AMC 10 · 2004 · #17
Grade 6 arithmeticThe two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?
Pick an answer.
AMC 10 2004 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: Jack's age and Bill's age are both two-digit numbers built from the same two digits, just written in reverse order (so if one is $10x+y$, the other is $10y+x$). Five years from now, Jack's age will be exactly twice Bill's age at that time. Find the difference between their ages right now.
Givens: Jack's age and Bill's age use the same two digits, with the tens and ones digits swapped.; In five years, Jack's age will be exactly twice Bill's age then.; Answer choices: (A) $9$, (B) $18$, (C) $27$, (D) $36$, (E) $45$.
Unknowns: The difference between Jack's and Bill's current ages.
Understand
Restated: Jack's age and Bill's age are both two-digit numbers built from the same two digits, just written in reverse order (so if one is $10x+y$, the other is $10y+x$). Five years from now, Jack's age will be exactly twice Bill's age at that time. Find the difference between their ages right now.
Givens: Jack's age and Bill's age use the same two digits, with the tens and ones digits swapped.; In five years, Jack's age will be exactly twice Bill's age then.; Answer choices: (A) $9$, (B) $18$, (C) $27$, (D) $36$, (E) $45$.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #6 Guess and Check, #3 Eliminate Possibilities
The two ages are locked together by their digits, so Tool #4 (Introduce a Variable) is the natural start: call the tens digit of one age $a$ and the ones digit $b$, and the reversal writes itself as $10a+b$ and $10b+a$. The 'twice as old in five years' sentence then becomes a single equation. That equation, $8a=19b+5$, has two digit-unknowns but only one equation — so instead of grinding algebra, Tool #6 (Guess and Check) walks $b$ through $0,1,2,\dots$ and Tool #3 (Eliminate Possibilities) throws out every value that fails to leave $a$ a single digit. Only one pair survives.
Execute — Answer: B
6.EE.B.6 Step 1 Name the two digits
- Let $a$ be the tens digit and $b$ the ones digit of Jack's age, so Jack is $10a+b$ years old.
- Bill has the same digits reversed, so Bill is $10b+a$ years old.
- Because Jack is the older one, $a$ is bigger than $b$.
💡 A two-digit number is just its tens digit times ten plus its ones digit, so swapping the digits swaps those two roles.
6.EE.B.7 Step 2 Turn the words into an equation
- In five years each person is five years older, so Jack will be $10a+b+5$ and Bill will be $10b+a+5$.
- 'Jack will be twice as old as Bill' means Jack's future age equals two times Bill's future age.
💡 'Twice as old' is a multiply-by-two link between the two future ages, which is exactly what an equation records.
6.EE.B.7 Step 3 Simplify to one clean relation
- Expand the right side to $20b+2a+10$, then gather the digits.
- Subtract $2a$ and $b$ from both sides and subtract $5$: $8a-19b=5$, or $8a=19b+5$.
- This single relation ties the two digits together.
💡 Collecting like terms boils the whole story down to one tidy link between $a$ and $b$.
4.OA.A.3 Step 4 Test digit values for b
- Both $a$ and $b$ must be single digits, so try $b=0,1,2,\dots$ and see when $19b+5$ is $8$ times a digit.
- $b=0$ gives $8a=5$ (no).
- $b=1$ gives $8a=24$, so $a=3$ — a real digit.
- $b=2$ gives $8a=43$ (not a multiple of $8$), and from $b=3$ onward $19b+5$ already passes $8\times 9=72$, forcing $a$ above $9$.
- So $a=3,\,b=1$ is the only fit: Jack is $31$ and Bill is $13$.
💡 Since digits only run $0$ through $9$, a short march through the choices quickly leaves just one that keeps both digits legal.
5.NBT.A.1 Step 5 Take the difference
- The difference in their current ages is $31-13=18$.
- This also matches the shortcut that a two-digit number minus its reverse is $(10a+b)-(10b+a)=9(a-b)=9(3-1)=18$.
- The answer is (B).
💡 Reversing the digits trades tens for ones, so the gap is always nine times the digit difference.
6.EE.B.6 Let $a$ be the tens digit and $b$ the ones digit of Jack's age, so Jack is $10a+ 6.EE.B.7 In five years each person is five years older, so Jack will be $10a+b+5$ and Bil 6.EE.B.7 Expand the right side to $20b+2a+10$, then gather the digits. Subtract $2a$ and 4.OA.A.3 Both $a$ and $b$ must be single digits, so try $b=0,1,2,\dots$ and see when $19b 5.NBT.A.1 The difference in their current ages is $31-13=18$. This also matches the shortc Review
Reasonableness: Check the ages against the story. In five years Jack is $31+5=36$ and Bill is $13+5=18$, and $36=2\times 18$, exactly 'twice as old.' The two ages, $31$ and $13$, really are the same digits reversed, and their difference $18$ sits right on choice (B). It even makes sense that every answer choice is a multiple of $9$, since any reversed-digit pair differs by $9(a-b)$ — the test is which multiple actually satisfies the age condition.
Alternative: Use the fact that an age difference never changes over time. Five years from now Jack is twice Bill, so the difference (Jack minus Bill) equals Bill's future age itself: $\text{difference}=2\,\text{Bill}_{+5}-\text{Bill}_{+5}=\text{Bill}_{+5}$. The difference is also $9(a-b)$, a multiple of $9$, so Bill's age in five years is a multiple of $9$. Testing the reversed pairs where the tens digit is smaller ($13,24,35,\dots$), only Bill $=13$ gives a future age $18$ that is a multiple of $9$ and doubles to Jack's $36$. Difference $=18$ again.
CCSS standards used (min grade 6)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Letting the digits be $a$ and $b$ and writing the two ages as $10a+b$ and $10b+a$.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Turning 'twice as old in five years' into $10a+b+5=2(10b+a+5)$ and simplifying it to $8a=19b+5$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Testing digit values $b=0,1,2,\dots$ to find the only pair $a=3,\,b=1$ that keeps both digits single.)5.NBT.A.1Recognize that a digit in one place represents ten times as much as to its right (Seeing that a two-digit number minus its reverse equals $9(a-b)$, giving the difference $18$.)
⭐ Write the two ages as $10a+b$ and $10b+a$, turn 'twice as old in five years' into one equation, and testing single digits leaves only $31$ and $13$ — a difference of $18$.
⭐ Write the two ages as $10a+b$ and $10b+a$, turn 'twice as old in five years' into one equation, and testing single digits leaves only $31$ and $13$ — a difference of $18$.
More like this
Same archetype — closest grade level first.