Each of the following equations contains one incorrect number. The correct numbers appear in the box below.
Your task is to make substitutions so that every equation is correct. If you do everything right, each number in the box will be used exactly once.
Use correct order of operations in evaluating the expressions.
(1) 10 - 7 = 6
(2) 8 + 5 = 14
(3) 12 + 9 - 4 = 15
(4) 4 + 3 x 5 = 16
(5) 4 / 2 + 16 / 4 = 8
(6) 17 - 4 = 15
(7) 20 - 12 / 3 = 13
(8) 8 - 18 / 3 = 6
1 9 4 13 6 17 7 19 |
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Solution to the Problem:
There are several solutions, but here is one set of answers:
(1) 10 - 4 = 6
(2) 8 + 5 = 13
(3) 12 + 7 - 4 = 15
(4) 1 + 3 x 5 = 16
(5) 4 / 2 + 16 / 4 = 6
(6) 19 - 4 = 15
(7) 17 - 12 / 3 = 13
(8) 8 - 18 / 9 = 6
To solve the problem, begin by determining all the possible values that would make each equation true.
The numbers that would work for each equation are:
(1) 1, 4, 13
(2) 6, 9, 13
(3) 6, 7, 10, 13
(4) 1, 4, 19
(5) 1, 6, 8, 24
(6) 2, 13, 19
(7) 16, 17, 21
(8) 2, 6, 9 12
You may discard all the underlined numbers because they do not appear in the box above.
There are two places that you could begin.
In equation #7, there is only one number remaining (17), so it must be the answer.
In equation #3, the number 7 is only used there, so it must be the answer.
Continue solving by using this elimination process.
The first: (1) 13 - 7 = 6 (2) 8 + 6 = 14 (3) 12 + 7 - 4 = 15 (4) 4 + 3 x 4 = 16 (5) 4 / 1 + 16 / 4 = 8 (6) 19 - 4 = 15 (7) 17 - 12 / 3 = 13 (8) 8 - 18 / 9 = 6 The second: (1) 13 - 7 = 6 (2) 9 + 5 = 14 (3) 12 + 7 - 4 = 15 (4) 4 + 3 x 4 = 16 (5) 4 / 1 + 16 / 4 = 8 (6) 19 - 4 = 15 (7) 17 - 12 / 3 = 13 (8) 8 - 18 / 9 = 6 The third: (1) 10 - 4 = 6 (2) 8 + 6 = 14 (3) 12 + 7 - 4 = 15 (4) 4 + 3 x 5 = 19 (5) 4 / 1 + 16 / 4 = 8 (6) 17 - 4 = 13 (7) 17 - 12 / 3 = 13 (8) 8 - 18 / 9 = 6 The fourth: (1) 10 - 4 = 6 (2) 9 + 5 = 14 (3) 12 + 7 - 4 = 15 (4) 4 + 3 x 5 = 19 (5) 4 / 1 + 16 / 4 = 8 (6) 17 - 4 = 13 (7) 17 - 12 / 3 = 13 (8) 8 - 6 / 3 = 6
Correctly solved by:
1. John Funk | Ventura, California |
2. Magdy Essafty (5 solutions) | Alexandria, Egypt |
3. Richard K. Johnson | La Jolla, California |
4. Tristan Collins (3 solutions) | Virginia Tech Blacksburg, Virginia |
5. K. Sengupta | Calcutta, India |
6. David and Judy Dixon | Bennettsville, South Carolina |
7. Stefan Gatachiu | Romania |