Important information:

- "five boxes" -- name them A.B.C.D.E
**from lightest to heaviest** -
"Each weighs a different
**integer**amount" - "weigh the boxes in
**pairs"**; "each box is weighed with every other box. "

The **ten** pairs are:

A B

A C

A D

A E

B C

B D

B E

C D

C E

D E

The of pairs are given in **sorted order:** 110, 112, 113, 114, 115, 116, 117, 118, 120, 121

Looking at the list of weights, we observe that there must be both sequential and non-sequential **integers **in the sorted solution list. So, we don't know which integers ore omitted, but we sorted ABCDE to be lowest to greatest, so we know:

A + B = 110

D + E = 121

[but we don't know the other pairs, yet]

Also, 4A+4B+4C=4D+4E =1156

so, A+B+C+D+E = 289

(110) + C + (121) = 289

** C = 58**

** **

A+C = 112 [the second smallest total\

** A = 54**

** **

C + E = 120 [the second largest total]

** E = 62**

A + B = 110

54 + B = 110

** B = 56**

54 + 56 + 58 + D + 62 = 289

** D = 59**

**To check:**

Edward A.

02/16/20