AMC 10 · 2003 · #25
Grade 4 number-theoryPick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Only two digits are free — the thousands digit and the hundreds digit — so name them a and b (Tool #4). The divisibility rule for 3 says a number is a multiple of 3 exactly when its digits add to a multiple of 3. That turns the whole problem into one condition on a+b. To count the digits b that fit, look at the repeating remainder pattern of 0,1,…,9 when divided by 3 (Tool #5): the pattern is what makes the count exact instead of guesswork. Because the needed remainder for b depends on a, split the nine values of a into three cases by their own remainder (Tool #7) and count each case with a short list (Tool #2). Add the three case-counts for the total.
Turn the number into a digit-sum condition
Write the number as ab23. It is a multiple of 3 exactly when its digit sum a + b + 2 + 3 = a + b + 5 is.
For divisibility by 3 only the digit sum matters, so the fixed 2 and 3 just add a constant 5.
For a test by three only the digit sum matters, so the fixed digits just add a constant.
▸ Why?
Every place value is one more than a multiple of three, so only the digit sum survives the division.
▸ Why?
A number is its digits weighted by their places, so peeling those weights apart is always legitimate.
Find the remainder the free digits must give
Since 5 leaves remainder 2, the free digits must supply the missing 1: a + b leaves remainder 1 when divided by 3.
The constant 5 already contributes a remainder of 2, so the two free digits must supply the missing remainder of 1.
4.NBT.B.6Introduce A VariableCount the b-digits for each remainder
Sort b from 0 to 9 by remainder: remainder 0 has {0,3,6,9}, so 4 choices, while remainders 1 and 2 have 3 choices each.
Ten digits split 4-3-3 by remainder because 0 through 9 isn't a whole number of full groups of 3.
4.OA.C.5Look For A PatternSplit the a-digits into three cases
Group a from 1 to 9 the same way: a≡1 forces b≡0, giving 3 × 4 = 12; a≡2 and a≡0 each give 3 × 3 = 9.
Each remainder-group of a pins down exactly which remainder b needs, so each case is a clean multiply.
4.OA.B.4Identify SubproblemsAdd the three cases
Each a lands in exactly one group, so the three cases never overlap and simply add: 12 + 9 + 9 = 30, choice (B).
The three remainder cases don't overlap, so their counts simply add.
4.OA.A.3Identify SubproblemsDivisibility by 3 only cares about the digit sum, so freeze the fixed digits and count how the two free digits can land in the right remainder group.
- Turn the number into a digit-sum condition
- Find the remainder the free digits must give
- Count the b-digits for each remainder
- Split the a-digits into three cases
- Add the three cases