AMC 10 · 2007 · #24
Grade 6 number-theoryLet n denote the smallest positive integer that is divisible by both 4 and 9, and whose base-10 representation consists of only 4's and 9's, with at least one of each. What are the last four digits of n?
Pick an answer.
AMC 10 2007 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: Build the smallest positive whole number whose digits are only 4's and 9's (using at least one 4 and at least one 9) that is a multiple of both 4 and 9, then report its last four digits.
Givens: Every digit of the number is either 4 or 9.; The number contains at least one 4 and at least one 9.; The number is divisible by 4 and by 9.
Unknowns: The smallest such number.; Its last four digits, matched to one answer choice.
Understand
Restated: Build the smallest positive whole number whose digits are only 4's and 9's (using at least one 4 and at least one 9) that is a multiple of both 4 and 9, then report its last four digits.
Givens: Every digit of the number is either 4 or 9.; The number contains at least one 4 and at least one 9.; The number is divisible by 4 and by 9.
Plan
Primary tool: #14 Extreme Principle
Secondary: #3 Eliminate Possibilities, #4 Introduce a Variable
The two divisibility rules act as filters that lock down the number's shape, and the word 'smallest' is a minimize signal, so the plan is: use the divisibility-by-4 rule to eliminate every ending except one, name the counts of each digit with a variable to force the divisibility-by-9 rule, then push toward the fewest digits and the smallest arrangement.
Execute — Answer: C
4.OA.B.4 Step 1 Divisibility by 4 fixes the ending
- A number is divisible by 4 exactly when its last two digits form a multiple of 4.
- With only 4's and 9's, the possible endings are 44, 49, 94, and 99.
- Checking each: 44 = 4 x 11 works, while 49, 94, and 99 are not multiples of 4.
- So the number must end in 44.
💡 Only the last two digits decide divisibility by 4, so I just test the four possible endings.
6.NS.B.4 Step 2 Divisibility by 9 forces nine 4's
- A number is divisible by 9 when its digit sum is a multiple of 9.
- Let the number have a copies of the digit 4 and b copies of the digit 9.
- The digit sum is 4a + 9b.
- Since 9b is already a multiple of 9, the whole sum is a multiple of 9 only when 4a is.
- Because 4 and 9 share no common factor, 4a is a multiple of 9 only when a itself is a multiple of 9.
- So the count of 4's must be 9, 18, 27, and so on.
💡 The 9's contribute nothing to the leftover, so the 4's alone must reach a multiple of 9, and it takes nine of them.
4.NBT.A.2 Step 3 Fewest digits: nine 4's and one 9
- To make the number as small as possible, first make it have as few digits as possible.
- The count of 4's must be at least 9, and there must be at least one 9, so the shortest number has exactly nine 4's and one 9 — ten digits in all.
- Any valid number with fewer digits is impossible.
💡 A number with fewer digits is always smaller, so start by minimizing the digit count.
4.NBT.A.2 Step 4 Arrange for the smallest value
- For ten digits made of nine 4's and one 9, the smallest value keeps the big digit as far right as possible while still ending in 44 (needed for divisibility by 4).
- Put 4's in front and slide the single 9 to the last spot that is not one of the final two 4's.
- That gives 4444444944.
- Its last four digits are 4944, which is choice (C).
💡 A larger digit hurts less when it sits in a lower place value, so bury the lone 9 as far right as the rules allow.
4.OA.B.4 A number is divisible by 4 exactly when its last two digits form a multiple of 4 6.NS.B.4 A number is divisible by 9 when its digit sum is a multiple of 9. Let the number 4.NBT.A.2 To make the number as small as possible, first make it have as few digits as pos 4.NBT.A.2 For ten digits made of nine 4's and one 9, the smallest value keeps the big digi Review
Reasonableness: Check the winner 4444444944 directly: the last two digits are 44 = 4 x 11, so it is divisible by 4; the digit sum is nine 4's plus one 9 = 36 + 9 = 45 = 9 x 5, so it is divisible by 9; it uses both digits and has ten digits, the fewest possible. Its last four digits 4944 match (C). Choice (B) 4494 ends in 94, which is not divisible by 4, so it could never be the ending — a quick sanity check that the divisibility filter is doing real work.
Alternative: Instead of building the number, screen the answer choices. Only endings that are multiples of 4 survive, which rules out (B) 4494 immediately. Among the rest, reason that the smallest number wants small leading digits, so its final block should be 4944 rather than 9444 or 9944, and 4444 would require the single required 9 to sit even further left, which forces a longer number — leaving (C).
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Testing divisibility by 4 by checking whether the last two digits form a multiple of 4.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Recognizing that because 4 and 9 have no common factor, 4a is a multiple of 9 only when a is a multiple of 9.)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Comparing candidate numbers to find the one with the fewest digits and the smallest value.)
⭐ Let the divisibility rules pin down the shape of the number, then make it small by using as few digits as you can and hiding any big digit on the right.
⭐ Let the divisibility rules pin down the shape of the number, then make it small by using as few digits as you can and hiding any big digit on the right.
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