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In a decreasing sequence of seven consecutive even integers

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by Jay@ManhattanReview » Sun Mar 12, 2017 12:50 am
vikkimba17 wrote:In a decreasing sequence of seven consecutive even integers, the sum of the first four integers is 68. What is the product of the last three integers in the sequence?

A)1,000
B)960
C)925
D)30
E)25
Hi vikkimba17,

We have a decreasing sequence of seven consecutive even integers.

Since the sum of the first four integers is 68, the average of them = 68/4 = 17. '17' is the integer that lies between the 6th and the 5th integer. Thus, the 5th integer would be 17 - 1 = 16 and the 6th one would be 17 + 1 = 18. Since the sequence of seven numbers is a sequence of consecutive even integers, the decreasing sequence would be: 20, 18, 16, 14, 12, 10, 8.

The product of the last three integers in the sequence = 12*10*8 = 960.

The correct answer: B

Hope this helps!

Relevant book: Manhattan Review GMAT Number Properties Guide

-Jay
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by vikkimba17 » Sun Mar 12, 2017 3:19 am
Thanks for the answer & explanation.
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by Brent@GMATPrepNow » Sun Mar 12, 2017 7:37 am
vikkimba17 wrote:In a decreasing sequence of seven consecutive even integers, the sum of the first four integers is 68. What is the product of the last three integers in the sequence?

A)1,000
B)960
C)925
D)30
E)25
We can also solve the question algebraically.

In a decreasing sequence of seven consecutive even integers....
Let x = the 1st number
So, x-2 = the 2nd number (since we have descending EVEN integer)
And x-4 = the 3rd number
And x-6 = the 4th number
And x-8 = the 5th number
And x-10 = the 6th number
And x-12 = the 7th number

...the sum of the first four integers is 68
We get: x + (x-2) + (x-4) + (x-6) = 68
Simplify: 4x - 12 = 68
Solve: x = 20
So, the 1st number is 20, which means all 7 numbers are 20, 18, 16, 14, 12, 10, 8

What is the product of the last three integers in the sequence?
(12)(10)(8) = 960

Answer: B

Cheers,
Brent
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by vikkimba17 » Sun Mar 12, 2017 7:40 am
Hi Brent, sounds great with this method. Thanks
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by [email protected] » Sun Mar 12, 2017 9:50 am
Hi vikkimba17,

There are some built-in Number Properties in this question that we can take advantage of. To start, since we're taking the product of 3 CONSECUTIVE EVEN integers, the product MUST be even - so we can eliminate Answers C and E immediately.

Next, we can think in terms of the what the product COULD be (regardless of the what the first 4 integers in the sequence were). Since the remaining three answers all end in a 0, one of the three even integers MUST be 10....

(10)(8)(6) = 480

Notice how that's exactly HALF of one of the answers. What if we changed the 6 into a 12...

(12)(10)(8) = 960

That's a match for one of the options; and since Answer A isn't a product that we can get to with 3 consecutive evens, 960 MUST be the answer.

Final Answer: B

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by Jeff@TargetTestPrep » Thu Mar 16, 2017 12:16 pm
vikkimba17 wrote:In a decreasing sequence of seven consecutive even integers, the sum of the first four integers is 68. What is the product of the last three integers in the sequence?
A)1,000
B)960
C)925
D)30
E)25
The seven even integers in the sequence are as follows:

x, x - 2, x - 4, x - 6, x - 8, x - 10, x - 12

Since the sum of the first 4 integers is 68, we have:

x + x - 2 + x - 4 + x - 6 = 68

4x = 80

x = 20

Since our last 3 integers are x - 8, x - 10, and x - 12, they are 12, 10, and 8, and thus their product is 12 x 10 x 8 = 960.

Answer: B

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by Matt@VeritasPrep » Thu Mar 16, 2017 7:58 pm
But there's a much easier way! Cheat the answers: find one that actually IS the product of three consecutive evens!

C and E are odd, so they're out.

D is easy: 2 * 3 * 5. Nope!

A is easy too: 10 * 10 * 10. No luck!

So it must be B. It divides by 10 and is around 1000, so 10 is almost certainly one of the evens. From there the others are easy to find: 8 * 10 * 12.
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by Matt@VeritasPrep » Thu Mar 16, 2017 8:00 pm
There's also another quick hack: exploiting the fact that your first four integers are evenly spaced. Given that, the average of the four must also be THE MEDIAN of the four. Since the average is 17, the median is also 17, and the two middle integers are 16 and 18.

From there, our set is 20, 18, 16, 14, 12, 10, 8, and we're set.
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by Matt@VeritasPrep » Thu Mar 16, 2017 8:01 pm
But I definitely think the problem is designed to respond to my first solution: if you spot the Easter egg, you save a lot of time, but there's a nice, intuitive, timewasting trap built in.
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