Difference between revisions of "1990 AHSME Problems/Problem 18"
(Created page with "== Problem == First <math>a</math> is chosen at random from the set <math>\{1,2,3,\cdots,99,100\}</math>, and then <math>b</math> is chosen at random from the same set. The prob...") |
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== Solution == | == Solution == | ||
| − | <math>\fbox{C}</math> | + | The units digits of the powers of <math>3</math> and <math>7</math> both cycle through <math>1,3,9,7</math> in opposite directions, and as <math>4\mid 100</math> each power's units digit is equally probable. There are <math>16</math> ordered pairs of units digits, and three of them <math>(1,7),(7,1),(9,9)</math> have a sum with units digit <math>8</math>. |
| + | |||
| + | Thus the probability is <math>\frac3{16}</math> which is <math>\fbox{C}</math> | ||
== See also == | == See also == | ||
Latest revision as of 11:52, 4 February 2016
Problem
First 
is chosen at random from the set 
, and then 
is chosen at random from the same set. The probability that the integer 
has units digit 
is
Solution
The units digits of the powers of 
and 
both cycle through 
in opposite directions, and as 
each power's units digit is equally probable. There are 
ordered pairs of units digits, and three of them 
have a sum with units digit .
Thus the probability is 
which is 
See also
| 1990 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
