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If a, b, and c are consecutive integers, where a < b < c, which of the
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13 Nov 2019, 00:54
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Competition Mode Question If a, b, and c are consecutive integers, where \(a < b < c\), which of the following cannot be the value of \(c^2  (a^2 + b^2)\)? (A) 77 (B) 32 (C) 21 (D) 10 (E) 0 Are You Up For the Challenge: 700 Level Questions
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If a, b, and c are consecutive integers, where a < b < c, which of the
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14 Nov 2019, 09:09
Bunuel wrote: Competition Mode Question If a, b, and c are consecutive integers, where \(a < b < c\), which of the following cannot be the value of \(c^2  (a^2 + b^2)\)? (A) 77 (B) 32 (C) 21 (D) 10 (E) 0 Are You Up For the Challenge: 700 Level Questions Convert \(c^2  (a^2 + b^2)\) in terms of b...... \((b+1)^2  ((b1)^2 + b^2)=b^2+2b+1(b^22b+1+b^2)=4bb^2=b(4b)=(b)(b4)\) Look for the choices that can be converted in the form \((b)(b4)\), that is product of numbers that differ by 4 *()(A) 77 = \((11)(7)\) (B) 32 = \((8)(4)\) (C) 21 = \((7)(3)\) (D) 10 = \((10)(1)=(5)(2)\)...NO(E) 0 = \((4)(0)\) D
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Re: If a, b, and c are consecutive integers, where a < b < c, which of the
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13 Nov 2019, 01:24
If a, b, and c are consecutive integers, where a<b<c, which of the following cannot be the value of c2−(a2+b2)?
(A) 77 (B) 32 (C) 21 (D) 10 (E) 0
Let the values of a = n  1, b = n & c = n + 1 > c^2−(a^2+b^2) = (n + 1)^2  [(n  1)^2 + n^2] > n^2 + 2n + 1  2n^2 + 2n  1 > n^2 + 4n > (n^2  4n + 4  4) > (n  2)^2 + 4
(A) 77 = (9)^2 + 4 > Possible (B) 32 = (6)^2 + 4 > Possible (C) 21 = (5)^2 + 4 > Possible (D) 10 > Not Possible (E) 0 = (2)^2 + 4 > Possible
IMO Option D



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Re: If a, b, and c are consecutive integers, where a < b < c, which of the
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13 Nov 2019, 04:53
Quote: If a, b, and c are consecutive integers, where a<b<c, which of the following cannot be the value of \(c^2−(a^2+b^2)\)?
(A) 77 (B) 32 (C) 21 (D) 10 (E) 0 \(c^2−(a^2+b^2)\) \(c^2=(a+2)^2=(a^2+4+4a)\) \(a^2+b^2=a^2+(a+1)^2=a^2+a^2+1+2a=2a^2+2a+1\) \(c^2−(a^2+b^2)=(a^2+4+4a)(2a^2+2a+1)=a^2+2a+3\) \((A)a^2+2a+3=77…a^22a80=0…(a10)(a+8)=0:a=integer=valid\) \((B)a^2+2a+3=32…a^22a35=0…(a7)(a+5)=0:a=integer=valid\) \((C)a^2+2a+3=21…a^22a24=0…(a6)(a+4)=0:a=integer=valid\) \((E)a^2+2a+3=0…a^22a3=0…(a3)(a+1)=0:a=integer=valid\) \((D)a^2+2a+3=10…a^22a13=0…13=prime:a=not.integer=invalid\) Ans (D)



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Re: If a, b, and c are consecutive integers, where a < b < c, which of the
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13 Nov 2019, 06:45
If a, b, and c are consecutive integers, where a<b<c, which of the following cannot be the value of \(c^2−(a^2+b^2)\)? (A) 77 (B) 32 (C) 21 (D) 10 (E) 0 \(c^2−(a^2+b^2)\) suggests that a,b and c can take any integer values where b = a + 1 and c = a + 2 Now, converting the equation in terms of lowest variable i.e. 'a' \(c^2−(a^2+b^2)\) = \((a+2)^2  (a^2 + (a + 1)^2)\) = \(3 + 2a  a^2\) For sake of ease checking values starting with a = 1 gives \(3 + 2a  a^2\) = 4 if a = 3, \(3 + 2a  a^2\) = 0 a = 6, \(3 + 2a  a^2\) = 21 a = 7, \(3 + 2a  a^2\) = 32 a = 10, \(3 + 2a  a^2\) = 77 Though by observing we can see that \(3 + 2a  a^2\) would give a negative value after giving a value of '0' for a = 3. We can stop at a = 4 which gives 21 thus 10 cannot be the value of \(3 + 2a  a^2\). IMO Answer D.
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Re: If a, b, and c are consecutive integers, where a < b < c, which of the
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13 Nov 2019, 22:23
Given that a,b, and c are consecutive integers such that a<b<c. We are to determine which of the following cannot be the value of c^2  (a^2 + b^2)
Since a,b, and c are consecutive numbers, and b is the number in the middle, then c=b+1 and a=b1 so c^2  (a^2 + b^2) = (b+1)^2  [(b^2) + (b1)^2] b^2 + 2b +1 [b^2 + b^2 2b + 1] 4b  b^2 = b(4b) from this we know that when b=4, then b(4b)=0, hence option E is out. when b=5, 5(1) = 5 when b=6, 6(2)=12 Since 10 lies between 5 and 12, there is no way of getting an integer value of b which yields 10 in the given expression above. The answer, therefore, has to be 10.
A further check is to equate b(4b) to 10 and find the determinant in order to confirm if it leads to a perfect square. 4bb^2=10 b^24b10=0 The determinant of the above quadratic equation is [164(1)(10)] = 56 Since 56 is not a perfect square to begin with, there is no way we can get integral roots of the above equation.
The answer is, therefore, option D.



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Re: If a, b, and c are consecutive integers, where a < b < c, which of the
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13 Nov 2019, 22:59
D C^2(b^2+a^2)=(a+2)^2(a+1)^2a^2 =4(a1)^2 =4(int)^2 is never 10
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Re: If a, b, and c are consecutive integers, where a < b < c, which of the
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13 Nov 2019, 23:02
Answer is 10 . Option D Consider the integers : 10, 11, 12 12^2(10^2+11^2) = 144221 = 77 Consider integers : 7,8,9 9^2(8^2+7^2) = 81113 = 32 Consider integers : 6,7,8 8^2(7^2 + 6^2 ) = 6485 = 21 Consider integers : 3,4,5 5^2(3^2+4^2) =0
Even if the numbers are negative , since we are squaring the values it will become positive.
Hence, answer is 10
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Re: If a, b, and c are consecutive integers, where a < b < c, which of the
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13 Nov 2019, 23:09
Ans.is D C^2(b^2+a^2)=(a+2)^2(a+1)^2a^2 =4(a1)^2 =4(int)^2 is never 10
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Re: If a, b, and c are consecutive integers, where a < b < c, which of the
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14 Nov 2019, 00:33
The answer is D. Using pattern, we figure out that as the numbers increase , the difference in the squares also increase. For numbers 3,4,5, The given expression is zero When you try 4,5,6 , the given expression is 6 When you try 5,6,7, the given expression is 12 As you can see 10 is not possible Therefore D Posted from GMAT ToolKit




Re: If a, b, and c are consecutive integers, where a < b < c, which of the
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