Difference between revisions of "1990 AJHSME Problems/Problem 10"
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==Solution== | ==Solution== | ||
− | Let the date behind <math>C</math> be <math>x</math>. Now the date behind <math>A</math> is <math>x+1</math>, and after looking at the calendar, the date behind <math>B</math> is <math>x+13</math>. Now we have <math>x+1+x+13=x+y</math> for some date <math>y</math>, and we desire for <math>y</math> to be <math>x+14</math>. Now we find that <math>y</math> is the date behind <math>P</math>, so the answer is <math>\boxed{(\text{A})}</math> | + | Let the date behind <math>C</math> be <math>x</math>. Now the date behind <math>A</math> is <math>x+1</math>, and after looking at the calendar, the date behind <math>B</math> is <math>x+13</math>. Now we have <math>x+1+x+13=x+y</math> for some date <math>y</math>, and we desire for <math>y</math> to be <math>x+14</math>. Now we find that <math>y</math> is the date behind <math>P</math>, so the answer is <math>\boxed{(\text{A})}</math> ~motorfinn |
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==See Also== | ==See Also== | ||
{{AJHSME box|year=1990|num-b=9|num-a=11}} | {{AJHSME box|year=1990|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 10:51, 8 September 2019
Problem
On this monthly calendar, the date behind one of the letters is added to the date behind . If this sum equals the sum of the dates behind and , then the letter is
Solution
Let the date behind be . Now the date behind is , and after looking at the calendar, the date behind is . Now we have for some date , and we desire for to be . Now we find that is the date behind , so the answer is ~motorfinn
See Also
1990 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.