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>
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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>.
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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 $a$ is chosen at random from the set $\{1,2,3,\cdots,99,100\}$, and then $b$ is chosen at random from the same set. The probability that the integer $3^a+7^b$ has units digit $8$ is

$\text{(A) } \frac{1}{16}\quad \text{(B) } \frac{1}{8}\quad \text{(C) } \frac{3}{16}\quad \text{(D) } \frac{1}{5}\quad \text{(E) } \frac{1}{4}$

Solution

The units digits of the powers of $3$ and $7$ both cycle through $1,3,9,7$ in opposite directions, and as $4\mid 100$ each power's units digit is equally probable. There are $16$ ordered pairs of units digits, and three of them $(1,7),(7,1),(9,9)$ have a sum with units digit $8$.

Thus the probability is $\frac3{16}$ which is $\fbox{C}$

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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