Difference between revisions of "1994 AJHSME Problems/Problem 15"
Mrdavid445 (talk | contribs) (Created page with "==Problem== If this path is to continue in the same pattern: <asy> unitsize(24); draw((0,0)--(1,0)--(1,1)--(2,1)--(2,0)--(3,0)--(3,1)--(4,1)--(4,0)--(5,0)--(5,1)--(6,1)); draw(...") |
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label("(E)",(47/3,1/3),W); | label("(E)",(47/3,1/3),W); | ||
</asy> | </asy> | ||
+ | |||
+ | ==Solution== | ||
+ | Notice the pattern from <math>0</math> to <math>4</math> repeats for every four arrows. Any number that has a remainder of <math>0</math> when divided by <math>4</math> corresponds to <math>0</math>. | ||
+ | |||
+ | The remainder when <math>425</math> is divided by <math>4</math> is <math>1</math>. The arrows from point <math>425</math> to point <math>427</math> correspond to points <math>1</math> and <math>3</math>, which have the same pattern as <math>\boxed{\text{(A)}}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AJHSME box|year=1994|num-b=14|num-a=16}} | ||
+ | {{MAA Notice}} |
Latest revision as of 10:52, 27 June 2023
Problem
If this path is to continue in the same pattern:
then which sequence of arrows goes from point to point ?
Solution
Notice the pattern from to repeats for every four arrows. Any number that has a remainder of when divided by corresponds to .
The remainder when is divided by is . The arrows from point to point correspond to points and , which have the same pattern as .
See Also
1994 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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.