How to Solve Math Problems by Working Backwards

Some problems hand you the ending and ask for the beginning. You get the final amount, or the last term, or the state after every step, and the starting value is the missing piece. Guessing a start and pushing it forward is slow. The stronger move is to stand at the ending and rewind, undoing one step at a time until you reach the start. This last lesson of the course is about that rewind.

Problem

A number trick goes like this. Think of a number, double it, then add $9$. Priya follows the steps and lands on $41$. What number did she think of?

Show a hint

Undo the steps from the end. The last thing done was adding $9$, so the first undo is subtracting $9$.

Show the full solution

Undo the add $9$ first, $41 - 9 = 32$. Then undo the doubling, $32 \div 2 = \boxed{16}$.

Check it forward, $16$ doubled is $32$, and $32 + 9 = 41$.

Problem

Another trick has three steps. Subtract $4$, multiply by $5$, then add $7$. Marcus finishes on $52$. What number did he start with?

Show a hint

Peel the steps in reverse order. The add $7$ came last, so it is undone first, then the multiply, then the subtract.
Each undo swaps the operation for its opposite. Adding becomes subtracting, multiplying becomes dividing.

Show the full solution

Undo the add $7$, $52 - 7 = 45$. Undo the multiply by $5$, $45 \div 5 = 9$. Undo the subtract $4$, $9 + 4 = \boxed{13}$.

Forward, $13 - 4 = 9$, times $5$ is $45$, plus $7$ is $52$.

Problem

Theo bought a rare trading card. He later sold it to Uma, taking a loss of $5$ tokens. Uma then sold it to Vik for $32$ tokens, making a profit of $8$ tokens on her purchase. How many tokens did Theo originally pay?

Show a hint

Start at the end. Uma sold for $32$ with a profit of $8$, so what she paid comes from undoing that profit.
What Uma paid is exactly what Theo received. Undo Theo's loss of $5$ from there.

Show the full solution

Uma sold for $32$ at a profit of $8$, so she paid $32 - 8 = 24$. That $24$ is what Theo got for the card, and it left him down $5$, so he paid $24 + 5 = \boxed{29}$.

A seller with a profit paid less than the sale price, so undoing a profit subtracts and undoing a loss adds.

Problem

A rain barrel goes through the same routine each day. It loses half its water to the garden, and then the gardener pours $6$ liters back in. After two full days of this, the barrel holds $21$ liters. How many liters did it hold at the start?

Show a hint

Rewind day two first. Undo the $6$-liter pour, then undo the halving by doubling.
That gives the level after day one, and the same two undos again reach the start.

Show the full solution

Rewind day two. Before the $6$ liters went in the barrel held $21 - 6 = 15$, and before losing half it held $30$. So day one ended at $30$.

Rewind day one the same way, $30 - 6 = 24$, then double to get $\boxed{48}$.

Inside each day the pour happens last, so it is the first thing you undo.

Problem

A sourdough starter doubles in size every day, and it completely fills its jar on day $10$. On which day was the jar exactly one quarter full?

Show a hint

Run the doubling backward. One day earlier the jar was half full.
One quarter is half of half.

Show the full solution

Working back from day $10$, day $9$ held half a jar, and day $8$ held half of that, which is one quarter.

$$\boxed{8}$$

Day $10$ is the only amount you are given, so stepping backward from it beats hunting for the starting size.

Problem

In a certain list, every number is the sum of the two numbers just before it. The list ends with $\ldots, \; a, \; b, \; 19, \; 31, \; 50$. Find the value of $a$.

Show a hint

The rule runs forward as addition, so it rewinds as subtraction. The number before $19$ and $31$ satisfies $b + 19 = 31$.
Once $b$ is known, the same subtraction one step earlier finds $a$.

Show the full solution

Each number is the sum of the two before it, so $b + 19 = 31$, giving $b = 12$. One step earlier, $a + 12 = 19$, so $a = \boxed{7}$.

Forward, the list reads $7, 12, 19, 31, 50$, and each term is the sum of the two before it.

Problem

A bacteria colony in a dish triples every hour. At $5$ in the afternoon the dish holds $405$ colonies. How many colonies did it hold at $1$ in the afternoon?

Show a hint

Four hours separate the two clock times, so rewind four triplings.
Each rewind divides by $3$.

Show the full solution

Rewinding one hour divides by $3$, and four hours pass between $1$ and $5$.

$$405 \div 3 \div 3 \div 3 \div 3 = 405 \div 81 = \boxed{5}$$

Forward check, $5, 15, 45, 135, 405$, one tripling per hour.

Problem

Two juice pitchers sit on a counter. First, pitcher A pours into pitcher B exactly as much juice as B already contains, doubling B. Then pitcher B pours into pitcher A exactly as much as A contains at that moment, doubling A. Now each pitcher holds $36$ centiliters. How many centiliters did pitcher A hold at the start?

Show a hint

Undo the second pour first. It doubled A to $36$, so before it A held $18$, and the poured amount came out of B.
Then undo the first pour, which had doubled B. Half of B's amount at that moment had come from A.

Show the full solution

Rewind the second pour. It doubled A to $36$, so A held $18$ before it, and the $18$ that arrived came from B, which held $36 + 18 = 54$.

Rewind the first pour. It doubled B to $54$, so B held $27$, and that $27$ came from A, which started with $18 + 27 = \boxed{45}$.

Forward, $45$ and $27$ become $18$ and $54$, then $36$ and $36$.

Problem

On a game show, a contestant spends half of her tokens in round one, then pays a $7$-token entry fee for round two. In round two she spends half of what she has left, then pays a $3$-token exit fee. She walks away with $11$ tokens. How many tokens did she start with?

Show a hint

Rewind the four events in reverse. Undo the exit fee, undo the round-two halving, undo the entry fee, undo the round-one halving.
Undoing a fee means adding it back, and undoing a halving means doubling.

Show the full solution

Rewind from $11$. Add back the exit fee to get $14$, double to get $28$, add back the entry fee to get $35$, and double again to get $\boxed{70}$.

Forward, half of $70$ leaves $35$, the entry fee leaves $28$, half leaves $14$, and the exit fee leaves $11$.

Practice these ideas

Practice

A number is multiplied by $4$, then $9$ is subtracted, giving $39$. What was the number?

Show the solution

Adding back the $9$ gives $48$, and dividing by $4$ gives $\boxed{12}$. Forward, $12 \times 4 - 9 = 39$.

Practice

A number has $8$ added to it, the result is divided by $3$, and that result is multiplied by $5$, giving $45$. What was the original number?

Show the solution

From $45$, dividing by $5$ gives $9$, multiplying by $3$ gives $27$, and subtracting $8$ gives $\boxed{19}$. Forward, $(19+8) \div 3 \times 5 = 45$.

Practice

Rosa bought a bicycle and later sold it to Sam at a $12$-token profit. Sam then sold it for $87$ tokens, taking a $9$-token loss. How many tokens did Rosa pay for the bicycle?

Show the solution

Sam paid $87 + 9 = 96$. Rosa received that $96$ at a $12$-token profit, so she had paid $96 - 12 = \boxed{84}$.

Practice

A candy jar is visited by two siblings, one after the other. Each takes half the candies in the jar and then, feeling guilty, puts $2$ back. After both visits the jar holds $14$ candies. How many were in the jar before the first visit?

Show the solution

Rewind the second sibling, $14 - 2 = 12$, and doubling gives $24$. Rewind the first, $24 - 2 = 22$, and doubling gives $\boxed{44}$.

Forward, $44 \to 22 + 2 = 24 \to 12 + 2 = 14$.

Practice

A video's view count doubles every day. On Friday it reaches $96{,}000$ views. How many days earlier did it stand at $12{,}000$ views?

Show the solution

Backward, $96{,}000 \to 48{,}000 \to 24{,}000 \to 12{,}000$, three halvings, so $\boxed{3}$ days earlier.

Practice

In a list where every number is the sum of the two before it, the final three numbers are $13, 21, 34$. What number sits four places before the $34$?

Show the solution

Rewinding, before $13$ sits $21 - 13 = 8$, and before that sits $13 - 8 = 5$. The list runs $5, 8, 13, 21, 34$, and four places before the $34$ is $\boxed{5}$.

Practice

A stamp collection's value has tripled twice over the years and now stands at $486$ tokens. What was its value before either tripling?

Show the solution

Two rewinds give $486 \div 3 = 162$ and $162 \div 3 = \boxed{54}$.

Practice

A car enters a parking garage on some level, drives up $5$ levels, down $8$, and up $2$, ending on level $6$. On which level did it enter?

Show the solution

Undo the moves in reverse, $6 - 2 = 4$, then $4 + 8 = 12$, then $12 - 5 = \boxed{7}$.

Forward, $7 + 5 - 8 + 2 = 6$.

Practice

After giving away half of his stickers and then $4$ more, Leo has $26$ stickers left. How many did he start with?

Show the solution

Adding the $4$ back gives $30$, and doubling gives $\boxed{60}$. Forward, half of $60$ is $30$, minus $4$ is $26$.

Practice

Four integers form a chain in which each number after the first is twice the previous number minus $1$. The fourth number is $41$. What is the first?

Show the solution

Rewinding one link means adding $1$ and halving. From $41$, the third number is $(41+1) \div 2 = 21$, the second is $(21+1) \div 2 = 11$, and the first is $(11+1) \div 2 = \boxed{6}$.

Forward, $6 \to 11 \to 21 \to 41$.

Work Backwards

Practice these ideas