AMC 10 · 2003 · #15
Grade 7 probabilityWhat is the probability that an integer in the set {1,2,3,...,100} is divisible by 2 and not divisible by 3?
Pick an answer.
AMC 10 2003 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: You pick one integer at random from $\{1,2,3,\dots,100\}$, with every integer equally likely. Find the probability that the integer is divisible by $2$ but not divisible by $3$.
Givens: The set is $\{1,2,3,\dots,100\}$, so there are $100$ integers in all; Each integer is equally likely to be picked; You want the integer to be divisible by $2$ and, at the same time, not divisible by $3$; Answer choices: (A) $\frac{1}{6}$, (B) $\frac{33}{100}$, (C) $\frac{17}{50}$, (D) $\frac{1}{2}$, (E) $\frac{18}{25}$
Unknowns: The probability that a randomly chosen integer is divisible by $2$ but not by $3$
Understand
Restated: You pick one integer at random from $\{1,2,3,\dots,100\}$, with every integer equally likely. Find the probability that the integer is divisible by $2$ but not divisible by $3$.
Givens: The set is $\{1,2,3,\dots,100\}$, so there are $100$ integers in all; Each integer is equally likely to be picked; You want the integer to be divisible by $2$ and, at the same time, not divisible by $3$; Answer choices: (A) $\frac{1}{6}$, (B) $\frac{33}{100}$, (C) $\frac{17}{50}$, (D) $\frac{1}{2}$, (E) $\frac{18}{25}$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #7 Identify Subproblems, #2 Make a Systematic List
The condition has two parts glued together: "divisible by $2$" AND "not divisible by $3$." Counting both at once is awkward, so use tool #16 (Change Focus / Count the Complement): first count all the even numbers, then throw out the ones that are also multiples of $3$. Tool #7 (Identify Subproblems) splits the work into three clean counts — how many multiples of $2$, how many multiples of $6$, and the difference. Tool #2 (Make a Systematic List) is the trustworthy way to count multiples in a range: they march in a steady sequence, so counting them never needs guesswork.
Execute — Answer: C
4.OA.B.4 Step 1 Count the even numbers
- The multiples of $2$ in the range are $2, 4, 6, \dots, 100$.
- They step up by $2$ each time, so their count is $100$ divided by $2$.
- That gives $50$ even numbers between $1$ and $100$.
💡 Multiples of $2$ are evenly spaced, so how many fit in $1$ to $100$ is just $100$ shared into groups of $2$.
6.NS.B.4 Step 2 Divisible by 2 and 3 means multiple of 6
- A number divisible by both $2$ and $3$ must be divisible by their least common multiple.
- Since $2$ and $3$ share no factors, that least common multiple is $2\times 3 = 6$.
- So the even numbers you must remove are exactly the multiples of $6$.
💡 Being in the $2$-times table and the $3$-times table at once means being in the $6$-times table.
4.OA.B.4 Step 3 Count the multiples of 6
- The multiples of $6$ in the range are $6, 12, 18, \dots, 96$.
- They step up by $6$, so their count is $100$ divided by $6$, dropping the leftover.
- That gives $16$ multiples of $6$ (the largest is $96 = 6\times 16$).
💡 Counting multiples of $6$ is the same steady-step idea: $100$ shared into groups of $6$, keeping only whole groups.
7.SP.C.8 Step 4 Subtract to get the favorable count
- Start with the $50$ even numbers, then take away the $16$ that are also multiples of $3$.
- What is left is even and not a multiple of $3$.
- That leaves $50 - 16 = 34$ integers that fit the condition.
💡 "Even but not a multiple of $3$" is just the even numbers with the bad ones (multiples of $6$) removed.
7.SP.C.7 Step 5 Write and simplify the probability
- Every integer is equally likely, so the probability is the favorable count over the total: $\dfrac{34}{100}$.
- Divide top and bottom by $2$ to get $\dfrac{17}{50}$.
- So the probability is $\dfrac{17}{50}$, which is choice (C).
💡 With equally likely outcomes, probability is the favorable count divided by the total count, then reduced.
4.OA.B.4 The multiples of $2$ in the range are $2, 4, 6, \dots, 100$. They step up by $2$ 6.NS.B.4 A number divisible by both $2$ and $3$ must be divisible by their least common m 4.OA.B.4 The multiples of $6$ in the range are $6, 12, 18, \dots, 96$. They step up by $6 7.SP.C.8 Start with the $50$ even numbers, then take away the $16$ that are also multiple 7.SP.C.7 Every integer is equally likely, so the probability is the favorable count over Review
Reasonableness: The answer $\frac{17}{50} = 0.34$ sits sensibly between $0$ and $1$. Half of the numbers are even ($0.5$), and roughly a third of those even numbers get knocked out for being multiples of $3$, leaving about two-thirds of a half — close to $0.33$–$0.34$, which matches. It is also clearly bigger than $\frac{1}{6}$ (too small, that's just the multiples of $6$) and smaller than $\frac{1}{2}$ (that would be all the evens), so choices (A) and (D) are ruled out.
Alternative: Use the repeating pattern of the last steps (tool #5). In any block of $6$ consecutive integers, exactly the numbers ending the residues $2$ and $4$ (mod $6$) are even and not multiples of $3$ — that is $2$ out of every $6$, a rate of $\frac{2}{6}=\frac{1}{3}$. From $1$ to $96$ there are $16$ full blocks giving $16\times 2 = 32$ such numbers; then $97,98,99,100$ add $98$ and $100$, two more, for $34$ total. Same count, same probability $\frac{34}{100}=\frac{17}{50}$.
CCSS standards used (min grade 7)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Counting how many multiples of $2$ (and of $6$) lie in the range $1$ to $100$.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Recognizing that "divisible by both $2$ and $3$" means divisible by $\operatorname{lcm}(2,3)=6$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Combining the two conditions by subtracting the multiples of $6$ from the multiples of $2$ to get $34$ favorable integers.)7.SP.C.7Develop probability models and use them to find probabilities of events (Writing the probability as favorable over total, $\frac{34}{100}$, under equally likely outcomes and reducing it to $\frac{17}{50}$.)
⭐ For "divisible by this but not that," count everything divisible by the first number, then subtract the ones also divisible by the second, and put that over the total.
⭐ For "divisible by this but not that," count everything divisible by the first number, then subtract the ones also divisible by the second, and put that over the total.
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