AMC 10 · 2021 · #4

Grade 6 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
divisibility-rulesprime-numbersdigit-sumsystematic-enumeration systematic-enumerationdigit-constraints ↑ Prerequisites: divisibility-rules
📏 Medium solution 💡 2 insights
Problem
A six-digit number is written with the fixed digits two, zero, two, one, zero followed by an unknown units digit. Exactly one choice of that digit makes the number prime. Find that digit.

Pick an answer.

(A)
1
(B)
3
(C)
5
(D)
7
(E)
9
How to solve
Strategy Eliminate Possibilities

Proving a six-digit number prime the direct way means trial-dividing by every prime up to its square root, which is far too slow to do five times. But the reverse job is cheap: to show a number is not prime you only need to produce one divisor. Tool #3 (Eliminate Possibilities) exploits that asymmetry together with the problem's promise that exactly one candidate is prime, so knocking out four candidates identifies the fifth without ever certifying it. Tool #2 (Make a Systematic List) first writes the five candidate numbers down so nothing is missed. Tool #5 (Look for a Pattern) then supplies the cheap divisor hunts: the units-digit test for 5, the digit-sum test for 3, and the alternating-digit test for 11 each read a factor straight off the digits, and between them they finish four of the five candidates.

1STEP 1

List the five candidate numbers

Only odd digits are candidates.

202100+A for A ∈ {1,3,5,7,9} gives 202101, 202103, 202105, 202107, 202109
2STEP 2

Remove the number ending in 5

Remove the one ending in five.

202105 = 5 × 40421
3STEP 3

Remove the two multiples of 3

The digit sum removes the multiples of three.

2+0+2+1+0+A = 5+A, and 5+1 = 6, 5+7 = 12 are the multiples of 3
4STEP 4

Separate the last two using 11

A test for eleven separates the last two.

3-0+1-2+0-2 = 0 → 202103 = 11 × 18373, while 9-0+1-2+0-2 = 6
5STEP 5

One candidate is left

What remains is 9.

A = 9, and 202109 is prime
Answer
9
Check that 202109 really is prime instead of leaning only on the problem's promise. If 202109 were composite it would factor as 202109 = m × n with 1 ＜ m ≤ n, which forces m ≤ √(202109). Since 449² = 201601 and 450² = 202500, that bound puts m below 450, so only the primes up to 449 need testing. Every one of them leaves a nonzero remainder, so no divisor exists and 202109 is prime. The four rejections also hold up when multiplied back out: 202101 = 3 × 67367, 202103 = 11 × 18373, 202105 = 5 × 40421, and 202107 = 3 × 67369. Exactly one candidate is prime, matching what the problem says.
💡Key takeaway

Proving a number is prime is slow, but proving one is not prime takes a single divisor, so on a multiple-choice question hunt for factors and let the last survivor win.

  • List the five candidate numbers
  • Remove the number ending in 5
  • Remove the two multiples of 3
  • Separate the last two using 11
  • One candidate is left

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