2017 AMC 8 Problem 17

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To help you master 2017 AMC 8 Problem 17, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

Key ideas, useful formulas, and takeaways you can apply to similar problems

Related problems from past AMC exams, plus new practice problems that build the same skills

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2017 AMC8 Problem 17

Solution

Solution 1

Variable-and-Equation Method

We can translate the two given scenarios into a system of algebraic equations. By letting one variable represent the number of chests and another represent the total number of coins, we can set the two expressions for the coins equal to each other.

When a word problem describes two different ways to group the same set of items, assign variables to the unknown quantities and write an algebraic equation for each scenario.

Step 1: Define the variables

Let $t$ represent the total number of treasure chests, and let $c$ represent the total number of gold coins.

Step 2: Translate the first scenario into an equation

If we put 9 coins in each chest, 2 chests are left empty. This means exactly $t - 2$ chests receive 9 coins, and there are no extra coins.

c = 9 (t - 2)

Step 3: Translate the second scenario into an equation

If we put 6 coins in each chest, every chest receives 6 coins, and there are 3 coins left over.

c = 6 t + 3

Step 4: Equate the two expressions and solve for $t$

Since both expressions represent the total number of coins $c$ , we can set them equal to each other. Expand the left side and solve for $t$ by grouping the terms.

9 (t - 2) = 6 t + 3

9 t - 18 = 6 t + 3

3 t = 21

t = 7

Step 5: Calculate the total number of coins

Substitute $t = 7$ back into either of our original equations to find $c$ . Let’s use the second equation.

c = 6 (7) + 3

c = 42 + 3

c = 45

Correct answer: C. $45$
The total number of gold coins is

45

, which corresponds to choice C. We can verify this using the first scenario:

9 \times (7 - 2) = 9 \times 5 = 45

, which matches perfectly.

Solution 2

Logical Redistribution Method

Instead of algebra, we can use logic by imagining we physically move the coins. We can start with the 6-coins-per-chest setup and move coins around to create the 9-coins-per-chest setup, keeping track of how many coins we hold in our hands.

The Difference Method (or Redistribution): When items are distributed in two different ways, you can mentally shift from one arrangement to the other by removing and reassigning the items, keeping track of the surplus or deficit.

Step 1: Start with the second scenario

Imagine we have placed 6 gold coins into every single chest, and we hold 3 leftover coins in our hand.

Step 2: Create the empty chests

The first scenario requires exactly 2 empty chests. To achieve this, we take all the coins out of 2 chests. Each of these chests had 6 coins, so we collect 12 coins.

2 \times 6 = 12

Step 3: Count the available coins to redistribute

We now have the 12 coins we just removed, plus the 3 original leftover coins. These are all in our hand, ready to be redistributed to the other chests.

12 + 3 = 15

Step 4: Determine how many coins each remaining chest needs

The first scenario requires the non-empty chests to have 9 coins. Since they currently have 6 coins, each of these remaining chests needs exactly 3 more coins.

9 - 6 = 3

Step 5: Distribute the coins and find the total

We have 15 coins to give out, and each chest needs 3 coins. We divide to find that 5 chests receive the upgraded amount. We then find the total number of coins by multiplying these 5 full chests by 9.

15 \div 3 = 5

5 \times 9 = 45

Correct answer: C. $45$
The total number of gold coins is

45

, matching choice C. We can quickly check the total number of chests: 5 full chests plus 2 empty chests equals 7 chests. If we put 6 coins in 7 chests, that uses 42 coins, leaving exactly 3 leftover from 45. Everything is consistent.

Solution 3

Multiple-Choice Shortcut — Testing the Answers

Since we have a list of possible total coin amounts, we can work backwards from the answer choices. The correct answer must yield a whole, identical number of treasure chests in both scenarios.

Back-Solving: In multiple-choice questions where the answer represents a total quantity, you can substitute the answer choices back into the problem conditions to see which one creates a logically consistent scenario without fractions.

Step 1: Express the number of chests using the first scenario

From the first scenario, if we know the total number of coins $c$ , the number of non-empty chests is $c \div 9$ , and we add the 2 empty chests to find the total chests $t$ .

t = \frac{c}{9} + 2

Step 2: Express the number of chests using the second scenario

From the second scenario, if we subtract the 3 leftover coins from $c$ , the remaining coins are evenly divided among the $t$ chests, with 6 coins per chest.

t = \frac{c - 3}{6}

Step 3: Test the middle answer choice

A highly effective multiple-choice strategy is to start by testing the middle value. Let’s test choice C, where the total number of coins is 45.

Step 4: Check if the two scenarios produce the same number of chests

Calculate the number of chests for each scenario assuming $c = 45$ . If they produce the same whole number of chests, we have found the correct answer.

t = \frac{45}{9} + 2 = 5 + 2 = 7

t = \frac{45 - 3}{6} = \frac{42}{6} = 7

Correct answer: C. $45$
Since both scenarios result in exactly 7 chests,

45

is the correct number of coins. The answer is

45

, which is choice C.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Variable Assignment for Two Scenarios:

When a problem provides two different conditions describing the same total amount, choose variables for the base components (e.g., number of chests) rather than just the final requested quantity.

Applies to distribution problems, seating arrangements, or any scenario where items are grouped differently. This keeps the equations simple and linear, as you can express the total quantity in two different ways and equate them.

The Redistribution (Difference) Logic:

Total surplus items gathered = (Number of groups emptied) × (Items per group). Number of groups upgraded = (Total surplus items) ÷ (Additional items needed per group).

Applies when changing from one uniform distribution to another uniform distribution. It tracks the conservation of items: any extra items given to upgraded groups must come from the groups that were emptied plus any initial leftovers.

Key Takeaway: Remember This Strategy for Next Time

A word problem describing two different ways to group or distribute the exact same items.

Try: Assign a variable to the number of groups (like treasure chests) and write two different expressions for the total number of items. Set the two expressions equal to each other.

Why: Since the total number of items doesn't change, setting the expressions equal eliminates one variable and allows you to quickly solve for the number of groups.

A problem involves rearranging items from an initial uniform distribution to a new one.

Try: Use the Redistribution Method (Difference Method): imagine physically moving the items. Calculate how many items you gain from emptying groups, and divide by how many extra items are needed for the remaining groups.

Why: This bypasses algebra entirely by directly balancing the supply of items you removed with the demand of the items you need to add.

A multiple-choice question where the options represent a total starting quantity.

Try: Try testing the answer choices, often starting with the middle value (Choice C), to see which one works.

Why: If the problem describes a logically strict scenario (like requiring a whole number of chests), incorrect options will quickly produce contradictions, such as fractional chests.

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