Difference between revisions of "1988 AHSME Problems/Problem 12"
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| − | We can draw a sample space diagram, and find that of the <math>9^2 = 81</math> possibilities, <math>9</math> of them give a sum of <math>0</math>, and each other sum mod <math>10</math> (from <math>1</math> to <math>9</math>) is given by <math>8</math> of the possibilities (and indeed we can check that <math>9 + 8 \times 9 = 81</math>). Thus <math>0</math> is the most likely, so the answer is <math>A</math>. | + | We can draw a sample space diagram, and find that of the <math>9^2 = 81</math> possibilities, <math>9</math> of them give a sum of <math>0</math>, and each other sum mod <math>10</math> (from <math>1</math> to <math>9</math>) is given by <math>8</math> of the possibilities (and indeed we can check that <math>9 + 8 \times 9 = 81</math>). Thus <math>0</math> is the most likely, so the answer is <math>\boxed{\text{A}}</math>. |
Revision as of 17:15, 26 February 2018
Problem
Each integer 
through 
is written on a separate slip of paper and all nine slips are put into a hat.
Jack picks one of these slips at random and puts it back. Then Jill picks a slip at random.
Which digit is most likely to be the units digit of the sum of Jack's integer and Jill's integer?
Solution
We can draw a sample space diagram, and find that of the 
possibilities, of them give a sum of 
, and each other sum mod 
(from to ) is given by 
of the possibilities (and indeed we can check that 
). Thus is the most likely, so the answer is 
.
See also
| 1988 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
