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When a certain perfect square is increased by 148, the result is anoth [#permalink]
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22 Apr 2015, 03:20
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Re: When a certain perfect square is increased by 148, the result is anoth [#permalink]
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22 Apr 2015, 04:18
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. \(x^2 + 148 = y^2\) (YX)(Y+X)=148 2×74=148 (3836)(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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22 Apr 2015, 04:36
a^2 +148= b^2 148=b^2  a^2 148=(b  a)*(b +a) 2*74=(3836) *(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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22 Apr 2015, 05:53
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)(xy) = 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) (3836) = 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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22 Apr 2015, 11:11
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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27 Apr 2015, 04:00
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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07 Mar 2016, 20:10
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)(yx) possible options: x+y=74 yx=2 > x=36, y=38
or x+y = 37 yx=4 can't be, since we would get y a noninteger.
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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07 Mar 2016, 23:16
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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01 Jun 2016, 18:53
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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01 Jun 2016, 22:39
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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26 Sep 2016, 11:00
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? Posted from my mobile device



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Re: When a certain perfect square is increased by 148, the result is anoth [#permalink]
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26 Sep 2016, 12:23
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 = 144 4 B. 1369 1369 + 148 = 151 7C. 1681 1681 + 148 = 182 9D. 1764 1764 + 148 = 191 2E. 2500 2500 + 148 = 264 8Now, 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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Updated on: 21 May 2017, 15:01
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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When a certain perfect square is increased by 148, the result is anoth [#permalink]
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22 Apr 2017, 20:35
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(n2)^2=4(n1) 4(n1)=148 n=38 n2=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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07 Feb 2018, 12:28
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 110, 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: GMAT assassins aren't born, they're made, Rich
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Re: When a certain perfect square is increased by 148, the result is anoth [#permalink]
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21 Feb 2018, 03:59
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]
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23 Apr 2018, 07:30
Let the numbers be x and y. y^2+148=x^2. ; or x^2y^2=148 ; (x+y)*(xy)=74*2 ; Solve for y ; you will get y=36. y^2 is 1296 or option A.
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Re: When a certain perfect square is increased by 148, the result is anoth [#permalink]
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24 Apr 2018, 11:32
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 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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