AMC 10 · 2010 · #18
Grade 7 probabilityPositive integers a, b, and c are randomly and independently selected with replacement from the set {1,2,3,…,2010}. What is the probability that abc+ab+a is divisible by 3?
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 three numbers $a$, $b$, $c$ independently at random, each equally likely to be any of $1, 2, 3, \dots, 2010$. Find the probability that $abc + ab + a$ is a multiple of $3$.
Givens: $a$, $b$, $c$ are each chosen uniformly and independently from $\{1, 2, \dots, 2010\}$; The quantity of interest is $abc + ab + a$; We want the chance that this quantity is divisible by $3$; Answer choices: (A) $\dfrac{1}{3}$, (B) $\dfrac{29}{81}$, (C) $\dfrac{31}{81}$, (D) $\dfrac{11}{27}$, (E) $\dfrac{13}{27}$
Unknowns: The probability that $abc + ab + a \equiv 0 \pmod{3}$
Understand
Restated: Pick three numbers $a$, $b$, $c$ independently at random, each equally likely to be any of $1, 2, 3, \dots, 2010$. Find the probability that $abc + ab + a$ is a multiple of $3$.
Givens: $a$, $b$, $c$ are each chosen uniformly and independently from $\{1, 2, \dots, 2010\}$; The quantity of interest is $abc + ab + a$; We want the chance that this quantity is divisible by $3$; Answer choices: (A) $\dfrac{1}{3}$, (B) $\dfrac{29}{81}$, (C) $\dfrac{31}{81}$, (D) $\dfrac{11}{27}$, (E) $\dfrac{13}{27}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem
Divisibility by $3$ depends only on remainders mod $3$, so Tool #9 (Solve an Easier Related Problem) shrinks the giant set $\{1,\dots,2010\}$ down to just three equally likely remainders $0$, $1$, $2$. Tool #7 (Identify Subproblems) factors the expression into $a(bc + b + 1)$, splitting the question into 'is $a$ a multiple of $3$?' and 'is $bc + b + 1$ a multiple of $3$?'. Tool #2 (Make a Systematic List) then counts the tiny finite table of remainder cases exactly, with no guessing.
Execute — Answer: E
7.SP.C.7 Step 1 Only remainders mod 3 matter
- Whether a number is divisible by $3$ depends only on its remainder when divided by $3$.
- Since $2010 = 3 \times 670$, exactly $670$ of the numbers $1$ to $2010$ leave remainder $0$, another $670$ leave remainder $1$, and $670$ leave remainder $2$.
- So each remainder is equally likely, with probability $\tfrac{670}{2010} = \tfrac{1}{3}$, and the three picks are independent.
💡 Because $2010$ splits evenly into three, each remainder is a perfectly fair one-in-three outcome.
6.EE.A.3 Step 2 Factor the expression
- Every term $abc$, $ab$, $a$ shares a factor of $a$.
- Pull it out: $abc + ab + a = a(bc + b + 1)$.
- A product is divisible by $3$ exactly when at least one of its factors is divisible by $3$.
- So the quantity is a multiple of $3$ when $a$ is a multiple of $3$, OR when $bc + b + 1$ is a multiple of $3$.
💡 Factoring turns one messy sum into a product, and a product is a multiple of $3$ as soon as one piece is.
4.OA.B.4 Step 3 Handle the easy case: a is a multiple of 3
- If $a \equiv 0 \pmod 3$, then $a(bc + b + 1)$ is automatically a multiple of $3$, no matter what $b$ and $c$ are.
- This happens with probability $\tfrac{1}{3}$.
- The only work left is the other case, where $a$ is not a multiple of $3$ (probability $\tfrac{2}{3}$); there we need $bc + b + 1 \equiv 0 \pmod 3$.
💡 If one factor is already a multiple of $3$, the whole product is, and $b, c$ do not matter.
7.SP.C.8 Step 4 List all remainder cases for b and c
- Now compute $bc + b + 1 \pmod 3$ for every pair of remainders $b, c \in \{0, 1, 2\}$ — nine equally likely pairs.
- If $b = 0$: value is $0 + 0 + 1 = 1$ for all $c$ (never $0$).
- If $b = 1$: value is $c + 2$, which is $0$ only when $c = 1$.
- If $b = 2$: value is $2c + 3 \equiv 2c$, which is $0$ only when $c = 0$.
- So exactly $2$ of the $9$ pairs work: $(b, c) = (1, 1)$ and $(2, 0)$.
💡 With only three remainders each, the whole space is a $3\times 3$ table you can check by hand.
7.SP.C.8 Step 5 Combine the two cases
- The expression is divisible by $3$ if $a \equiv 0$ (probability $\tfrac{1}{3}$), OR if $a \not\equiv 0$ AND $bc + b + 1 \equiv 0$ (probability $\tfrac{2}{3} \times \tfrac{2}{9} = \tfrac{4}{27}$).
- These two cases do not overlap, so add them: $\tfrac{1}{3} + \tfrac{4}{27} = \tfrac{9}{27} + \tfrac{4}{27} = \tfrac{13}{27}$.
- The answer is $\textbf{(E)}\ \tfrac{13}{27}$.
💡 Add the chances of the separate, non-overlapping ways the product can be a multiple of $3$.
7.SP.C.7 Whether a number is divisible by $3$ depends only on its remainder when divided 6.EE.A.3 Every term $abc$, $ab$, $a$ shares a factor of $a$. Pull it out: $abc + ab + a = 4.OA.B.4 If $a \equiv 0 \pmod 3$, then $a(bc + b + 1)$ is automatically a multiple of $3$ 7.SP.C.8 Now compute $bc + b + 1 \pmod 3$ for every pair of remainders $b, c \in {0, 1, 7.SP.C.8 The expression is divisible by $3$ if $a \equiv 0$ (probability $\tfrac{1}{3}$), Review
Reasonableness: The result $\tfrac{13}{27} \approx 0.481$ is a bit under one half, which is sensible: divisibility by $3$ alone would give $\tfrac{1}{3}$, and the extra factor structure only adds chances, so the answer should exceed $\tfrac{1}{3}$ but stay well below $1$. Cross-check by direct counting of remainder triples: over all $3^3 = 27$ equally likely remainder patterns for $(a,b,c)$, the $9$ with $a \equiv 0$ all work, and among the $18$ with $a \not\equiv 0$ exactly the ones with $(b,c) \in \{(1,1),(2,0)\}$ work, giving $2 \times 2 = 4$ more. Total $9 + 4 = 13$ out of $27$, confirming $\tfrac{13}{27}$.
Alternative: Tool #16 (Change Focus / Count the Complement): find the probability the expression is NOT divisible by $3$. This needs $a \not\equiv 0$ (prob $\tfrac{2}{3}$) and $bc + b + 1 \not\equiv 0$ (prob $\tfrac{7}{9}$ from the same table), giving $\tfrac{2}{3}\cdot\tfrac{7}{9} = \tfrac{14}{27}$. Subtracting from $1$ gives $1 - \tfrac{14}{27} = \tfrac{13}{27}$, the same answer.
CCSS standards used (min grade 7)
7.SP.C.7Develop probability models and use them to find probabilities of events (Establishing that each remainder $0$, $1$, $2$ mod $3$ is equally likely with probability $\tfrac{1}{3}$ because $2010$ divides evenly into three classes.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring $abc + ab + a$ into the product $a(bc + b + 1)$.)4.OA.B.4Find factors and multiples and recognize multiples of a whole number (Using the rule that a product is a multiple of $3$ exactly when one of its factors is, splitting into the $a \equiv 0$ case and the $bc + b + 1 \equiv 0$ case.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Listing the $3\times 3$ table of $(b,c)$ remainders to count the $2$ favorable pairs, then combining the independent cases into $\tfrac{1}{3} + \tfrac{2}{3}\cdot\tfrac{2}{9} = \tfrac{13}{27}$.)
⭐ For divisibility by $3$, only remainders mod $3$ matter — factor the expression, then check the tiny table of remainder cases instead of the huge original set.
⭐ For divisibility by $3$, only remainders mod $3$ matter — factor the expression, then check the tiny table of remainder cases instead of the huge original set.
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