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Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
If a, b, c and d are four consecutive integers, and a^b = c^
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01 Oct 2019, 12:04
01 Oct 2019, 12:04
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Question Stats:
40% (02:35) correct
60% (02:21) wrong based on 60 sessions
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If \(a\), \(b\), \(c\) and \(d\) are four consecutive integers (not necessarily in that order), and \(a^b = c^d\), what is the least possible value of \(a+b+c+d\) ?
A) -8
B) -6
C) -2
D) 6
E) 12
A) -8
B) -6
C) -2
D) 6
E) 12
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If a, b, c and d are four consecutive integers, and a^b = c^
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02 Oct 2019, 05:31
02 Oct 2019, 05:31
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GreenlightTestPrep wrote:
If \(a\), \(b\), \(c\) and \(d\) are four consecutive integers (not necessarily in that order), and \(a^b = c^d\), what is the least possible value of \(a+b+c+d\) ?
A) -8
B) -6
C) -2
D) 6
E) 12
A) -8
B) -6
C) -2
D) 6
E) 12
This question takes some playing around with numbers.
It also helps to recognize the following properties:
\(1^k = 1\) for all values of k
\((-1)^k = 1\) for all EVEN integer values of k
\(k^0 = 1\) for all non-zero values of k
If \(a = -1\), \(b = -2\), \(c=-3\) and \(d = 0\), then the equation \(a^b = c^d\) becomes \((-1)^{-2} = (-3)^0\)
Simplify to get: \(1= 1\)....PERFECT!
In this case, \(a+b+c+d=(-1)+(-2)+(-3)+0=-6\)
So, it's possible to get a sum of -6 (answer choice B).
HOWEVER, perhaps it's possible to get a sum of -8 (answer choice A)
Well, it turns out that we CAN'T get a sum of -8. Here's why:
We know that the 4 numbers are CONSECUTIVE INTEGERS.
So, if we let x = the smallest integer, then
x+1 = the next integer
x+2 = the next integer
And x+3 = the next integer
So, the sum of these 4 consecutive integers = x + (x+1) + (x+2) + (x+3)
= 4x + 6
= 4x + 4 + 2
= 4(x+1) + 2
Notice that 4(x+1) is a multiple of 4
So, 4(x+1) + 2 is 2 greater than some multiple of 4
Since the sum of the four integers must be 2 greater than some multiple of 4, we
can eliminate answer choice A, since -8 is a multiple of 4
Answer: B
Cheers,
Brent
Re: If a, b, c and d are four consecutive integers, and a^b = c^
[#permalink]
02 Oct 2019, 06:44
02 Oct 2019, 06:44
this was tough but good q
