AMC 10 · 2010 · #21
Grade 7 probabilitynumber-theoryPick an answer.
AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Naming the two free digits turns every palindrome into one algebraic expression. That expression splits into two pieces, and checking each piece for divisibility by 7 is far easier than testing 90 numbers one by one. Counting the winners and dividing by the total then gives the probability.
Write the palindrome with digits
A 4-digit palindrome abba equals 1001a + 110b, with a from 1 to 9 and b from 0 to 9, so there are 90 of them.
One formula with two dials, a and b, stands in for all 90 palindromes at once.
6.EE.A.2Introduce A VariableSplit the divisibility into two parts
Since 1001 = 7 x 11 x 13, the part 1001a is always a multiple of 7, so only 110b decides divisibility.
The a-part is a free pass for 7, so only the b-part decides the outcome.
The outer-digit part is a free pass, so only the inner digit decides the outcome.
▸ Why?
A number is its digits weighted by their places, so each digit's contribution can be checked apart.
▸ Why?
A part that is already a multiple leaves the remainder untouched, so it drops out of the test.
Find which inner digits work
110b leaves the same remainder as 5b mod 7, so 5b is a multiple of 7 only at b = 0 or b = 7: 2 of the 10 inner digits.
Only the remainder of b matters, and just two digits land on a multiple of 7.
4.OA.B.4Make A Systematic ListCount winners over total
The outer digit a stays free, so 9 x 2 = 18 of the 90 palindromes work: 18/90 = 1/5, choice (E).
Free outer digit times 2 good inner digits, divided by all 90, gives a clean 1/5.
7.SP.C.7Change Focus Count The ComplementBecause 1001 is already a multiple of 7, only the middle digit decides the answer, and 2 of its 10 choices work, so the chance is 1/5.
- Write the palindrome with digits
- Split the divisibility into two parts
- Find which inner digits work
- Count winners over total