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AMC 8 · 2007 · #18

Grade 5 number-theory
Archetype Lattice / Grid Pattern from Small Cases
multi-digit-arithmeticplace-valuepattern-recognitionunits-digit-tracking pattern-recognitionidentify-subproblemseasier-related-problem ↑ Prerequisites: multi-digit-arithmeticplace-value
📏 Short solution 💡 2 insights
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Problem

The product of the two 9999-digit numbers

303,030,303,...,030,303303,030,303,...,030,303 and 505,050,505,...,050,505505,050,505,...,050,505

has thousands digit AA and units digit BB. What is the sum of AA and BB?

(A) 3(B) 5(C) 6(D) 8(E) 10\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10

Pick an answer.

(A)
3
(B)
5
(C)
6
(D)
8
(E)
10
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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

#16 Change Focus / Count the Complement 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.
$$\text{last 4 digits of } N_1 \times N_2 = \text{last 4 digits of } (\text{last 4 digits of } N_1) \times (\text{last 4 digits of } N_2)$$

💡 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.

#16 Change Focus / Count the Complement 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 $0505$.
$$\text{last 4 of } N_1 = 0303 = 303,\quad \text{last 4 of } N_2 = 0505 = 505$$

💡 Reading the rightmost four digits of a multi-digit number is exactly Grade 5 place-value identification.

#9 Solve an Easier Related Problem 5.NBT.B.5 Step 3

Multiply the two small numbers using the partial-products method.

$$303 \times 505 = 300 \times 500 + 300 \times 5 + 3 \times 500 + 3 \times 5 = 150000 + 1500 + 1500 + 15 = 153015$$

💡 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.

#16 Change Focus / Count the Complement 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.
$$\text{last 4 of } 153015 = 3015 \;\Rightarrow\; A = 3,\; B = 5 \;\Rightarrow\; A + B = 8 \;\Rightarrow\; \textbf{(D)}$$

💡 Picking out the thousands digit and the units digit from a written numeral is the Grade 5 place-value definition in action.

[1] #16 5.NBT.A.1 Switch focus from the whole product to just its last four digits. The thousands
[2] #16 5.NBT.A.1 Read the last four digits off each factor. $N_1$ uses the pattern $3,0,3,0,3,\do
[3] #9 5.NBT.B.5 Multiply the two small numbers using the partial-products method.
[4] #16 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.1 Understand 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.5 Fluently 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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