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

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If a, b, and c are consecutive integers, where a < b < c, which of the  [#permalink]

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New post 13 Nov 2019, 00:54
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If a, b, and c are consecutive integers, where a < b < c, which of the  [#permalink]

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New post 14 Nov 2019, 09:09
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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......
\((b+1)^2 - ((b-1)^2 + b^2)=b^2+2b+1-(b^2-2b+1+b^2)=4b-b^2=b(4-b)=-(b)(b-4)\)

Look for the choices that can be converted in the form \(-(b)(b-4)\), 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  [#permalink]

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New post 13 Nov 2019, 01:24
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1
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  [#permalink]

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New post 13 Nov 2019, 04:53
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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


\(c^2−(a^2+b^2)\)
\(c^2=(a+2)^2=(a^2+4+4a)\)
\(a^2+b^2=a^2+(a+1)^2=a^2+a^2+1+2a=2a^2+2a+1\)
\(c^2−(a^2+b^2)=(a^2+4+4a)-(2a^2+2a+1)=-a^2+2a+3\)

\((A)-a^2+2a+3=-77…a^2-2a-80=0…(a-10)(a+8)=0:a=integer=valid\)
\((B)-a^2+2a+3=-32…a^2-2a-35=0…(a-7)(a+5)=0:a=integer=valid\)
\((C)-a^2+2a+3=-21…a^2-2a-24=0…(a-6)(a+4)=0:a=integer=valid\)
\((E)-a^2+2a+3=0…a^2-2a-3=0…(a-3)(a+1)=0:a=integer=valid\)
\((D)-a^2+2a+3=-10…a^2-2a-13=0…13=prime:a=not.integer=invalid\)

Ans (D)
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Re: If a, b, and c are consecutive integers, where a < b < c, which of the  [#permalink]

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New post 13 Nov 2019, 06:45
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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  [#permalink]

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New post 13 Nov 2019, 22:23
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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=b-1
so c^2 - (a^2 + b^2) = (b+1)^2 - [(b^2) + (b-1)^2]
b^2 + 2b +1 -[b^2 + b^2 -2b + 1]
4b - b^2 = b(4-b)
from this we know that when b=4, then b(4-b)=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(4-b) to -10 and find the determinant in order to confirm if it leads to a perfect square.
4b-b^2=-10
b^2-4b-10=0
The determinant of the above quadratic equation is [16-4(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  [#permalink]

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New post 13 Nov 2019, 22:59
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D
C^2-(b^2+a^2)=(a+2)^2-(a+1)^2-a^2
=4-(a-1)^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  [#permalink]

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New post 13 Nov 2019, 23:02
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Answer is -10 . Option D
Consider the integers : 10, 11, 12
12^2-(10^2+11^2) = 144-221 = -77
Consider integers : 7,8,9
9^2-(8^2+7^2) = 81-113 = -32
Consider integers : 6,7,8
8^2-(7^2 + 6^2 ) = 64-85 = -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  [#permalink]

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New post 13 Nov 2019, 23:09
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Ans.is D
C^2-(b^2+a^2)=(a+2)^2-(a+1)^2-a^2
=4-(a-1)^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  [#permalink]

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New post 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

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Re: If a, b, and c are consecutive integers, where a < b < c, which of the   [#permalink] 14 Nov 2019, 00:33
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