Competition · AMC preparation · step 4 of 4
AMC 8 · 2017 · #8
Grade 4 number-theorylogicPick an answer.
AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
We are told exactly one of four statements is false. The cleanest way in is Tool #3 (Eliminate Possibilities): test each statement as the candidate "false" one and rule out the cases that lead to a contradiction. Statement (1) "prime" clashes with (2) "even" (only the prime 2 is even, but 2 has one digit) and with (3) "divisible by 7" (only the prime 7 is divisible by 7, but 7 has one digit). So (1) has to be the false statement, forcing (2), (3), (4) to be true. From there Tool #2 (Systematic List) takes over: list every two-digit multiple of lcm(2,7) = 14 in order and keep only the one whose digits include a 9.
Find the first contradiction
A two-digit number that is both prime and even would have to be 2, a single digit — so (1) and (2) can't both be true.
Knowing that 2 is the only even prime is exactly the Grade 4 "prime or composite / factor pairs" skill.
4.OA.B.4Eliminate PossibilitiesFind the second contradiction
Likewise a prime that is also a multiple of 7 must be 7 itself, one digit — so (1) and (3) can't both be true.
Multiples of 7 that are also prime is again the Grade 4 "recognize multiples / determine prime" idea.
4.OA.B.4Eliminate PossibilitiesPin down the false statement
If (1) were true, (2) and (3) would both fail — two falses, not allowed — so (1) is the false statement and (2), (3), (4) are true.
Eliminating "(1) is true" leaves "(1) is false" — pure process of elimination on a finite set of cases.
The single false statement has to be statement (1), "it is prime"; so statements (2), (3), and (4) are all true.
▸ Why?
Only one of the four statements is allowed to be false. If statement (1) 'prime' were the true one, it would force two of the other statements — 'even' and 'divisible by 7' — to be false at the same time, which is one false too many. So statement (1) itself must be the false one, leaving (2), (3), and (4) true.
▸ Why?
A two-digit number cannot be both prime and even, so a true 'prime' would make 'even' false. The only even prime is 2: a prime's whole prime factorization is just itself, so if 2 divided it, 2 would have to be that single prime factor, forcing the number to equal 2 — a one-digit number.
▸ Why?
A two-digit number cannot be both prime and divisible by 7, so a true 'prime' would also make 'divisible by 7' false. The only prime divisible by 7 is 7: a prime's whole prime factorization is just itself, so if 7 divided it, 7 would have to be that single prime factor, forcing the number to equal 7 — a one-digit number.
List the multiples of 14
Even and divisible by 7 means a multiple of lcm(2,7) = 14, so list the two-digit multiples: 14, 28, 42, 56, 70, 84, 98.
Listing the multiples of 14 within 10-99 is a Grade 4 "recognize multiples" exercise.
4.OA.B.4Make A Systematic ListUse the digit 9 clue
Only 98 in that list has a digit 9, so the house number is 98 and its units digit is 8 — choice (D).
Identifying the units (ones) digit of a two-digit number is the Grade 1 place-value standard.
1.NBT.B.2Make A Systematic ListThis AMC 8 problem only needs Grade 4 ideas about primes and multiples you already know!
- Find the first contradiction
- Find the second contradiction
- Pin down the false statement
- List the multiples of 14
- Use the digit 9 clue
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