2025 AMC 12A Problems/Problem 18: Difference between revisions
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<imath>\textbf{(A)}~36 \qquad \textbf{(B)}~84 \qquad \textbf{(C)}~186 \qquad \textbf{(D)}~336 \qquad \textbf{(E)}~486</imath> | <imath>\textbf{(A)}~36 \qquad \textbf{(B)}~84 \qquad \textbf{(C)}~186 \qquad \textbf{(D)}~336 \qquad \textbf{(E)}~486</imath> | ||
== Solution 1 == | == Solution 1 == | ||
let 0<=x<y<z<=8; | |||
x cannot be 0 because it makes xy>z --> 0>z; | |||
x cannot be 1 because it makes xy>z ---> y>z; | |||
x=2, y=3, z can be 4, 5 but not others; | |||
x=2, y=4, z can be 5, 6, 7; | |||
x=2, y=5, z can be 6, 7, 8; | |||
x=2, y=6, z can be 7, 8; | |||
x=2, y=7, z can be 8; | |||
for x=2, total 11 cases; | |||
similarly, for x=3,y=4, 5, 6, 7, total 10 cases; for x=4, y =5, 6, 7, total 6 cases; x=5, y=6, 7, 3 cases; x=6, y=7, z=8 1 cases; | |||
total = 11 + 10 +6 +3 +1 = 31. permutate x, y, z for ordered triple, it is 31*6=186, C. | |||
Revision as of 17:17, 6 November 2025
Problem 19
How many ordered triples 
of different positive integers less than or equal to 
satisfy 
, 
, and 
?
Solution 1
let 0<=x<y<z<=8; x cannot be 0 because it makes xy>z --> 0>z; x cannot be 1 because it makes xy>z ---> y>z;
x=2, y=3, z can be 4, 5 but not others; x=2, y=4, z can be 5, 6, 7; x=2, y=5, z can be 6, 7, 8; x=2, y=6, z can be 7, 8; x=2, y=7, z can be 8; for x=2, total 11 cases;
similarly, for x=3,y=4, 5, 6, 7, total 10 cases; for x=4, y =5, 6, 7, total 6 cases; x=5, y=6, 7, 3 cases; x=6, y=7, z=8 1 cases;
total = 11 + 10 +6 +3 +1 = 31. permutate x, y, z for ordered triple, it is 31*6=186, C.
