Problem 23 · 2011 Math Kangaroo
Hard
Logic & Word Problems
magic-squaresum-constraint
Johannes wrote the numbers 6, 7 and 8 in the circles as shown. He wants to write the numbers 1, 2, 3, 4 and 5 in the remaining circles so that the sum of the numbers along each side of the square is 13. What will be the sum of the numbers in the grey circles?
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Answer: E — 16
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Hint 1 of 3
The grey circles are the four corners, and the white circles are the four middles.
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Hint 2 of 3
Add up the four sides together: each corner sits on two sides, so it gets counted twice, while each middle is counted once.
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Hint 3 of 3
You also know all eight numbers (the corners plus the middles) are 1 to 8.
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Approach: add all four sides at once
- Add the four side-totals together: 13 + 13 + 13 + 13 = 52. In that big sum each corner is counted twice (it touches two sides) and each middle is counted once.
- All eight numbers 1, 2, 3, 4, 5, 6, 7, 8 add up to 36, so the corners and middles together make 36.
- Take the 52 away from the 36 idea: 52 is one extra copy of every corner above the full 36, so the grey corners add to 52 − 36 = 16, answer E.
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