Difference between revisions of "2010 AMC 12B Problems/Problem 11"

Revision as of 00:36, 22 December 2020

The following problem is from both the 2010 AMC 12B #11 and 2010 AMC 10B #21, so both problems redirect to this page.

Problem 11

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

View the palindrome as some number with form (decimal representation): $a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$ . But because the number is a palindrome, $a_3 = a_0, a_2 = a_1$ . Recombining this yields $1001a_3 + 110a_2$ . 1001 is divisible by 7, which means that as long as $a_2 = 0$ , the palindrome will be divisible by 7. This yields 9 palindromes out of 90 ( $9 \cdot 10$ ) possibilities for palindromes. However, if $a_2 = 7$ , then this gives another case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to $18/90 = \boxed {\frac{1}{5} } = \boxed {E}$

Addendum (Alternate)

$7\mid 1001a^3+110b^2$ and $1001 \equiv 0 (\textrm{mod} 7)$ . Knowing that $a$ does not factor (pun intended) into the problem, note 110's prime factorization and $7\mid b$ . There are only 10 possible digits for b (0-9), but $7\mid b$ only holds if $b=0, 7$ . This is 2 of the 10 digits, so $\frac{2}{10}=\boxed{\textbf{E)}\frac{1}{5}}$

~BJHHar

Video Solution

https://youtu.be/ZfnxbpdFKjU

~IceMatrix

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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 12 Problems and Solutions

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 21: / Line 21: @@
 == See also ==
 {{AMC12 box|year=2010|num-b=10|num-a=12|ab=B}}
+{{AMC10 box|year=2010|num-b=20|num-a=22|ab=B}}
 [[Category:Introductory Combinatorics Problems]]
 {{MAA Notice}}

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