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Re: If a, b, c and d are four consecutive integers, and a^b = c^ [#permalink]
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I first listed some consecutive integers to see what sums they can make:
-4, -3, -2, -1, 0, 1, 2, 3, 4
Here are the sums I get looking at groupings of 4: ...-10, -6, -2, 6, 10...
So right off the bat, I can eliminate answer choice A = -8, since no consecutive integers will add to it.

Next, I looked at the next lowest sum of integers to see if I can arrange them in a way that works:
-3, -2, -1, 0 add up to -6
I came up with (-1)^(-2) = (-3)^0 = 1
It worked!

So the answer choice is B = -6.
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Re: If a, b, c and d are four consecutive integers, and a^b = c^ [#permalink]
I would strongly recommend downloading the Ultimate GRE Quant Cheat Sheet for this.

It has very thorough chapters on exponent rules and sequence properties (both arithmetic and geometric, actually). This is a tough question, but without recalling certain formulas and rules, students will struggle to answer this in a reasonable time.

Recognise that the sequence is an arithmetic sequence following the pattern (n+1). So, let's write this sequence as

(n)(n+1)(n+2)(n+3)

the formula to find the sum of the terms in an arithmetic sequence is

Sum = middle term * number of terms

The middle term is the arithmetic mean of two terms that are equidistant from the median term.

The formula can be re-written as

Sum = (n/2) * (a1 + an)

Where n is the number of terms in the sequence and an is the nth term in the sequence.

If you know the common difference in the arithmetic sequence, there is another formula you could use. Given we are told that the sequence is made of consecutive integers, we know the common difference is +1. We won't waste time on that formula here, but I thought it's worth mentioning,

Anyway... getting back into it...

Sum = (n/2) * (a1 + an)

Let's call Sum "S". So in this case,

S = (4/2) * ((n) + (n + 3))
S = 2 * (2n + 3)
S = 4n + 6
S = 4n + 4 + 2
S = 4(n+1) + 2

So, we know that the sum of the terms is 2 greater than a multiple of 4.

Eliminate (A) and (E).

Now, we know that, generally speaking, the sum of the terms in an arithmetic sequence is least when the values in the sequence are least. In other words, our job is to minimise (n).

Let's try -6.

-6 = 4(n+1) + 2
-8 = 4(n+1)
(-8/4) = (n+1)
-2 = (n+1)

therefore, n = -3

Will that satisfy the restriction (a^b) = (c^d) given by the question?

Here's our sequence, assuming n = 3
(n)(n+1)(n+2)(n+3) is,
-3, -2, -1, 0

Remember some key exponent rules:

0^x = 1

(-1)^negative even integer = 1
(-1)^negative odd integer = -1

now... RTFQ! a, b, c, d, are not necessarily in order!

So, let's say:

(-1)^(−2)= 1

and

(-3)^(0) = 1

Bingo!
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Re: If a, b, c and d are four consecutive integers, and a^b = c^ [#permalink]
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Ultimate GRE Quant Cheat Sheet

It would be better if you could provide the students with a useful link. Or are you not able to link to it due to a few posts?
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Re: If a, b, c and d are four consecutive integers, and a^b = c^ [#permalink]
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Carcass wrote:
Ultimate GRE Quant Cheat Sheet

It would be better if you could provide the students with a useful link. Or are you not able to link to it due to a few posts?


yes, at the moment we cannot share a link.
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