Problem 40 worked answer

U + Me = Us: The Screen Test

Designed for Grade 4 · Grade 5 · Uses Middle Grade 4 math · About 5–12 minutes

Complete to collectIndicoliteLevel 3

At a glance

Part 1: The five equations are $8 + 72 = 80$ , $8 + 73 = 81$ , $8 + 74 = 82$ , $8 + 75 = 83$ , and $8 + 76 = 84$ .
Part 2: These are all the possibilities because the addition must carry 1 ten, which forces $𝑀 = 7$ . Checking every remaining possible value of $𝐸$ leaves exactly $𝐸 = 2, 3, 4, 5,$ or $6$ .

Key idea: the carry connects the columns

The addend $𝑈$ is the one-digit number 8. Adding a one-digit number can change the tens digit of $𝑀 𝐸$ only when the ones column makes a new ten.

Because different letters must have different digits, the tens digit cannot stay $𝑀$ and also become $𝑈$ . The ones column must therefore carry 1 ten. That one carry links the two columns and turns the puzzle into a short, complete list.

1. Find every finished equation

Start with the tens column. The carried 1 is added to $𝑀$ , and the result's tens digit is $𝑈 = 8$ . Therefore

𝑀 + 1 = 8,

so $𝑀 = 7$ .

Now use the ones column. The sum $8 + 𝐸$ must make a new ten. Therefore $𝐸$ must be at least 2. Digits 7 and 8 are already used by $𝑀$ and $𝑈$ , so the only values still worth checking are $𝐸 = 2, 3, 4, 5, 6,$ and $9$ .

If $𝐸 = 2$ , then $8 + 2 = 10$ , so $𝑆 = 0$ : $8 + 72 = 80$ .
If $𝐸 = 3$ , then $8 + 3 = 11$ , so $𝑆 = 1$ : $8 + 73 = 81$ .
If $𝐸 = 4$ , then $8 + 4 = 12$ , so $𝑆 = 2$ : $8 + 74 = 82$ .
If $𝐸 = 5$ , then $8 + 5 = 13$ , so $𝑆 = 3$ : $8 + 75 = 83$ .
If $𝐸 = 6$ , then $8 + 6 = 14$ , so $𝑆 = 4$ : $8 + 76 = 84$ .
If $𝐸 = 9$ , then $8 + 9 = 17$ , so $𝑆$ would be 7. But $𝑀$ already uses 7, so this case breaks the different-letter rule.

The five listed equations are the valid finished screens.

2. Prove that the list is complete

Every digit from 0 through 9 falls into one of these groups:

$𝐸 = 0$ or $𝐸 = 1$ makes no carry. Then the result's tens digit would still be $𝑀$ , forcing $𝑀 = 𝑈$ , which is not allowed.
$𝐸 = 2, 3, 4, 5,$ or $6$ gives one of the five valid equations above.
$𝐸 = 7$ or $𝐸 = 8$ repeats a digit already used by $𝑀$ or $𝑈$ .
$𝐸 = 9$ makes $𝑆 = 7$ , repeating the digit used by $𝑀$ .

Those groups cover every possible digit for $𝐸$ , so no other equation can work. Each accepted value of $𝐸$ is different and determines exactly one value of $𝑆$ , so no equation is repeated.

Check

Each listed equation is a true addition equation. In all five, $𝑈 = 8$ , $𝑀 = 7$ , and the four letters use four different digits:

$8 + 72 = 80$ uses the digits 8, 7, 2, and 0.
$8 + 73 = 81$ uses the digits 8, 7, 3, and 1.
$8 + 74 = 82$ uses the digits 8, 7, 4, and 2.
$8 + 75 = 83$ uses the digits 8, 7, 5, and 3.
$8 + 76 = 84$ uses the digits 8, 7, 6, and 4.

The different-letter rule, the two-digit-number rule, and the equation all hold in every case.

Technical fit and rating

MJ 4.5 · C3 · W2 Stretch challenge · Moderate workload

Discrete mathematics and logic

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