Eight Integers

Algebra Level 1

8 consecutive even integers have a sum of 104. What is the product of the smallest and largest integers?

138 120 248 140 230 134 200 121

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Wesley Tam by wesley tam , 20, USA

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2 solutions

James Cochrane
May 27, 2015

104=(8/2) * (2 * smallest + (8-1) * 2))

smallest=6

104=(8/2)*(6+largest)

largest=20

smallest*largest=120

I'm just using two versions of the sum of arithmetic series formulae.

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Moderator note:

Neat trick! This is the solution I'm looking for.

can you explain me how you did it with a.p. ?

Tootie Frootie - 6 years ago
Wesley Tam
May 26, 2015

Using algebra,we can find the answer so...

x+(x+2)+(x+4)+(x+6)+(x+8)+(x+10)+(x+12)+(x+14)=104

8x=160

x=20

As 20 is the biggest integer of the set and they are only even consecutive integers you count down by 2s.since there are only eight integers you stop by the 8th consecutive integer(this includes 20).Counting down 20,18,16,14,12,10,8,6 you find out 6 id the smallest integer in the group.So6*20=120

To check your answer, you can add the integers together to see if they add too 104 so.

6+8=14,14+10=24,24+12=36,36+14=50,50+16=66,66+18=84,and 84+20=104

Doing this you know you know you have the correct integers and knowing that 20*6=120, you have your answer

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