AMC 8 · 2007 · #18
Grade 5 number-theoryProblem
The product of the two 99-digit numbers
303,030,303,...,030,303 and 505,050,505,...,050,505
has thousands digit A and units digit B. What is the sum of A and B?
Pick an answer.
AMC 8 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: Multiply two $99$-digit numbers — $N_1 = 303{,}030{,}303{,}\dots{,}030{,}303$ and $N_2 = 505{,}050{,}505{,}\dots{,}050{,}505$. Call the thousands digit of the product $A$ and the units digit $B$. Find $A + B$.
Givens: $N_1$ has $99$ digits in the repeating pattern $3,0,3,0,3,\dots,0,3$ (last four digits are $0303$); $N_2$ has $99$ digits in the repeating pattern $5,0,5,0,5,\dots,0,5$ (last four digits are $0505$); $A$ = thousands digit of $N_1 \times N_2$; $B$ = units digit of $N_1 \times N_2$; Answer choices: (A) $3$, (B) $5$, (C) $6$, (D) $8$, (E) $10$
Unknowns: The sum $A + B$
Understand
Restated: Multiply two $99$-digit numbers — $N_1 = 303{,}030{,}303{,}\dots{,}030{,}303$ and $N_2 = 505{,}050{,}505{,}\dots{,}050{,}505$. Call the thousands digit of the product $A$ and the units digit $B$. Find $A + B$.
Givens: $N_1$ has $99$ digits in the repeating pattern $3,0,3,0,3,\dots,0,3$ (last four digits are $0303$); $N_2$ has $99$ digits in the repeating pattern $5,0,5,0,5,\dots,0,5$ (last four digits are $0505$); $A$ = thousands digit of $N_1 \times N_2$; $B$ = units digit of $N_1 \times N_2$; Answer choices: (A) $3$, (B) $5$, (C) $6$, (D) $8$, (E) $10$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #9 Solve an Easier Related Problem
Tool #16 (Change Focus) is the key move: instead of computing the whole $198$-digit product, focus only on what the question actually needs — the last four digits. Standard column multiplication shows that the last four digits of a product depend only on the last four digits of each factor; everything to the left feeds into higher places and never comes back down. Tool #9 (Easier Problem) confirms the shortcut by trying a baby version (much shorter $303\dots$ and $505\dots$ numbers) and checking that the last four digits of the product are unchanged.
Execute — Answer: D
5.NBT.A.1 Step 1 - Switch focus from the whole product to just its last four digits.
- The thousands and units digits both live in the last four places, so any other digits of the product are irrelevant.
- In long multiplication, each column of the product is built from columns at or to the right of the same place in the factors — digits far to the left can never affect the ones, tens, hundreds, or thousands column.
💡 Place-value thinking from Grade 5: a digit in the ten-thousands place or higher cannot land in the ones, tens, hundreds, or thousands column of the answer.
5.NBT.A.1 Step 2 - Read the last four digits off each factor.
- $N_1$ uses the pattern $3,0,3,0,3,\dots,0,3$ (it starts and ends with $3$), so its last four digits are $0303$.
- $N_2$ uses the pattern $5,0,5,0,5,\dots,0,5$, so its last four digits are $0505$.
💡 Reading the rightmost four digits of a multi-digit number is exactly Grade 5 place-value identification.
5.NBT.B.5 Step 3 Multiply the two small numbers using the partial-products method.
💡 Replacing two $99$-digit numbers with two $3$-digit numbers is the Easier Problem move. The product $303 \times 505$ is a clean Grade 5 multi-digit multiplication.
5.NBT.A.1 Step 4 - Take the last four digits of $153015$ to recover the last four digits of the full product.
- Then read off the thousands digit ($A$) and units digit ($B$) and add them.
💡 Picking out the thousands digit and the units digit from a written numeral is the Grade 5 place-value definition in action.
5.NBT.A.1 Switch focus from the whole product to just its last four digits. The thousands 5.NBT.A.1 Read the last four digits off each factor. $N_1$ uses the pattern $3,0,3,0,3,\do 5.NBT.B.5 Multiply the two small numbers using the partial-products method. 5.NBT.A.1 Take the last four digits of $153015$ to recover the last four digits of the ful Review
Reasonableness: Sanity check with shorter versions: take $N_1' = 30303$ and $N_2' = 50505$ (the same repeating patterns, but only $5$ digits each). Then $N_1' \times N_2' = 30303 \times 50505 = 1{,}530{,}458{,}415$, whose last four digits are $8415$ — wait, that does not match. Recompute carefully: $30303 \times 50505 = 30303 \times 50000 + 30303 \times 505 = 1{,}515{,}150{,}000 + 15{,}303{,}015 = 1{,}530{,}453{,}015$. The last four digits are $3015$, exactly the same $3015$ we obtained from $303 \times 505$. So the thousands digit is $3$, the units digit is $5$, and $A + B = 8$, matching answer (D). The reason the longer factors give the same last four digits is precisely the place-value shortcut from step 1.
Alternative: Tool #9 (Easier Problem) on its own: solve the same problem with much shorter versions $30303 \times 50505$ (or even $303 \times 505$). The product ends in $\dots 3015$ in every case, because the digits beyond position four in either factor only push value into the ten-thousands column and higher. So the thousands digit is $3$, the units digit is $5$, and $A + B = 3 + 5 = 8$ — answer (D).
CCSS standards used (min grade 5)
5.NBT.A.1Understand the place value system (Recognizing that the thousands and units digits of a product depend only on the last four digits of each factor, and reading $A$ (thousands) and $B$ (units) off the numeral $3015$.)5.NBT.B.5Fluently multiply multi-digit whole numbers using the standard algorithm (Computing $303 \times 505 = 153015$ via partial products to obtain the last four digits of the full product.)
⭐ When you only need the last few digits of a giant product, throw away every digit to the left — Grade 5 place value reduces this AMC 8 problem to a single tidy multiplication.
⭐ When you only need the last few digits of a giant product, throw away every digit to the left — Grade 5 place value reduces this AMC 8 problem to a single tidy multiplication.
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