Difference between revisions of "1988 AJHSME Problems/Problem 10"
(New page: ==Problem== Chris' birthday is on a Thursday this year. What day of the week will it be <math>60</math> days after her birthday? <math>\text{(A)}\ \text{Monday} \qquad \text{(B)}\ \text...) |
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<math>\text{(A)}\ \text{Monday} \qquad \text{(B)}\ \text{Wednesday} \qquad \text{(C)}\ \text{Thursday} \qquad \text{(D)}\ \text{Friday} \qquad \text{(E)}\ \text{Saturday}</math> | <math>\text{(A)}\ \text{Monday} \qquad \text{(B)}\ \text{Wednesday} \qquad \text{(C)}\ \text{Thursday} \qquad \text{(D)}\ \text{Friday} \qquad \text{(E)}\ \text{Saturday}</math> | ||
| − | ==Solution== | + | ==Solutions== |
| − | 7 days after her birthday would be a Thursday, as would 14, 21, 28, 35, 42, 49, and 56. Therefore the 60th would be four days after a Thursday, which is a <math>Monday\Rightarrow \mathrm{(A)}</math>. | + | ===Solution 1=== |
| + | 7 days after her birthday would be a Thursday, as would 14, 21, 28, 35, 42, 49, and 56. Therefore the 60th would be four days after a Thursday, which is a <math>\text{Monday} \Rightarrow \mathrm{(A)}</math>. | ||
| + | ===Solution 2=== | ||
| + | Note that <math>60\equiv4\pmod7</math>. We count 4 days past Thursday, and arrive at Monday. <math> \mathrm{(A)}</math> | ||
==See Also== | ==See Also== | ||
| − | [[ | + | {{AJHSME box|year=1988|num-b=9|num-a=11}} |
| + | [[Category:Introductory Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 22:55, 4 July 2013
Problem
Chris' birthday is on a Thursday this year. What day of the week will it be 
days after her birthday?
Solutions
Solution 1
7 days after her birthday would be a Thursday, as would 14, 21, 28, 35, 42, 49, and 56. Therefore the 60th would be four days after a Thursday, which is a 
.
Solution 2
Note that 
. We count 4 days past Thursday, and arrive at Monday. 
See Also
| 1988 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
