AMC 10 · 2025 · #22
Grade 7 probabilityA seven-digit positive integer is chosen at random. What is the probability that the number is divisible by 11, given that the sum of its digits is 61?
Pick an answer.
AMC 10 2025 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: A random seven-digit number has digit sum 61. Among all such numbers, what fraction are divisible by 11?
Givens: The number has exactly seven digits, written $\overline{a_1a_2a_3a_4a_5a_6a_7}$.; The digits add up to $a_1+a_2+\cdots+a_7 = 61$.; Each digit satisfies $0 \le a_i \le 9$ (and $a_1 \ne 0$).
Unknowns: The probability that such a number is divisible by 11, i.e. (count divisible by 11) / (count with digit sum 61).
Understand
Restated: A random seven-digit number has digit sum 61. Among all such numbers, what fraction are divisible by 11?
Givens: The number has exactly seven digits, written $\overline{a_1a_2a_3a_4a_5a_6a_7}$.; The digits add up to $a_1+a_2+\cdots+a_7 = 61$.; Each digit satisfies $0 \le a_i \le 9$ (and $a_1 \ne 0$).
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #7 Identify Subproblems, #4 Introduce a Variable, #14 Extreme Principle, #2 Make a Systematic List
The digit sum 61 is only 2 short of the maximum 63, so instead of counting big digits directly, count how much each digit falls short of 9. That complement swap $d_i = 9 - a_i$ turns a hard 'sum to 61' count into an easy 'sum to 2' count, and it works again for the divisibility case. Split the job into a denominator (all numbers with digit sum 61) and a numerator (those also divisible by 11).
Execute — Answer: A
7.SP.C.8 Step 1 Split into numerator and denominator
- Because the digit sum is fixed at 61, every valid seven-digit number is equally likely.
- The probability is the number of these numbers that are also divisible by 11, divided by the total number of these numbers.
💡 A conditional probability is just favorable outcomes over all the outcomes that meet the condition.
6.EE.B.7 Step 2 Flip digits to their shortfall
- The digits are all near 9, so measure how far each falls short.
- Let $d_i = 9 - a_i$, so $a_i = 9 - d_i$ with $0 \le d_i \le 9$.
- Substituting into the digit-sum equation, the seven 9's give 63, and the shortfalls must remove 2.
💡 Counting the small gap to 9 is far easier than counting the large digits themselves.
7.SP.C.8 Step 3 Count all numbers with digit sum 61
- Now count nonnegative integer solutions of $d_1+\cdots+d_7 = 2$.
- Since the total is only 2, the cap $d_i \le 9$ never bites, so stars and bars applies directly.
- Every solution gives digits $\ge 7$, so the leading digit is automatically nonzero.
💡 Distributing 2 identical units among 7 slots is a plain stars-and-bars count.
5.NBT.A.1 Step 4 Write the divisibility-by-11 condition
- The rule for 11 uses alternating positions: subtract the even-position digits from the odd-position digits and require a multiple of 11.
- Call the odd-position sum $O = a_1+a_3+a_5+a_7$ and the even-position sum $E = a_2+a_4+a_6$.
- Then $O - E = 11k$, while $O + E = 61$.
💡 Place value makes 11 divide a number exactly when the alternating digit sum is a multiple of 11.
6.EE.B.5 Step 5 Pin down k with the extreme principle
- Adding the two equations gives $2O = 61 + 11k$, so $61 + 11k$ must be even, forcing $k$ odd.
- But $E \le 3\times 9 = 27$ and $O \le 4\times 9 = 36$.
- If $k \le -1$ then $E \ge 36 > 27$, impossible; if $k \ge 3$ then $O \ge 47 > 36$, impossible.
- The only odd value left is $k = 1$.
💡 The digit sums can only stretch so far, so only one value of k survives the boundary check.
7.SP.C.8 Step 6 Count the divisible cases
- The odd-position sum $O = 36$ uses four digits, so each must be its maximum 9: only one way.
- The even-position sum $E = 25$ uses three digits; apply the same shortfall trick, $d_i = 9 - a_i$, giving $d_2+d_4+d_6 = 27-25 = 2$.
- Stars and bars counts these.
💡 Forcing the four odd digits to 9 leaves just a tiny gap of 2 to spread over the three even slots.
7.SP.C.5 Step 7 Form the probability
- There are 6 favorable numbers out of 28 total, so the probability is $6/28$, which reduces to $3/14$.
- That matches choice (A).
💡 Reduce the favorable-over-total fraction to lowest terms to read off the answer.
7.SP.C.8 Because the digit sum is fixed at 61, every valid seven-digit number is equally 6.EE.B.7 The digits are all near 9, so measure how far each falls short. Let $d_i = 9 - a 7.SP.C.8 Now count nonnegative integer solutions of $d_1+\cdots+d_7 = 2$. Since the total 5.NBT.A.1 The rule for 11 uses alternating positions: subtract the even-position digits fr 6.EE.B.5 Adding the two equations gives $2O = 61 + 11k$, so $61 + 11k$ must be even, forc 7.SP.C.8 The odd-position sum $O = 36$ uses four digits, so each must be its maximum 9: o 7.SP.C.5 There are 6 favorable numbers out of 28 total, so the probability is $6/28$, whi Review
Reasonableness: The probability $3/14 \approx 0.214$ is between 0 and 1, as any probability must be, and it is below $1/4$, which fits the fact that only a special alternating-sum alignment (all four odd-position digits equal to 9) lets a near-maximal number be divisible by 11. Both counts are cross-checkable by listing digit patterns: digit sum 61 means either five 9's and two 8's ($\binom{7}{2}=21$ ways) or six 9's and one 7 ($7$ ways), totaling 28; of these, exactly the ones with the 7 or both 8's in even positions work, giving 6.
Alternative: Instead of the complement substitution, enumerate the digit multisets directly. Digit sum 61 forces either {five 9's, two 8's} or {six 9's, one 7}. The first has $\binom{7}{2}=21$ arrangements, the second has $7$, totaling 28. For divisibility by 11 the odd positions must sum to 36 (all four are 9), so the non-9 digits sit in the three even positions: 3 ways to place the single 7 among even slots, and 3 ways to place the two 8's among even slots, giving $6$. Again $6/28 = 3/14$.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting the 28 total numbers and the 6 divisible numbers as organized outcome counts, then forming the probability.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Turning the digit-sum condition into the equation $d_1+\cdots+d_7 = 2$ after the shortfall substitution.)5.NBT.A.1Recognize that a digit in one place represents ten times as much as to its right (Justifying the divisibility-by-11 rule through place value and the alternating sum of digits.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Bounding the odd- and even-position sums to show the multiplier k can only equal 1.)7.SP.C.5Understand that the probability of a chance event is between 0 and 1 (Interpreting and sanity-checking the final ratio $3/14$ as a valid probability.)
⭐ When digits are packed near 9, count how far each falls short of 9 instead of counting the digits themselves.
⭐ When digits are packed near 9, count how far each falls short of 9 instead of counting the digits themselves.
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