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 10:52, 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 | |
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All AHSME Problems and Solutions |
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