AMC 10 · 2006 · #20

Grade 7 probability

Archetype Modular Cycle Frequency Count └ Sub-archetype Floor Count Across Residue Bins

modular-arithmeticprobability-basicsystematic-enumeration logical-deduction ↑ Prerequisites: modular-arithmetic

📏 Medium solution 💡 2 insights

Problem

You pick six different whole numbers at random, each from 1 to 2006. Find the probability that at least one pair among them has a difference that is a multiple of 5.

Pick an answer.

(A)

$\frac{1}{2}$

(B)

$\frac{3}{5}$

(C)

$\frac{2}{3}$

(D)

$\frac{4}{5}$

(E)

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Extreme Principle

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.

1STEP 1

Rewrite the condition with remainders

Two numbers differ by a multiple of 5 exactly when they leave the same remainder after dividing by 5, so a good pair means equal remainders.

a-b ≡ 0 (mod 5) ⇔ a ≡ b (mod 5)

2STEP 2

There are only five remainder groups

Dividing any whole number by 5 leaves a remainder of 0, 1, 2, 3, or 4 and nothing else, so there are only five remainder groups.

remainder ∈ {0,1,2,3,4} → 5 groups

3STEP 3

Six numbers into five groups must repeat

Spread them out as far as you can: five groups take one number each, so the sixth is forced to share — a good pair, every single time.

6 ＞ 5 → two numbers share a remainder group

4STEP 4

Read off the probability

A good pair appears for every possible choice, so the event is certain and its probability is 1 — choice (E).

P(good pair) = 1 → (E)

Answer

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 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.

💡Key takeaway

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.

Rewrite the condition with remainders
There are only five remainder groups
Six numbers into five groups must repeat
Read off the probability

Rewrite the condition with remainders

There are only five remainder groups

Six numbers into five groups must repeat

Read off the probability

More like this