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Difference between revisions of "1994 AJHSME Problems/Problem 15"
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==Solution==
 
==Solution==
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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>.
+
<math>425</math> is divisible by 5 and <math>427</math> is divisible by 7. So the pattern from <math>5</math> to <math>7</math> would have to be the exact same as the pattern from <math>425</math> to <math>427</math>. So the only possible answer would be <math>\boxed{\text{(A)}}</math>.
−
 
+
- Kandy_17
−
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==
 
==See Also==
 
{{AJHSME box|year=1994|num-b=14|num-a=16}}
 
{{AJHSME box|year=1994|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:33, 25 November 2021

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((2/3,1/5)--(1,0)--(2/3,-1/5)); draw((4/5,2/3)--(1,1)--(6/5,2/3)); draw((5/3,6/5)--(2,1)--(5/3,4/5)); draw((9/5,1/3)--(2,0)--(11/5,1/3)); draw((8/3,1/5)--(3,0)--(8/3,-1/5)); draw((14/5,2/3)--(3,1)--(16/5,2/3)); draw((11/3,6/5)--(4,1)--(11/3,4/5)); draw((19/5,1/3)--(4,0)--(21/5,1/3)); draw((14/3,1/5)--(5,0)--(14/3,-1/5)); draw((24/5,2/3)--(5,1)--(26/5,2/3)); draw((17/3,6/5)--(6,1)--(17/3,4/5)); dot((0,0)); dot((1,0)); dot((1,1)); dot((2,1)); dot((2,0)); dot((3,0)); dot((3,1)); dot((4,1)); dot((4,0)); dot((5,0)); dot((5,1)); label("$0$",(0,0),S); label("$1$",(1,0),S); label("$2$",(1,1),N); label("$3$",(2,1),N); label("$4$",(2,0),S); label("$5$",(3,0),S); label("$6$",(3,1),N); label("$7$",(4,1),N); label("$8$",(4,0),S); label("$9$",(5,0),S); label("$10$",(5,1),N); label("$\vdots$",(5.85,0.5),E); label("$\cdots$",(6.5,0.15),S); [/asy]

then which sequence of arrows goes from point $425$ to point $427$?

[asy] unitsize(24); dot((0,0)); dot((0,1)); dot((1,1)); draw((0,0)--(0,1)--(1,1)); draw((-1/5,2/3)--(0,1)--(1/5,2/3)); draw((2/3,6/5)--(1,1)--(2/3,4/5)); label("(A)",(-1/3,1/3),W); dot((4,0)); dot((5,0)); dot((5,1)); draw((4,0)--(5,0)--(5,1)); draw((14/3,1/5)--(5,0)--(14/3,-1/5)); draw((24/5,2/3)--(5,1)--(26/5,2/3)); label("(B)",(11/3,1/3),W); dot((8,1)); dot((8,0)); dot((9,0)); draw((8,1)--(8,0)--(9,0)); draw((39/5,1/3)--(8,0)--(41/5,1/3)); draw((26/3,1/5)--(9,0)--(26/3,-1/5)); label("(C)",(23/3,1/3),W); dot((12,1)); dot((13,1)); dot((13,0)); draw((12,1)--(13,1)--(13,0)); draw((38/3,6/5)--(13,1)--(38/3,4/5)); draw((64/5,1/3)--(13,0)--(66/5,1/3)); label("(D)",(35/3,1/3),W); dot((17,1)); dot((17,0)); dot((16,0)); draw((17,1)--(17,0)--(16,0)); draw((84/5,1/3)--(17,0)--(86/5,1/3)); draw((49/3,1/5)--(16,0)--(49/3,-1/5)); label("(E)",(47/3,1/3),W); [/asy]

Solution

$425$ is divisible by 5 and $427$ is divisible by 7. So the pattern from $5$ to $7$ would have to be the exact same as the pattern from $425$ to $427$. So the only possible answer would be $\boxed{\text{(A)}}$. - Kandy_17

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

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