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Grades 7–8
4 sections · 4 questions

When computing will not help

Four questions a calculator is no use for — too big, nothing to compute from, and two that cannot be done at all, each for a different reason.

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What's on this test

Section 1 asks for the last digit of a number with eighty-five digits. The arithmetic is perfectly real and completely out of reach, so the only way in is to notice that the last digits repeat. Section 2 gives two purchases and three prices and asks about a total; half of these questions the facts genuinely do not decide, and the honest answer is CANNOT TELL with two price lists to prove it. Section 3 turns over exactly two cups at a time and asks for the fewest turns — and sometimes no number of turns will ever do it, which a count's parity proves. Section 4 asks whether dominoes can cover a board, and when they cannot, it is a chessboard coloring rather than a parity that says so.

How the paper is put together

The reasons escalate, which is why they are in this order. Too big to compute. Nothing to compute from. And then two questions where no amount of computing could ever settle it, because what is being established is that something is impossible — first by counting parity, then by coloring — and only an argument does that. A child who works down the page meets a steadily weaker version of the tool they arrived with.

When computing will not help

Name: Date:

1. Challenge puzzlesThe number is far too big to write. Find what repeats, and count carefully.1 question

1Today is Friday. A comet will be overhead in 183 days' time.What day of the week will that be?Why

2. Challenge puzzlesNot every price can be found. Answer what you can, or write CANNOT TELL.1 question

2A market stall labels none of its fruit. Fact 1: 2 apples, 2 pears and 3 plums come to 196c. Fact 2: 2 pears and 3 plums come to 80c.What does one apple cost?Why

When computing will not help

3. Challenge puzzlesSome of these cannot be done. Say which, and say how you know.1 question

3A panel has 10 switches and 2 of them are on. You may throw exactly 2 of them at a time, never more and never fewer.What is the fewest throws that leaves every switch on? Write NEVER if it cannot be done.Why

When computing will not help

4. Challenge puzzlesDecide whether dominoes can cover the board exactly, and say how you know.1 question

4An 8 by 4 grid of lockers has the two corner lockers at the same end walled off — the two along the top edge, left and right.Can the rest be covered exactly by double lockers, each covering two squares that share an edge, with none left over and none overlapping? Say YES or NO, and say how you know.Why

Think about it

None of these could be settled by calculating. What did you do instead?

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Answer key

Challenge puzzles

If they are stuck: It cannot be computed, so look at the small cases. Do they settle down?

1. Days of the week repeat every 7, so only the leftover matters. 26 whole weeks is 26 × 7 = 182 days, and 183 − 182 = 1 left over. So the comet arrives 1 day after Friday: Saturday.

Challenge puzzles

If they are stuck: Do not look for the prices; they cannot be found. What CAN be built from the two facts?

2. Fact 1 less fact 2: 196 − 80 = 116c, and that buys 2 apples — two lots of what the question asks for. So one lot is 116 ÷ 2 = 58c. The answer came from the two totals alone and never from a price: a pear could be 1c or 4c and both facts would still hold.

Challenge puzzles

If they are stuck: Trying it will not settle this. Ask what ONE move does to the count.

3. There are 10 to work with. One throw changes the number switched on by at most 2, and there are 10 − 2 = 8 to gain, so fewer than 4 throws cannot possibly do it. And 4 can, counting the number switched on as you go: 2 → 4 → 6 → 8 → 10.

Challenge puzzles

If they are stuck: Do not try to cover it. Color it like a chessboard first — what does every single tile have to cover?

4. Color it like a chessboard first, because that is what rules the other one out. The two walled off squares are at the same end of one edge; on a board with an even side they are opposite colors, so what is left still has 15 dark and 15 light — equal, which is the test every covering must pass. And it can be done: that edge's row keeps an even run of squares once its two ends are gone, and every other row is a full even row, so lay the double lockers along each row in turn. It can be done.

When computing will not help

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