+0

0
597
3
+4

In a set of five consecutive integers, the largest integer is less than twice the average of the five integers. What is the smallest integer that could be in the set?

Aug 18, 2018

#1
+28533
+2

Let s be the smallest integer. Then the 5 numbers are: s, s+1, s+2, s+3, s+4

Their average is (5s + 10)/5 or s + 2

So: s + 4 < 2(s + 2)

s + 4 < 2s + 4

s >0

So the smallest integer greater than 0 is 1.

Edited to correct mistake.  Thanks to Guest below for noting this.

Aug 18, 2018
edited by Alan  Aug 19, 2018
edited by Alan  Aug 19, 2018
#2
0

If s is the smallest of the five integers, the mean is $$(5s+10)/5=s+2$$ (not (s+5) as previously suggested). We therefore require $$s+4<2(s+2)$$, which reduces to $$s>0$$. Thus the smallest integer must be 1 or greater.

As a reality check note that if s=1 we have the set 1,2,3,4,5 with average 3 and $$2\times3$$ is indeed greater than 5, but with s=0 we have 0,1,2,3,4 with mean 2 and since $$2\times2=4$$ the inequality just fails.

Aug 18, 2018
#3
+28533
+1

Thank you, you are correct Guest.  I made an arithmetical mistake!  I've now edited my reply to correct the error.

Alan  Aug 19, 2018