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

Revision as of 10:26, 23 September 2025

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

Problem

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 1

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 $18/90 = \boxed {\frac{1}{5} } = \boxed {E}$

Solution (Divisibility Rules)

We can notice the palindrome is of the form $\overline{abba}$ . Then, by the divisibility rule of $7$ , $7$ must divide $100a+11b-2a = 98a+11b.$ This nicely simplifies to the fact that $7 \mid 4b,$ so $b$ is clearly $0$ or $7$ . This gives us $9 \cdot 2$ total choices for the palindrome divisible by $7$ , divided by $9 \cdot 10$ total choices for $\overline{abba}$ , giving us an answer of $\boxed{\text{(E)}} \ \dfrac{1}{5}$ .

~icecreamrolls8

Addendum (Alternate)

$7\mid 1001a^3+110b^2$ and $1001 \equiv 0 \pmod 7$ . 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 through 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

Solution 4

The palindrome is in the form of $\overline{abba}$ . There are $9$ possible values of $a$ and 10 possible values for $b$ . Thus, there are $90$ four-digit palindromes. The smallest palindrome is $1001$ , which is a multiple of $7$ . Note how each palindrome with the same thousands and ones digit increase by $110$ (i.e., $1111-1001=110$ , $1221-1111=110$ , etc). $110 \equiv 5$ (mod $7$ ). Since $5$ and $7$ is coprime, to get another palindrome that is a multiple of $7$ , we need to increase by $770$ . $1001+770=1771$ . Next, when the thousands place is $2$ , we start with $2002$ , which is a multiple of $1001$ , so $2002 \equiv 0$ (mod $7$ ). Therefore, $2772$ is also a satisfactory palindrome. Similarly, for every thousands place, there are $2$ satisfactory palindromes, which means there are a total of $2 \cdot 9 = 18$ satisfactory palindromes. Thus, the answer is $\frac{18}{90} = \frac{1}{5}$ or $\boxed{\text{(E)} \frac{1}{5}}$ .

Video Solution

https://youtu.be/ZfnxbpdFKjU

~IceMatrix

@@ Line 22: / Line 22: @@
 == Solution 4 ==
-The palindrome is in the form of <math>\overline{abba}</math>. There are <math>9</math> possible values of <math>a</math> and 10 possible values for <math>b</math>. Thus, there are <math>90</math> four-digit palindromes. The smallest palindrome is <math>1001</math>, which is a multiple of <math>7</math>. Note how each palindrome with the same thousands and ones digit increase by <math>110</math> (i.e., <math>1111-1001=110</math>, <math>1221-1111=110</math>, etc). <math>110 \equiv 5</math> (mod <math>7</math>). Since <math>5</math> and <math>7</math> is coprime, to get another palindrome that is a multiple of <math>7</math>, we need to increase by <math>770</math>. <math>1001+770=1771</math>. Next, when the thousands place is <math>2</math>, we start with <math>2002</math>, which is a multiple of <math>1001</math>, so <math>2002 \equiv 0</math> (mod <math>7</math>). Therefore, <math>2772</math> is also a satisfactory palindrome. Similarly, for every thousands place, there are <math>2</math> satisfactory palindromes, which means there are a total of <math>2 \cdot 9 = 18</math> satisfactory palindromes. Thus, the answer is <math>\frac{18}{90} = \frac{1}{5}</math> or <math>\boxed{(E) \frac{1}{5}}</math>.
+The palindrome is in the form of <math>\overline{abba}</math>. There are <math>9</math> possible values of <math>a</math> and 10 possible values for <math>b</math>. Thus, there are <math>90</math> four-digit palindromes. The smallest palindrome is <math>1001</math>, which is a multiple of <math>7</math>. Note how each palindrome with the same thousands and ones digit increase by <math>110</math> (i.e., <math>1111-1001=110</math>, <math>1221-1111=110</math>, etc). <math>110 \equiv 5</math> (mod <math>7</math>). Since <math>5</math> and <math>7</math> is coprime, to get another palindrome that is a multiple of <math>7</math>, we need to increase by <math>770</math>. <math>1001+770=1771</math>. Next, when the thousands place is <math>2</math>, we start with <math>2002</math>, which is a multiple of <math>1001</math>, so <math>2002 \equiv 0</math> (mod <math>7</math>). Therefore, <math>2772</math> is also a satisfactory palindrome. Similarly, for every thousands place, there are <math>2</math> satisfactory palindromes, which means there are a total of <math>2 \cdot 9 = 18</math> satisfactory palindromes. Thus, the answer is <math>\frac{18}{90} = \frac{1}{5}</math> or <math>\boxed{\text{(E)} \frac{1}{5}}</math>.
 ==Video Solution==

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

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