfect square is increased by 148, the result is anoth

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When a certain perfect square is increased by 148, the result is anoth [#permalink]

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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

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: When a certain perfect square is increased by 148, the result is anoth [#permalink]

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Bunuel 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

Kudos for a correct solution.



\(x^2 + 148 = y^2\)

(Y-X)(Y+X)=148
2×74=148
(38-36)(38+36)=148

So \(X^2=36^2\)

Ans A (as unit digit is 6)
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Re: When a certain perfect square is increased by 148, the result is anoth [#permalink]

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New post 22 Apr 2015, 03:36
a^2 +148= b^2
148=b^2 - a^2
148=(b - a)*(b +a)
2*74=(38-36) *(38+36)
which gives
b=38 and a=36

square of 36 is 1296
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Re: When a certain perfect square is increased by 148, the result is anoth [#permalink]

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Bunuel 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

Kudos for a correct solution.


x^2 + 148 = y^2
(x+y)(x-y) = 148

148 = 2 * 74 , By looking at options all the numbers given are above 30, since 30^2 = 900, so break down 74 into sum of 30's such that the difference is 2

74 = 38+36

(38+36) (38-36) = 148

Hence answer is 36^2 = 1296 + 148 = 1444 = 38^2
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Re: When a certain perfect square is increased by 148, the result is anoth [#permalink]

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Every perfect square ends with unit digit as one of the following -
1,4,5,6,9,0

add 148 in each option i.e. 8 in each options unit digit and check unit digit to discard options
A)6+8=4
B)9+8=7 .... Eliminated
C)1+8=9
D)4+8=2 .....Eliminated
E)0+8=8 ...... Eliminated

Now facotrize option A ) 1296+148=1444 this can be factorize as 361*4=19*19*2*2

Hence this is answer A ..
no need to check option C.
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Re: When a certain perfect square is increased by 148, the result is anoth [#permalink]

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Bunuel 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

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Let’s call the two perfect squares x^2 and y^2, respectively. Then the given information translates as x^2+148=y^2. Subtracting x^2 gives 148=y^2−x^2, a difference of squares. This, in turn, factors as (y+x)(y−x)=148.

The next step is tricky. It begins with factoring 148, which breaks down as 2∗2∗37. Since we’re dealing with perfect squares, x and y are positive integers, and (y+x) and (y−x) must be paired integer factors of 148. The options are 148∗1,74∗2, and 37∗4. But our number properties establish that (y+x) and (y−x) must be either both odd or both even, so only 74∗2 is an actual possibility. And because for any positive integers (y+x)>(y−x), we can conclude that y+x=74 and y–x=2. Solving by elimination, 2y=76, y=38, and x=36.

Finally, we just need to square 36. But rather than multiplying it out, note that 36^2 ends in 6 – as does only one answer, choice A. This answer must be the one we want.
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Re: When a certain perfect square is increased by 148, the result is anoth [#permalink]

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i mistakenly believed that perfect square is multiplied by 148...
so..we know x^2 +148 = y^2
148 = y^2 - x^2
148 = (x+y)(y-x)
possible options:
x+y=74
y-x=2 -> x=36, y=38

or
x+y = 37
y-x=4
can't be, since we would get y a non-integer.

now..it's either 36 or 38
look at the last digits..6^2 = 6..so last digit must be 6
8^2= 4 -> so last digit must be 4. We can eliminate B, C, and E.

36*36 = 1296 - A

38*38 = 1444 - don't have such an answer.

A
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Re: When a certain perfect square is increased by 148, the result is anoth [#permalink]

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Bunuel 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

Kudos for a correct solution.


It helps to know how to quickly calculate squares of numbers.

\(30^2 = 900, 31^2 = 961 ( = (30 + 1)^2 = 900 + 1 + 2*30)\)
\(40^2 = 1600, 41^2 = 1681 (= (40+1)^2 = 1600 + 1 + 2*40)\)

We know that the distance between consecutive squares increases as we go to greater numbers.
\(30^2\) to \(31^2\) is a difference of 61.
\(40^2\) to \(41^2\) is a difference of 81.

When you add 148 to a perfect square to get another in this range, it means the numbers must be 2 apart such as 33 and 35 etc. Also they must lie between 30 and 40 because twice of 61 is 122 and twice of 81 is 162. 148 lies somewhere in between.

(A) and (B) are the only two possible options.
Consider (A) - it ends in 6 so the square root ends in 6 too. You add 148 then it will end in 4. The perfect square of number ending in 8 will end in 4. So it works.
Consider (B) too - it ends in 9 so the square root ends in 3 or 7. When you add 148, it ends in 7. No perfect square ends in 7 so this is out.

Answer (A)
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Re: When a certain perfect square is increased by 148, the result is anoth [#permalink]

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New post 01 Jun 2016, 17:53
Bunuel wrote:
Bunuel 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

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Let’s call the two perfect squares x^2 and y^2, respectively. Then the given information translates as x^2+148=y^2. Subtracting x^2 gives 148=y^2−x^2, a difference of squares. This, in turn, factors as (y+x)(y−x)=148.

The next step is tricky. It begins with factoring 148, which breaks down as 2∗2∗37. Since we’re dealing with perfect squares, x and y are positive integers, and (y+x) and (y−x) must be paired integer factors of 148. The options are 148∗1,74∗2, and 37∗4. But our number properties establish that (y+x) and (y−x) must be either both odd or both even, so only 74∗2 is an actual possibility. And because for any positive integers (y+x)>(y−x), we can conclude that y+x=74 and y–x=2. Solving by elimination, 2y=76, y=38, and x=36.

Finally, we just need to square 36. But rather than multiplying it out, note that 36^2 ends in 6 – as does only one answer, choice A. This answer must be the one we want.


"But our number properties establish that (y+x) and (y−x) must be either both odd or both even"

Can someone please explain why this is the case?
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Re: When a certain perfect square is increased by 148, the result is anoth [#permalink]

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tae808 wrote:
"But our number properties establish that (y+x) and (y−x) must be either both odd or both even"

Can someone please explain why this is the case?


If y + x is Even, it is either of these two cases:
Even + Even = Even
Odd + Odd = Even

In both cases,
Even - Even = Even
Odd - Odd = Even

So y - x will be even when y + x is even.

Same holds for odd too.

If y + x is odd, it is either of these two cases:
Even + Odd = Odd
Odd + Even = Odd

In both cases,
Even - Odd = Odd
Odd - Even = Odd

So when y + x is odd, y - x must be odd too.
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When a certain perfect square is increased by 148, the result is anoth [#permalink]

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New post 26 Sep 2016, 10:00
VeritasPrepKarishma wrote:
Bunuel 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

Kudos for a correct solution.


It helps to know how to quickly calculate squares of numbers.

\(30^2 = 900, 31^2 = 961 ( = (30 + 1)^2 = 900 + 1 + 2*30)\)
\(40^2 = 1600, 41^2 = 1681 (= (40+1)^2 = 1600 + 1 + 2*40)\)

We know that the distance between consecutive squares increases as we go to greater numbers.
[m]30^2[/

When you add 148 to a perfect square to get another in this range, it means the numbers must be 2 apart such as 33 and 35 etc. Also they must lie between 30 and 40 because twice of 61 is 122 and twice of 81 is 162. 148 lies somewhere in between.

(A) and (B) are the only two possible options.
Consider (A) - it ends in 6 so the square root ends in 6 too. You add 148 then it will end in 4. The perfect square of number ending in 8 will end in 4. So it works.
Consider (B) too - it ends in 9 so the square root ends in 3 or 7. When you add 148, it ends in 7. No perfect square ends in 7 so this is out.

Answer (A)


Could you please explain the above?
Why does it have to be 2 digits apart? And why have you multiplied it by 2 to find the possible range?

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Re: When a certain perfect square is increased by 148, the result is anoth [#permalink]

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New post 26 Sep 2016, 11:23
Bunuel 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

Kudos for a correct solution.


\(x^2\) \(+\) \(148\) = \(y^2\)

Or, \(x^2\) = \(y^2\) \(-\) \(148\)

We know , Units digit of a square number must be - { 0 , 1 , 4 , 5 , 6 , 9 }


We are asked to find the value of original number , so by rule Original Square Number + 148 must end in 0 or 1 or 4 or 5 or 6 or 9

Now, Check the options -

A. 1296

1296 + 148 = 1444

B. 1369

1369 + 148 = 1517

C. 1681

1681 + 148 = 1829

D. 1764

1764 + 148 = 1912

E. 2500

2500 + 148 = 2648

Now, you find only option (A) and (C) are possible... So we can check these individually

(A) \(\sqrt{1444}\) = 38

(C) \(\sqrt{1829}\) = 42.77


So, Correct answer must be (A)

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When a certain perfect square is increased by 148, the result is anoth [#permalink]

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New post 22 Apr 2017, 15:52
Only numbers that end with 1,4,5,6,9 & 0 can be perfect squares

Only choices A & C qualify after adding 148.

So for 1296 & 1681, the only perfect square that I can think of is 30^2 & 40^2
30^2 = 900 & 40^2 = 1600
Start with 1296: 1296 + 148 = 1444.
1296 is greater than 30^2 and the units digit has to be 4 of the final square.
So 1444 is either 32^2 or 38^2
38^2 = (30+8)^2 = (900 + 2*30*8 + 64) = 1444
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Last edited by colorblind on 21 May 2017, 14:01, edited 1 time in total.
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When a certain perfect square is increased by 148, the result is anoth [#permalink]

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New post 22 Apr 2017, 19:35
Bunuel 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

Kudos for a correct solution.


as 148 is even, assume square roots are consecutive odd or even integers
n^2-(n-2)^2=4(n-1)
4(n-1)=148
n=38
n-2=36
36^2=1296
A
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Re: When a certain perfect square is increased by 148, the result is anoth [#permalink]

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New post 07 Feb 2018, 11:28
Hi All,

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 Answers 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:
[Reveal] Spoiler:
A


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Re: When a certain perfect square is increased by 148, the result is anoth [#permalink]

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New post 21 Feb 2018, 02:59
I solved it this way.
No perfect square ends with digit 2,3,7,8.
So if we add 148 to any number and the result is perfect square, the result should not have unit digit 2,3,7,8.
Based on this we can strike B,D and E.
Between A and C. When you add 148 to C result is 1829.
40^2 = 1600, so C must be more than 40 square. I randomly checked for square of 43 and 42.
(40+3)^2 = 1600+9+2*40*3---> 1849.
So 1829 is not a perfect square. Answer A.

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Re: When a certain perfect square is increased by 148, the result is anoth   [#permalink] 21 Feb 2018, 02:59
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