← Sensim Lab Sensim Math

AMC 8 · 2007 · #18

Easy mode Grade 5
📗 View original problem →

Problem

Take two very long numbers, each with 9999 digits:

303,030,303,…,030,303303,030,303,\ldots,030,303 and 505,050,505,…,050,505505,050,505,\ldots,050,505

The first number is made by repeating the pattern "0303" all the way down. The second is made by repeating "0505." Both end with the digit 33 and the digit 55, 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 BB. Its thousands digit (the 44th digit from the right) is called AA.

What is A+BA + B?

(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
View mode:

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.

More like this

Same archetype — closest grade level first.