Difference between revisions of "2010 AMC 10B Problems/Problem 21"
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<math>a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0</math>. But because the number is a palindrome, <math>a_3 = a_0, a_2 = a_1</math>. Recombining this yields <math>1001a_3 + 110a_2</math>. 1001 is divisible by 7, which means that as long as <math>a_2 = 0</math>, the palindrome will be divisible by 7. This yields 9 palindromes out of 90 (<math>9 \cdot 10</math>) possibilities for palindromes. However, if <math>a_2 = 7</math>, 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 <math>\frac{18}{90}=\boxed{\textbf{(E)}\ \frac15}</math> | <math>a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0</math>. But because the number is a palindrome, <math>a_3 = a_0, a_2 = a_1</math>. Recombining this yields <math>1001a_3 + 110a_2</math>. 1001 is divisible by 7, which means that as long as <math>a_2 = 0</math>, the palindrome will be divisible by 7. This yields 9 palindromes out of 90 (<math>9 \cdot 10</math>) possibilities for palindromes. However, if <math>a_2 = 7</math>, 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 <math>\frac{18}{90}=\boxed{\textbf{(E)}\ \frac15}</math> | ||
| + | ==Another Solution == | ||
| + | |||
| + | It is known that the palindromes can be expressed as: <math>1000x+100y+10y+x </math>(as it is a four digit palindrome it must be of the form xyyx, where x and y are positive integers from [0,9]. | ||
| + | Using the divisibility rules of 7, 100x+10y+y-2x = 98x+11y \equiv 0 \mod{7} | ||
| + | |||
| + | The 98x is now irrelelvant | ||
| + | |||
| + | Thus we solve: | ||
| + | |||
| + | 11y \equiv 0 \mod{7} | ||
| + | |||
| + | Which has two solutions: 0 and 7 | ||
| + | |||
| + | There are thus, two options for y out of the 10, so 2/10 = 1/5. | ||
== See also == | == See also == | ||
{{AMC10 box|year=2010|ab=B|num-b=20|num-a=22}} | {{AMC10 box|year=2010|ab=B|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:35, 21 January 2014
Problem 21
A palindrome between 
and 
is chosen at random. What is the probability that it is divisible by 
?
Solution
View the palindrome as some number with form (decimal representation):
. But because the number is a palindrome, 
. Recombining this yields 
. 1001 is divisible by 7, which means that as long as 
, the palindrome will be divisible by 7. This yields 9 palindromes out of 90 (
) possibilities for palindromes. However, if 
, 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 
Another Solution
It is known that the palindromes can be expressed as: 
(as it is a four digit palindrome it must be of the form xyyx, where x and y are positive integers from [0,9].
Using the divisibility rules of 7, 100x+10y+y-2x = 98x+11y \equiv 0 \mod{7}
The 98x is now irrelelvant
Thus we solve:
11y \equiv 0 \mod{7}
Which has two solutions: 0 and 7
There are thus, two options for y out of the 10, so 2/10 = 1/5.
See also
| 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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