Solve shape-coded products in order

3.OA.A.43.OA.D.8 · take · grade 3

Archetype: Apply a Newly Defined Operation · step in a 5-type progression

Each shape stands for the same number wherever it appears. The shapes $\bullet$ , $\blacktriangle$ , and $\blacksquare$ are three different one-digit numbers. Using the equations below, find the value of $\bullet + \blacktriangle + \blacksquare$ .

$\bullet \times \bullet = 36 \qquad \blacktriangle \times \bullet = 48 \qquad \blacksquare \times 2 = 1\,\blacktriangle$

Here $1\,\blacktriangle$ is the two-digit number whose tens digit is $1$ and whose ones digit is $\blacktriangle$ .

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Understand

Three shapes stand for three different one-digit numbers. From the equations circle x circle = 36, triangle x circle = 48, and square x 2 = the two-digit number '1 triangle', find the sum circle + triangle + square.

Givens

circle x circle = 36.
triangle x circle = 48.
square x 2 = 1(triangle), the two-digit number with tens digit 1 and ones digit triangle.
The three shapes are different one-digit numbers.

Unknowns

The values of circle, triangle, and square.
Their sum circle + triangle + square.

Constraints

Each shape is a single-digit whole number.
Different shapes are different numbers.

Plan

#3 Eliminate Possibilities · also uses: #6 Guess and Check

Solve in order from the equation that can be determined alone: the square of circle pins circle, then triangle follows, then square. Each step has only one valid one-digit value.

Execute

#3 Eliminate Possibilities 3.OA.A.4

circle x circle = 36, so circle is the number that times itself is 36. That is 6.

6 \times 6 = 36

Only 6 among one-digit numbers squares to 36, so circle = 6.

#3 Eliminate Possibilities 3.OA.A.4

triangle x circle = 48 and circle = 6, so triangle x 6 = 48, giving triangle = 8.

\blacktriangle \times 6 = 48 \Rightarrow \blacktriangle = 8

The missing factor of 48 with 6 is 8, a basic division fact.

#6 Guess and Check 3.OA.D.8

The two-digit number 1(triangle) is 18 since triangle = 8. So square x 2 = 18, giving square = 9. The shapes 6, 8, 9 are all different, so the sum is 6 + 8 + 9 = 23.

\blacksquare \times 2 = 18 \Rightarrow \blacksquare = 9,\quad 6+8+9 = 23

Doubling to reach 18 needs 9, and adding the three found values gives the total.

Answer: 23

Review

Check all three: 6 x 6 = 36, 8 x 6 = 48, and 9 x 2 = 18 = '1 triangle' with triangle = 8; the shapes 6, 8, 9 are distinct one-digit numbers, so the sum 23 is correct.

You could list one-digit values: only circle = 6 fits the first equation, which then forces triangle = 8 and square = 9, the same result.

Standards · min grade 3

3.OA.A.4 Determine unknown whole number in multiplication or division equation — Finding circle from circle x circle = 36 and triangle from triangle x 6 = 48.
3.OA.D.8 Solve two-step word problems using four operations within 100 — Using the two-digit clue to find square and summing the three shape values.

💡 Start with the shape you can pin down, then unlock the rest in order -- just Grade 3 multiplication facts!