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Strategies and gamesGrades 3–6

Weighing and pouring

With a 5-litre jug and a 3-litre jug, how do you measure exactly 4 litres? With a balance and no weights, how do you find the one light coin among many? In these tasks a child plans a short sequence of steps, and every rule counts: no marks on the jugs, no weights for the balance. It is a first meeting with algorithms, plans that work whatever happens.

The main move
Write down every state and look for the shortest way
What it teaches
To plan steps ahead, and to think through every outcome of a weighing
5 l  3 l
 5    0
 2    3
 2    0
 0    2
 5    2
 4    3

The rules of weighing and pouring

Three outcomes of a weighing

Each weighing has three outcomes, so it can split the coins into three groups. That decides how many coins a number of weighings can handle.

WeighingsOutcomesCoins they can handle
13up to 3
23 × 3 = 9up to 9
33 × 3 × 3 = 27up to 27

How to solve: a table of states

For pouring, make a table with a column for each jug and a row for each step, and try every allowed step from each row until the goal appears; the first time it appears is the fewest steps. For weighing, plan what to do after each of the three outcomes. The examples below go from the simplest to an olympiad's.

  1. Write down the state, what each jug or pan holds
  2. Try every allowed step from it
  3. Stop at the goal, and count the steps

Example 1 · Grades 3–4

One litre

A cook has an empty 3-litre jug, an empty 5-litre jug and a tap. The jugs have no marks. In one step she fills a jug from the tap, pours a jug out, or pours from one jug into the other until the first is empty or the second is full. What is the fewest steps to get exactly 1 litre in a jug?

  1. Fill the 3-litre jug and pour it into the 5-litre jug, which now holds 3 litres.
  2. Fill the 3-litre jug again and pour from it until the big jug is full: only 2 more litres fit.
  3. 1 litre is left in the small jug, after 4 steps: fill, pour, fill, pour.
3 l  5 l
 3    0
 0    3
 3    3
 1    5

Answer 4 steps.

Example 2 · Grades 3–4

Eight coins

There are 8 coins that look the same. One of them is fake and lighter than the others, which all weigh the same. What is the smallest number of weighings on a balance without weights that is sure to find the fake?

  1. Put 3 coins on each pan and keep 2 aside.
  2. If a pan rises, the fake is among its 3 coins: weigh 1 against 1, and the lighter one is the fake, or the third coin if they balance.
  3. If the pans balance, the fake is one of the 2 aside: weigh them against each other. Either way 2 weighings are enough, and 1 is not: it has only 3 outcomes for 8 coins.
[●●●] [●●●]  ●●

Answer 2 weighings.

Example 3 · Grades 5–6

A hundred coins

Among 100 coins that look the same, one is fake and lighter than the others, which all weigh the same. What is the smallest number of weighings on a balance without weights that is sure to find the fake?

  1. Each weighing has 3 outcomes, so 4 weighings tell apart at most 3 × 3 × 3 × 3 = 81 cases. That is fewer than 100 coins, so 4 weighings are not enough.
  2. Put 33 coins on each pan and keep 34 aside: whatever happens, at most 34 coins are left. In the same way 34 come down to at most 12, then 4, then 2, then 1.
  3. That makes 5 weighings, as the coins go from 100 to 34, 12, 4, 2 and 1.

Answer 5 weighings.

Where children go wrong

MathTrail explains every wrong answer by the trap it fell into. These are the commonest in this topic, with something you can say to your child.

How to help at home

  1. Jugs in the sink

    Take two cups of different sizes and try to measure an amount that neither holds. A sink is the best lab, and spilled water does no harm.

  2. Find the light one

    Hide one lighter object among a few that look the same and find it with a home balance, or a coat hanger with two bags. Count the weighings.

  3. Plan first

    Before pouring or weighing, write the plan as a table and check it step by step. Planning first is the skill, not luck.

Learn first

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