*Grades 9 & 10 5-Star Challenge #2*

Every possible pair within a set of five numbers is summed.

The sums are 0, 3, 11, 12, 20, 21, 23, 29, 32, 41. Find the five numbers.

Every possible pair within a set of five numbers is summed.

The sums are 0, 3, 11, 12, 20, 21, 23, 29, 32, 41. Find the five numbers.

In this problem, the answer may be obtained by trial and error, but there are some neat aspects to the problem if one works to solve it. Let us assume that the five numbers are distinct (as they would need to be if their ten pair-sums were distinct). Let us consider the sums as F < G < H < J < K < L < M < N < P < Q. Let the five numbers be a < b < c < d < e. The order of a, b, c, d, and e offers only an incomplete understanding of the order of the sums. We have

(a+b) < (a+c) < [(b+c),(a+d)] < [(a+e),(b+d)] < [(c+d),(b+e)] < (c+e) < (d+e).

We know that F=a+b and G=a+c, and that Q=e+d and P=c+e. We do not know if the third and fourth smallest sums represent a+d and b+c or b+c and a+d, respectively.

This problem can be approached by solving all of the little equations to establish which is greater of (b+c) and (a+d). Once b+c is known, the system of three equations a+b=F, a+c=G, b+c=(H or J) can be solved for the values of a, b, and c. The values of d and e can be found shortly thereafter by substitution.

There is a less tedious way to solve the problem. By identifying that each number is summed with each possible partner, we know that each number a < b < c < d < e appears four times in the sum F+G+H+J+K+L+M+N+P+Q, so the sum of the ten pair-sums must be equal to 4(a+b+c+d+e). By adding the first and last sum, we know that F+Q=a+b+d+e, and we can therefore find the value of c.

In our problem, the pair-sums are 0, 3, 11, 12, 20, 21, 23, 29, 32, 41, whose sum is 192, which is equal to 4(a+b+c+d+e). This means that 192/4=48=a+b+c+d+e.

We know that F+Q=0+41=a+b+d+e, so

**c=7**.

If c=7, we know that G=3=(a+c)=(a+7), so

**a=-4**. Then F=0=(a+b)=(-4+b) tells us that

**b=4**. P=32=(c+e)=(7+e), so

**e=25**. Since 48=a+b+c+d+e,

**d must be 16**.

Nate,

ReplyDeleteHere's a numbers joke I got from Reader's Digest: What did the zero say to the eight?

I'll let you mull that one over, you math whiz, you. Love,

Big Sis

"Nice belt!"

ReplyDeleteYeah, that's an old classic. I have a book mark there in my old book of 501 nerdy jokes that I read every night before I go to sleep.

Also, in the blank to type in the word below for the comment entry, there is a little icon of a guy in a wheelchair that I could click if I need extra assistance. Does somebody show up to my desk and help me? Do they wonder how I got this far into the comment process without them? Is this discriminatory against all of the people in wheelchairs who still can type? If I was in a wheelchair, it does not follow that I would be able to write a comment but not read the security word.

Well, perhaps sadly, it was a new joke to me. Probably the only one I'll remember for the next five years until I stumble across another old classic that's new to me.

ReplyDelete