Difference between revisions of "1988 AJHSME Problems/Problem 10"
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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== |
+ | ===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>. | 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== |
Revision as of 10:36, 20 August 2011
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 |