AMC 8 · 2007 · #18
Easy mode Grade 5Problem
Take two very long numbers, each with 99 digits:
303,030,303,…,030,303 and 505,050,505,…,050,505
The first number is made by repeating the pattern "03" all the way down. The second is made by repeating "05." Both end with the digit 3 and the digit 5, respectively.
Now multiply these two huge numbers together. Call the answer the product.
Look at the product. Its units digit (the very last digit on the right) is called B. Its thousands digit (the 4th digit from the right) is called A.
What is A+B?
Diagram
Both numbers repeat a 2-digit block; only the rightmost 4-digit window decides 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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