Problem
Find the hidden rule, then place numbers
Circles joined by segments each hold a number. A Rule box pairs each circle number with a second number. What the rule does is never stated. Find the rule, fill the blanks and place the numbers.
Reasoning · Grade 5-2 Number Arrangements
Your answer
Watch out The circle's own number seems to drive it, but the pair is the sum of its neighbours.
How to solve
Strategy Look for a Pattern — Part (1) is a rule hunt, so the natural move is to line up the cases that are already complete and ask what the picture and the arrow number have in common (tool 5). The strongest clue is the simplest circle: circle 2 touches only circle 4, and 2 → 4 -- a coincidence too tidy to ignore. Once the rule is out, part (2) is a placement puzzle on a finite set of seven numbers, and the way in is to count how many segments leave each circle (tool 1: mark the picture up) and match the biggest arrow numbers to the busiest circles. Trying all 5040 orders would be hopeless by hand, so I eliminate impossible placements one at a time (tool 3) until only one arrangement survives, then list all seven rule lines and check them (tool 2).
1STEP 1
Hunt for the rule using the simplest circle
The simplest circle shows the pair is the neighbours' sum.
1 → 3+5+4 = 12
A circle with a single neighbour is the cheapest test case there is, because its arrow number cannot be anything but that one neighbour.
4.OA.C.5Look For A Pattern2STEP 2
Confirm the rule on the remaining known cases
The rule fits the other known lines.
4 → 1+2+6 = 9, 6 → 5+4 = 9
A rule guessed from one example is a hunch; a rule that survives every example you can test is something you can build on.
2.OA.B.2Look For A Pattern3STEP 3
Fill in the two blanks of part (1)
The two blanks are 6 and 10.
3 → 1+5 = 6, 5 → 3+1+6 = 10
Totalling the arrow numbers two different ways -- circle by circle and segment by segment -- catches a missed connection, which is the usual slip in this kind of picture.
2.OA.B.2Look For A Pattern4STEP 4
Mark up the part (2) picture: count segments at each circle
In the next picture mark each circle's segment count.
1+2+3+4+5+6+7 = 28
Counting the segments at each circle turns a tangle of lines into seven plain numbers, and those numbers say straight away which circles can carry large arrow totals.
4.OA.A.3Draw A Diagram5STEP 5
Place the 3: only the busiest circle can reach 19
Only the busiest circle can reach 19.
7+6+5 = 18 ＜ 19
An extreme value is the best place to start: 19 is so large that only the circle with the most lines can produce it, which pins down the hardest number first.
4.OA.A.3Eliminate PossibilitiesOnly the circle with the most segments running into it can reach a total as large as 19, so the 3 has to go there.
Why?
A circle's total climbs with how many segments feed it, so a circle with fewer segments can never beat one with more.
🧱Transitivity of orderIf the first is smaller than the second and the second smaller than the third, the first is smaller than the third, so pairwise comparisons chain into one full order.
Why?
Every other circle falls short of 19 and is ruled out at once, leaving one place the number can sit.
🧱Rule out the rest and the last one is forcedIf every possibility but one has been ruled out, the survivor must be the answer — no further checking is needed.
6STEP 6
Place the 6 and the 1
The leftover sum fixes the 6 and the 1.
28 - 3 - 19 = 6
Once the busiest circle is known, the total 28 turns each remaining rule line into a small subtraction instead of a search.
4.OA.A.3Eliminate Possibilities7STEP 7
Place the 5 and the 7
The pair adding to 12 is 5 and 7.
28 - 1 - (E + G) = 15 → E + G = 12, 5 + 7 = 12
Two clues squeeze the same pair of circles -- a sum of 12 and a one-neighbour rule line -- and together they leave no choice at all.
4.OA.A.3Eliminate Possibilities8STEP 8
Place the last two numbers
The last two are 2 and 4.
D = 5 - 3 = 2, C = 4 with 1 + 3 = 4
The last number has nowhere else to go, so the only question left is whether it happens to fit -- and here it does, which is the sign that the whole chain was right.
2.OA.B.2Eliminate Possibilities9STEP 9
Check all seven rule lines
All seven lines match the rule.
15, 11, 19, 4, 3, 1, 5
Seven short additions is the whole verification, and doing all seven -- not just the ones you reasoned about -- is what turns a likely answer into a certain one.
4.OA.C.5Make A Systematic ListAnswer
6, 10 / 1, 6, 4, 2, 5, 3, 7
3 + 1 + 5 = 9
The arrow numbers should track how busy each circle is, and they do: the circle with five segments carries the biggest total, 19, while the two circles with a single segment carry the two smallest one-number totals, 3 and 1. A stronger check uses the whole picture at once: adding the seven arrow numbers gives 15+11+19+4+3+1+5 = 58, and adding the two endpoint numbers of each of the nine segments gives (1+6)+(1+4)+(1+2)+(1+3)+(4+3)+(2+3)+(2+7)+(5+3)+(3+7) = 7+5+3+4+7+5+9+8+10 = 58 as well, because each segment contributes to exactly two circles. The two totals agreeing means no segment was missed or double-counted. Every number 1 to 7 is used exactly once, as required, and the answers are plain counts with no units.
Takeaway
Start with the circle that has only one line coming out of it -- its arrow number has to be its single neighbour, and that one clue hands you the whole rule.
- Hunt for the rule using the simplest circle
- Confirm the rule on the remaining known cases
- Fill in the two blanks of part (1)
- Mark up the part (2) picture: count segments at each circle
- Place the 3: only the busiest circle can reach 19
- Place the 6 and the 1
- Place the 5 and the 7
- Place the last two numbers
- Check all seven rule lines
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