AMC 10 · 2010 · #21
Grade 7 probabilityA palindrome between 1000 and 10,000 is chosen at random. What is the probability that it is divisible by 7?
Pick an answer.
AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Pick a 4-digit palindrome at random (a number from 1000 to 9999 that reads the same forwards and backwards). Find the probability that it is a multiple of 7.
Givens: The number has 4 digits, so it is between 1000 and 9999.; It is a palindrome: the first digit equals the last, and the second equals the third.; Every such palindrome is equally likely to be chosen.
Unknowns: The probability that a randomly chosen 4-digit palindrome is divisible by 7.
Understand
Restated: Pick a 4-digit palindrome at random (a number from 1000 to 9999 that reads the same forwards and backwards). Find the probability that it is a multiple of 7.
Givens: The number has 4 digits, so it is between 1000 and 9999.; It is a palindrome: the first digit equals the last, and the second equals the third.; Every such palindrome is equally likely to be chosen.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #2 Make a Systematic List, #16 Change Focus / Count the Complement
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.
Execute — Answer: E
6.EE.A.2 Step 1 Write the palindrome with digits
- A 4-digit palindrome looks like abba, where a is the outer digit and b is the inner digit.
- Using place value, its value is 1000a + 100b + 10b + a.
- Group the like pieces to get 1001a + 110b.
- Here a runs from 1 to 9 (no leading zero) and b runs from 0 to 9, so there are 9 x 10 = 90 palindromes in all.
💡 One formula with two dials, a and b, stands in for all 90 palindromes at once.
4.OA.B.4 Step 2 Split the divisibility into two parts
- A sum is divisible by 7 exactly when the two parts fit together mod 7.
- Factor the first part: 1001 = 7 x 11 x 13, so 1001a is always a multiple of 7, no matter what a is.
- That means the whole number 1001a + 110b is divisible by 7 precisely when the leftover part 110b is divisible by 7.
💡 The a-part is a free pass for 7, so only the b-part decides the outcome.
4.OA.B.4 Step 3 Find which inner digits work
- Now check when 110b is a multiple of 7.
- Since 110 = 7 x 15 + 5, the number 110b leaves the same remainder as 5b when divided by 7.
- Testing b = 0, 1, 2, ..., 9, the value 5b is a multiple of 7 only when b = 0 (gives 0) or b = 7 (gives 35).
- So exactly 2 of the 10 possible inner digits work.
💡 Only the remainder of b matters, and just two digits land on a multiple of 7.
7.SP.C.7 Step 4 Count winners over total
- The outer digit a is completely free (any of its 9 values keeps the number divisible by 7), and the inner digit b has 2 winning choices.
- So there are 9 x 2 = 18 palindromes divisible by 7, out of 90 total.
- The probability is 18/90, which reduces to 1/5.
- The answer is (E).
💡 Free outer digit times 2 good inner digits, divided by all 90, gives a clean 1/5.
6.EE.A.2 A 4-digit palindrome looks like abba, where a is the outer digit and b is the in 4.OA.B.4 A sum is divisible by 7 exactly when the two parts fit together mod 7. Factor th 4.OA.B.4 Now check when 110b is a multiple of 7. Since 110 = 7 x 15 + 5, the number 110b 7.SP.C.7 The outer digit a is completely free (any of its 9 values keeps the number divis Review
Reasonableness: Because 1001 already carries the factor of 7, divisibility depends only on the single inner digit b, and 2 out of its 10 values work. A 2-in-10 chance is 1/5, matching choice (E). It also passes a spot check: 7007 (a=7, b=0) = 7 x 1001 and 1771 (a=1, b=7) = 7 x 253 are both palindromes divisible by 7.
Alternative: Instead of remainders, list the palindromes divisible by 7 directly. For each a from 1 to 9 you always get the b=0 palindrome (a00a = 1001a) and the b=7 palindrome, and no others, giving 2 per value of a, so 18 total over 90 = 1/5.
CCSS standards used (min grade 7)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Naming the digits a and b and writing the palindrome as 1001a + 110b.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Factoring 1001 = 7 x 11 x 13 and testing which values of b make 110b a multiple of 7.)7.SP.C.7Develop probability models and use them to find probabilities of events (Forming the probability as favorable palindromes over total palindromes, 18/90 = 1/5.)
⭐ Because 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.
⭐ Because 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.
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