2010 AMC 10B Problem 21

Below is the professionally curated solution for Problem 21 of the 2010 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 10B solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:palindromedivisibilitybasic probability

Difficulty rating: 1540

21.

A palindrome between 10001000 and 10,00010,000 is chosen at random. What is the probability that it is divisible by 7?7?

110\dfrac{1}{10}

19\dfrac{1}{9}

17\dfrac{1}{7}

16\dfrac{1}{6}

15\dfrac{1}{5}

Solution:

Note that we can express any 44 digit number as abcd.abcd. This can be expressed in long form as 103a+102b+10c+d. 10^3a + 10^2b + 10c + d. Since in a palindrome, we have that a=da = d and b=c.b = c. We can simplify this to get 1001a+110b. 1001a + 110b. Note that 10011001 is divisible by 7.7. This means that 110b110b must also be divisible by 7.7.

The only way for this to happen is if bb is 00 or 77 since 110110 is not divisible by 7.7.

There are 99 options for aa and 22 options for b,b, for a total of 92=189 \cdot 2 = 18 palindromes.

The total number of palindromes is 9109 \cdot 10 since there are 99 options for the thousands digit and 1010 options for the hundreds digit.

The desired probability is then 18910=15. \dfrac{18}{9 \cdot 10} = \dfrac{1}{5}.

Thus, E is the correct answer.

Problem 21 in Other Years