AMC 10 · 2011 · #5
Grade 4 arithmeticIn multiplying two positive integers a and b, Ron reversed the digits of the two-digit number a. His erroneous product was 161. What is the correct value of the product of a and b?
Pick an answer.
AMC 10 2011 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: Ron multiplied a two-digit number $a$ by another positive integer $b$, but he flipped the two digits of $a$ before multiplying. That wrong multiplication gave $161$. Using the correct value of $a$, find the true product $a\times b$.
Givens: $a$ is a two-digit positive integer and $b$ is a positive integer; Ron used the digit-reversal of $a$ instead of $a$ itself; His wrong product (reversed $a$)$\times b$ equals $161$; Answer choices: (A) $116$, (B) $161$, (C) $204$, (D) $214$, (E) $224$
Unknowns: The correct product $a\times b$
Understand
Restated: Ron multiplied a two-digit number $a$ by another positive integer $b$, but he flipped the two digits of $a$ before multiplying. That wrong multiplication gave $161$. Using the correct value of $a$, find the true product $a\times b$.
Givens: $a$ is a two-digit positive integer and $b$ is a positive integer; Ron used the digit-reversal of $a$ instead of $a$ itself; His wrong product (reversed $a$)$\times b$ equals $161$; Answer choices: (A) $116$, (B) $161$, (C) $204$, (D) $214$, (E) $224$
Plan
Primary tool: #11 Work Backwards
Secondary: #3 Eliminate Possibilities, #7 Identify Subproblems
The wrong product $161$ is the end of Ron's mistake, and we need the numbers he started from, so Tool #11 (Work Backwards) fits: factor $161$ to recover the two numbers he actually multiplied. Tool #3 (Eliminate Possibilities) then picks which factor is the reversed two-digit number — only one factor has two digits. Tool #7 (Identify Subproblems) breaks the rest into small steps: un-flip the digits to get $a$, then do the correct multiplication.
Execute — Answer: E
4.OA.B.4 Step 1 Factor the wrong product
- Ron's wrong product is (reversed $a$)$\times b = 161$, and both of those are whole numbers.
- So they form a factor pair of $161$.
- Break $161$ into primes: it is not even, not a multiple of $3$ or $5$, but $161 = 7\times 23$.
- Since $7$ and $23$ are both prime, the only ways to write $161$ as a product of two whole numbers are $1\times 161$ and $7\times 23$.
💡 Splitting the wrong answer into factors reveals the two numbers Ron actually multiplied.
1.NBT.B.2 Step 2 Pick the reversed two-digit factor
- Flipping the digits of a two-digit number still leaves a two-digit number, so the reversed $a$ must have two digits.
- In the pair $7\times 23$, only $23$ has two digits, so the reversed $a$ is $23$ and $b$ is $7$.
- The other split $1\times 161$ is impossible because neither $1$ nor $161$ is a two-digit reversal.
💡 A two-digit number can only be a two-digit factor, which eliminates every other choice.
1.NBT.B.2 Step 3 Un-flip the digits to get $a$
- Reversed $a$ is $23$, which is $2$ tens and $3$ ones.
- Ron flipped $a$ to get this, so flipping it back recovers the true $a$: the tens and ones digits swap, giving $3$ tens and $2$ ones, which is $32$.
- So $a = 32$ and $b = 7$.
💡 Reversing the reversal returns the original number.
4.NBT.B.5 Step 4 Multiply correctly
- Now use the real numbers.
- The correct product is $a\times b = 32\times 7$.
- Multiply: $30\times 7 = 210$ and $2\times 7 = 14$, so $32\times 7 = 210 + 14 = 224$.
- That matches choice (E).
- Note $161$ itself is choice (B), placed to catch anyone who stops at the wrong product.
💡 Once the true digits are back in place, the correct product is a plain multiplication.
4.OA.B.4 Ron's wrong product is (reversed $a$)$\times b = 161$, and both of those are who 1.NBT.B.2 Flipping the digits of a two-digit number still leaves a two-digit number, so th 1.NBT.B.2 Reversed $a$ is $23$, which is $2$ tens and $3$ ones. Ron flipped $a$ to get thi 4.NBT.B.5 Now use the real numbers. The correct product is $a\times b = 32\times 7$. Multi Review
Reasonableness: Check the story both ways. Reversed $a = 23$ times $b = 7$ gives $23\times 7 = 161$, exactly Ron's wrong product, so the factoring is right. The true $a = 32$ times $b = 7$ gives $224$, which is close to but a bit larger than $161$ — sensible, since $32 > 23$. The answer $224$ is choice (E), and the tempting $161$ (choice B) is only the mistaken product, not the correct one.
Alternative: Skip primes and just scan two-digit reversals whose reversal divides $161$. Testing $b = 7$ (the smallest prime factor of $161$) gives reversed $a = 161\div 7 = 23$, whose un-reversal $32$ times $7$ is $224$. You can also test each answer choice: only $224 = 32\times 7$ has the digit-reversal $23\times 7 = 161$, confirming (E).
CCSS standards used (min grade 4)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Factoring $161 = 7\times 23$ to find the whole numbers Ron multiplied and seeing that $7$ and $23$ are prime.)1.NBT.B.2Understand that the two digits of a two-digit number represent tens and ones (Recognizing that a two-digit number reverses to a two-digit number, and swapping the tens and ones of $23$ to recover $a = 32$.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Computing the correct product $32\times 7 = 224$.)
⭐ Break the wrong answer into its factors, flip the two-digit one back to the real number, then multiply again to get the true product.
⭐ Break the wrong answer into its factors, flip the two-digit one back to the real number, then multiply again to get the true product.
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