Difference between revisions of "2005 Alabama ARML TST Problems/Problem 4"
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(I can't seem to follow this; but I believe the answer is 17) |
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==Problem== | ==Problem== | ||
− | For how many [[ordered pair]]s of [[digit]]s <math> | + | For how many [[ordered pair]]s of [[digit]]s <math>(A,B)</math> is <math>2AB8</math> a [[multiple]] of <math>12</math>? |
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==Solution== | ==Solution== | ||
+ | A number is divisible by <math>12</math> if it is divisible by <math>3</math> and <math>4</math>. A number is divisible by <math>4</math> if its last two digits are divisible by <math>4</math>, so <math>4|\overline{B8} \Longrightarrow B = 0,2,4,6,8</math>. A number is divisible by <math>3</math> if the sum of its digits is <math>3</math>, so | ||
− | + | *<math>3| s(2A08) = 10 + A \Longrightarrow A = 2,5,8</math> - 3 solutions | |
+ | *<math>3| s(2A28) = 12 + A \Longrightarrow A = 0,3,6,9</math> - 4 solutions | ||
+ | *<math>3| s(2A48) = 14 + A \Longrightarrow A = 1,4,7</math> - 3 solutions | ||
+ | *<math>3| s(2A68) = 16 + A \Longrightarrow A = 2,5,8</math> - 3 solutions | ||
+ | *<math>3| s(2A88) = 18 + A \Longrightarrow A = 0,3,6,9</math> - 4 solutions | ||
− | <math> | + | These sum to <math>17</math>. |
− | + | ==See also== | |
− | + | {{ARML box|year=2005|state=Alabama|num-b=3|num-a=5}} | |
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− | + | [[Category:Intermediate Number Theory Problems]] | |
− | + | [[Category:Intermediate Combinatorics Problems]] | |
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Latest revision as of 20:34, 5 January 2008
Problem
For how many ordered pairs of digits is
a multiple of
?
Solution
A number is divisible by if it is divisible by
and
. A number is divisible by
if its last two digits are divisible by
, so
. A number is divisible by
if the sum of its digits is
, so
- 3 solutions
- 4 solutions
- 3 solutions
- 3 solutions
- 4 solutions
These sum to .
See also
2005 Alabama ARML TST (Problems) | ||
Preceded by: Problem 3 |
Followed by: Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |