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In a decreasing sequence of seven consecutive even integers, the sum o

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In a decreasing sequence of seven consecutive even integers, the sum o [#permalink]

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New post 16 Mar 2017, 09:43
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A
B
C
D
E

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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
[Reveal] Spoiler: OA

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Re: In a decreasing sequence of seven consecutive even integers, the sum o [#permalink]

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New post 16 Mar 2017, 10:14
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A quick trick, product of three consecutive even integers has to be a multiple of 4! or 24.

2*4*6 = 24
4*6*8 = 196
.
.
.
So on!

Here, we are asked for the product of three consecutive even integers.

Their product must be a multiple of 24. Only option satisfying this is 960.

Option B.


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Re: In a decreasing sequence of seven consecutive even integers, the sum o [#permalink]

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New post 16 Mar 2017, 11:21
Answer is B

to solve:

a+(a-2)+(a-4)+(a-6) ==> 4a-12=68 ==> 80/4 ==>20

so a =20

The number are

20
18
16
14
12
10
08

product of the last 3 numbers= 12*10*8 =960.
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Re: In a decreasing sequence of seven consecutive even integers, the sum o [#permalink]

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New post 17 Mar 2017, 01:59
let the digits be a, a-2, a-4, a-6, a-8, a-10, a-12

a +a-2 +a-4 +a-6 = 68
a = 20

therefore the last three numbers = 12, 10, 8
product of those three = 960
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Re: In a decreasing sequence of seven consecutive even integers, the sum o [#permalink]

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SajjadAhmad 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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Re: In a decreasing sequence of seven consecutive even integers, the sum o [#permalink]

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New post 02 Apr 2017, 22:59
2n+12, 2n+10, 2n+8, 2n+6, 2n+4, 2n+2,+ 2n

Sum of the first four integers= 2n+12+2n+10+2n+8+2n+6= 8n+36

8n+36=68
n= 4

Product of last three integers= 8*10*12= 960
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Re: In a decreasing sequence of seven consecutive even integers, the sum o [#permalink]

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New post 15 Jan 2018, 12:07
Hi All,

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:
[Reveal] Spoiler:
B


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Re: In a decreasing sequence of seven consecutive even integers, the sum o   [#permalink] 15 Jan 2018, 12:07
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