Pairs
Ann thought of 5 numbers and told Bob all the sums that could be made by adding the numbers in pairs. The list of sums is 6, 7, 8, 8, 9, 9, 10,10, 11, 12. Help Bob to find out which numbers Ann was thinking of.
[This is an exercise for practice from the Hungarian KöMaL magazine, September 2000 issue. You'll find many other challenging problems in English on the KöMaL website.]
Here is a solution done by Soh Yong Sheng, age 13, Raffles Institution, Singapore.
Ann thought of 5 numbers and told Bob all the sums that could be made by adding the numbers in pairs. The list of sums is 6, 7, 8, 8, 9, 9, 10,10, 11, 12.
Summing these numbers, we get 90 in the 10 pairs, that is 20 numbers.
We know that each number appear 4 times since 20/5=4.
90/4=22.5 which is the sum of these numbers.
The .5 is an important clue. This means there are 5 numbers with .5 at the back . It must be all 5 numbers since the sums are all integers. Also, note that there are no other possibilities.
Without the .5 at the back, we rewrite the whole sequence as 5,6,7,7,8,8,9,9,10,11.
The sum of the numbers is 20 and the average is 4.
Let's consider the 7s, 8s and 9s since they occur in pairs.
7=1+6, 2+5, 3+4
8=1+7, 2+6, 3+5, 4+4
9=1+8, 2+7, 3+6, 4+5
Let's experiment with 7 since it has only 3 pairs.
Case 1 1256...impossible since 1+2=3 which is too small
Case 2 1346...same as above 1+3=4
Case 3 2345...has to be the one since the above don't work at all
and 6 is the other number to pair off with 5 to get 11. Testing out
we can see this works:
5=2+3, 6=2+4, 7=3+4=2+5, 8=2+6=3+5, 9=3+6=4+5, 10=4+6, 11=5+6.
Now add .5 to the sequence again and we get 2.5, 3.5, 4.5, 5.5, 6.5
Fiona Watson, year 10, from Stamford High School and Louise Howes and Laura Searle, Y7, Bourne Grammar School also solved this, also using some logical thinking and trial and error.
You may like to try using algebra. Call the numbers a, b, c, d and e, in order of size, and write down 10 equations. They are very easy to solve.