Missing Digits

In this activity, the students use their knowledge of basic facts and number patterns to find missing digits in numbers. Most of the questions could be done using trial and improvement, but the students are likely to find it more efficient (and more satisfying) to first narrow the range of options as much as possible by studying the problem and seeing what they can deduce. You could suggest that they are learning to be number detectives.
You could introduce the activity to the whole class or a group using examples of your own. In this way, you leave all the problems in the book for the students to do themselves. Below are possible examples for group teaching or discussion.

Example 1: ☐☐3 + 3☐ = 44
• The last digit of the first number is 3; 3 + ☐ gives the final digit of the sum, 4.
• This means the missing digit must be 1 (because 3 + 1 = 4). Checking: 113 + 31 = 144.

Example 2: 1☐ x 3☐ = ☐2☐
• When you multiply this missing digit by itself, the answer (product) ends with that same digit.
• There are only three digits for which this is true: 1 (because 1 x 1 = 1), 5 (because5 x 5 = 25), and 6 (because 6 x 6 = 36).
• Try 1: 11 x 31 = 341. So it’s not 1. Try 5. If 5 doesn’t work, you would know it must be6. In this case, 5 does work: 15 x 35 = 525.

Example 3: ☐6 x ☐= 5☐
• The missing digit is even (because 6 is an even number, and whenever you multiply a whole number by an even number you get an even number).
• The digit must be 2, 4, 6, or 8 (the only even digits apart from 0, which wouldn’t work because anything multiplied by 0 is 0).
• 4, 6, or 8 wouldn’t do because the first number would have to be 46, 66, or 86, and if you multiplied these numbers by 4, 6, or 8, you’d get a number that was a lot greater than 50 something.
• This leaves 2 as the only possible even number. Checking: 26 x 2 = 52.

The dialogue alongside question 1 models how the students should be thinking. Question 2 asks them to write down the steps they used to solve each problem. For example, they could do this using numbered or bulleted points or in a table.
In question 3, the students need to think about patterns (or strategies) they may have used, such as:
• odd + odd = even (for example: 9 + 5 = 14)
• odd – odd = even (for example: 9 – 5 = 4)
• even + even = even
• even + odd = odd
• even – odd = odd
• odd – even = odd
• odd x odd = odd
• even x even = even
• odd x even = even.
One approach to question 4 is for the students to make up some simple number sentences in theformat “number +/–/x/÷ number = number” and then to look for a repeated digit that they could remove and replace with a box, thus creating a missing digit problem for a classmate to solve.
For example:
• Make up a subtraction sentence and solve it: 253 – 152 = 101
• Now create a missing digit problem by replacing the digit 2 with a box: ☐53 – 15☐ = 101;or the digit 5 with a box: 2☐3 – 1☐2 = 101; or the digit 1 with a box: 253 – ☐52 = ☐0☐
• Check these possibilities and see what is involved in solving them. Some will need to be discarded as too simple. Not all will have a unique solution (for example,2☐3 – 1☐2 = 101 works for any digit).
Some trial and improvement may be needed before each student has a small group of problems suitable for giving to a classmate.

Answers to Activity

1. ☐ = 5
2. Strategies will vary, but the answers are

3. Answers will vary. Possibilities include:
In addition problems, if the ☐ is in the ones place, the ones digit in the answer and in the other number will tell you what the ☐is.
The size of the answer in a multiplication problem can tell you how big the digit in the  ☐ must be (when ☐ each is the same digit). For example, in ☐6 x ☐ = 18☐, the ☐ could not be more than 4 because 50 x 5 = 250, which is more than 180.
4. Problems will vary.