Difference between revisions of "2010 AMC 10B Problems/Problem 21"

Revision as of 12:29, 23 July 2018

Problem 21

A palindrome between $1000$ and $10,000$ is chosen at random. What is the probability that it is divisible by $7$ ?

$\textbf{(A)}\ \dfrac{1}{10} \qquad \textbf{(B)}\ \dfrac{1}{9} \qquad \textbf{(C)}\ \dfrac{1}{7} \qquad \textbf{(D)}\ \dfrac{1}{6} \qquad \textbf{(E)}\ \dfrac{1}{5}$

Solution

The palindromes can be expressed as: $1000x+100y+10y+x$ (since it is a four digit palindrome, it must be of the form $xyyx$ , where x and y are integers from $[1, 9]$ and $[0, 9]$ , respectively.)

We simplify this to:

Because the question asks for it to be divisible by 7,

We express it as $1001x+110y \equiv 0 \pmod 7$ .

Because $1001 \equiv 0 \pmod 7$ ,

We can substitute $1001$ for $0$

We are left with $110y \equiv 0 \pmod 7$

Since $110 \equiv 5 \pmod 7$ we can simplify the $110$ in the expression to

$5y \equiv 0 \pmod 7$ .

In order for this to be true, $y \equiv 0 \pmod 7$ must also be true.

Thus we solve:

$y \equiv 0 \pmod 7$

Which has two solutions: $0$ and $7$

There are thus two options for $y$ out of the 10, so the answer is $2/10 = \boxed{\textbf{(E)}\ \frac15}$

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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-We simplify this to <math>1001x+110y</math>.
+We simplify this to:
+ <math>1001x+110y</math>.
 Because the question asks for it to be divisible by 7,

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Difference between revisions of "2010 AMC 10B Problems/Problem 21"

Revision as of 12:29, 23 July 2018

Problem 21

Solution

See also