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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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12 Nov 2019, 23:54
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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

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

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14 Nov 2019, 08:09
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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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13 Nov 2019, 00:24
2
2
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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13 Nov 2019, 03:53
1
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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13 Nov 2019, 05:45
1
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 $$3 + 2a - a^2$$.

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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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13 Nov 2019, 21:23
1
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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13 Nov 2019, 21:59
1
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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13 Nov 2019, 22:02
1
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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13 Nov 2019, 22:09
1
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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Joined: 31 Oct 2015
Posts: 95
Re: If a, b, and c are consecutive integers, where a < b < c, which of the  [#permalink]

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13 Nov 2019, 23:33
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]

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01 Mar 2020, 09:22
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

a, b, c are consecutive integer. So a = k, b = k+1, c= k+2

c^2 - a^2 - b^2 = (k+2)^2 - (k+1)^2 - k^2

= (2k+3) - k^2
= - [(k^2 - 2k +1) -4]
= - [(k-1)^2 - 2^2]
= -(k-3)(k+1)

So this equation can written as the -1* [product of two number a & b, in which a-b = 4]

I. -77 = -1 * 7 * 11. 11-7 =4. SUFF
II. -32 = -1 * 4 * 8. 8-4 =4. SUFF
III. -21 = -1 * 3 * 7. 7-3=4. SUFF
III. -10. INSUFF
IV. 0. SUFF

IMO D.
Re: If a, b, and c are consecutive integers, where a < b < c, which of the   [#permalink] 01 Mar 2020, 09:22
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# If a, b, and c are consecutive integers, where a < b < c, which of the

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