AMC 10 · 2006 · #20
Grade 7 probabilitySix distinct positive integers are randomly chosen between 1 and 2006, inclusive. What is the probability that some pair of these integers has a difference that is a multiple of 5?
Pick an answer.
AMC 10 2006 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 six different whole numbers, each from $1$ to $2006$. Find the probability that at least one pair among them has a difference that is a multiple of $5$.
Givens: Six numbers are chosen, all distinct; Each number is a whole number from $1$ to $2006$ inclusive; A "good" pair is any two of the six whose difference is a multiple of $5$; Answer choices: (A) $\frac{1}{2}$, (B) $\frac{3}{5}$, (C) $\frac{2}{3}$, (D) $\frac{4}{5}$, (E) $1$
Unknowns: The probability that some pair of the six numbers has a difference divisible by $5$
Understand
Restated: You pick six different whole numbers, each from $1$ to $2006$. Find the probability that at least one pair among them has a difference that is a multiple of $5$.
Givens: Six numbers are chosen, all distinct; Each number is a whole number from $1$ to $2006$ inclusive; A "good" pair is any two of the six whose difference is a multiple of $5$; Answer choices: (A) $\frac{1}{2}$, (B) $\frac{3}{5}$, (C) $\frac{2}{3}$, (D) $\frac{4}{5}$, (E) $1$
Plan
Primary tool: #14 Extreme Principle
Secondary: #16 Change Focus / Count the Complement, #2 Make a Systematic List, #3 Eliminate Possibilities
The phrase "difference is a multiple of $5$" is hard to test pair by pair, so Tool #16 (Change Focus) rewrites it as "the two numbers leave the same remainder when divided by $5$." Tool #2 (Make a Systematic List) then shows there are only five possible remainders. Tool #14 (Extreme Principle) finishes it: try the extreme case of spreading the numbers as far apart as possible so no two share a remainder — you run out of remainders after five, so the sixth number is forced to collide, making a good pair unavoidable no matter what you pick.
Execute — Answer: E
4.NBT.B.6 Step 1 Rewrite the condition with remainders
- Two numbers have a difference that is a multiple of $5$ exactly when they leave the same remainder after division by $5$.
- If $a$ and $b$ both leave remainder $r$, then $a-b$ has remainder $0$, so it is a multiple of $5$; and if their remainders differ, the difference cannot be a multiple of $5$.
- So a "good pair" is just two numbers in the same remainder group.
💡 Numbers with the same leftover after dividing by $5$ line up, so their gap is a clean multiple of $5$.
4.OA.B.4 Step 2 There are only five remainder groups
- When any whole number is divided by $5$, the remainder can only be $0$, $1$, $2$, $3$, or $4$ — five choices and no others.
- So every one of the six chosen numbers must fall into one of just five remainder groups.
💡 Dividing by $5$ can leave only five possible leftovers, so there are five bins to sort into.
4.NBT.B.6 Step 3 Six numbers into five groups must repeat
- Try the most spread-out case, where you fight to keep every number in a different group.
- You can place at most one number in each of the five groups — that uses up five numbers.
- The sixth number has nowhere new to go, so it must join a group that already holds one of the others.
- Those two share a remainder, so they form a good pair.
- This happens for every possible choice of six numbers.
💡 With more numbers than groups, at least two are forced to land in the same group.
7.SP.C.5 Step 4 Read off the probability
- Because a good pair appears no matter which six numbers are chosen, the event is certain.
- A certain event has probability $1$.
- So the probability that some pair has a difference that is a multiple of $5$ is $1$, which is choice (E).
💡 Something that must happen every single time has probability exactly $1$.
4.NBT.B.6 Two numbers have a difference that is a multiple of $5$ exactly when they leave 4.OA.B.4 When any whole number is divided by $5$, the remainder can only be $0$, $1$, $2$ 4.NBT.B.6 Try the most spread-out case, where you fight to keep every number in a differen 7.SP.C.5 Because a good pair appears no matter which six numbers are chosen, the event is Review
Reasonableness: The answer $1$ fits the setup: to avoid any good pair you would need all six numbers to sit in different remainder groups, but there are only five groups, so avoidance is impossible. Since no choice of six can fail, no answer below $1$ (like $\frac{4}{5}$) can be right — a smaller probability would mean some selections escape, and none do. The range $1$ to $2006$ is a red herring; it only guarantees each remainder group is available, and the count never mattered.
Alternative: Count the complement (Tool #16). The event fails only when all six numbers have different remainders. But six different remainders would need six remainder values, and only five exist, so the number of failing selections is $0$. The probability of failing is $0$, so the probability of success is $1 - 0 = 1$, again choice (E).
CCSS standards used (min grade 7)
4.NBT.B.6Find whole-number quotients and remainders (Grouping the numbers by their remainder after division by $5$ and seeing that a same-remainder pair has a difference divisible by $5$.)4.OA.B.4Find all factor pairs and recognize multiples (Listing the five possible remainders $0,1,2,3,4$ that come from dividing by $5$.)7.SP.C.5Understand that the probability of a chance event is between 0 and 1 (Reading a guaranteed event as probability exactly $1$.)
⭐ Sort by remainder: six numbers can only land in five remainder groups, so two must match and their difference is a multiple of $5$ every time.
⭐ Sort by remainder: six numbers can only land in five remainder groups, so two must match and their difference is a multiple of $5$ every time.
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