Problem
Largest and smallest line sum
A star has 11 circles, with five arms from the middle. Each arm is a line of three circles including the centre. The numbers 1 to 11 are used once each. Find every way to make the five line totals equal.
Reasoning · Grade 5-2 Foundations of Magic Squares
Your answer
Watch out Dividing 66 by five fails: the centre is counted five times.
How to solve
Strategy Change Focus / Count the Complement — Filling circles one at a time and hoping the five totals come out equal is a long road with a lot of rubbing out. The shortcut is to change what you look at: instead of one line, add all five line totals together. In that grand total every outer number appears once but the centre number appears five times, because the centre belongs to every line. That single observation ties the line total to the centre number, so the whole puzzle collapses into 'which centre numbers are allowed?'. From there the candidates form a short list to test (tools 2 and 6), and each surviving centre gives a filling that practically builds itself: pair the ten remaining numbers so each pair has the same sum. Drawing the star and labelling the five arms (tool 1) keeps the pairing honest.
1STEP 1
Add up all five line totals at once
The numbers 1 to 11 total 66.
1+2+…+11 = 66
Looking at all five lines together instead of one at a time makes the centre circle show its special role: it is the only number that gets counted more than once.
4.NBT.B.4Change Focus Count The ComplementAdding all five line totals at once counts the centre number five times and every arm number once.
Why?
The centre lies on every line, so it is counted once from each of them while an arm number belongs to one line only.
🧱Counted twice, so halveIf every item got counted from both of its two ends, the total is exactly double the truth, so halving fixes it.
Why?
The combined total is therefore all the numbers plus four extra copies of the centre, so the centre can be peeled off cleanly.
🧱Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.
2STEP 2
Turn that into a test on the centre number
Adding the arms counts the centre four extra times.
66 + 4 × 1 = 70, 66 + 4 × 6 = 90, 66 + 4 × 11 = 110
Only eleven numbers can go in the middle, so listing all eleven totals and checking which end in 0 or 5 settles the question completely -- no cleverness required.
4.OA.B.4Make A Systematic List3STEP 3
Read off the three line totals
Only 1, 6 and 11 divide exactly, giving line totals of 14, 18 and 22.
70 ÷ 5 = 14, 90 ÷ 5 = 18, 110 ÷ 5 = 22
The problem asks for three fillings with three different totals, and the arithmetic has just handed over exactly three possibilities, which is a strong sign the reasoning is on track.
4.OA.B.4Change Focus Count The Complement4STEP 4
See what each arm has to do
Taking the centre off leaves each arm's outer pair to make 13, 12 or 11.
14 - 1 = 13, 18 - 6 = 12, 22 - 11 = 11
Once the centre is fixed, the five lines stop interfering with one another -- each arm becomes its own tiny problem of finding two numbers with a given sum.
4.OA.A.3Draw A Diagram5STEP 5
Build the filling with centre 1
With centre 1, pair the rest into sums of 13.
2+11 = 3+10 = 4+9 = 5+8 = 6+7 = 13, 1 + 13 = 14
Ten numbers in a row pair up perfectly smallest-with-largest, because moving one number up by one and its partner down by one keeps the sum the same.
2.OA.B.2Guess And Check6STEP 6
Build the fillings with centre 6 and centre 11
Centres 6 and 11 also pair up.
6 + 12 = 18, 11 + 11 = 22
The same smallest-with-largest pairing works every time, because taking the centre number out of 1 to 11 always leaves ten numbers that fold neatly in half.
2.OA.B.2Guess And Check7STEP 7
Check all fifteen lines
So there are three ways.
Fifteen short additions is the whole check, and doing every one of them rules out the easy mistake of reusing a number on two different arms.
4.OA.A.3Make A Systematic ListAnswer
1, 6, 11
14, 18, 22
A sanity check on size: the smallest three numbers add to 1+2+3 = 6 and the largest three add to 9+10+11 = 30, so any line total must lie between 6 and 30. The totals found, 14, 18 and 22, sit comfortably inside that range and are evenly spaced 4 apart, which makes sense because each step up in the centre number (1 → 6 → 11) is 5, and 4 × 5 = 20 spread over 5 lines raises each line by 4. Another check: the average of 1 through 11 is 6, so a typical three-circle line should total about 3 × 6 = 18 -- and 18 is exactly the middle answer. The numbers are plain labels, so no units are involved. Finally, the argument shows these three are not just three answers but the only three, so no fourth line total exists to be missed.
Takeaway
Add all five lines at once: the centre number gets counted five times, so the middle circle alone decides the line total -- and only 1, 6 and 11 can sit there.
- Add up all five line totals at once
- Turn that into a test on the centre number
- Read off the three line totals
- See what each arm has to do
- Build the filling with centre 1
- Build the fillings with centre 6 and centre 11
- Check all fifteen lines