2017 AMC 8 Problem 24
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To help you master 2017 AMC 8 Problem 24, 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 24
24.
Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December , . On how many days during the next year did she not receive a phone call from any of her grandchildren?
Solution
Complement Counting with Inclusion-Exclusion
Instead of directly counting the days with no calls, it is easier to count the days when Mrs. Sanders DOES receive at least one call, and then subtract that from the 365 days in the year. We use the Inclusion-Exclusion Principle to ensure we do not double-count days when multiple grandchildren call.
The Inclusion-Exclusion Principle for 3 sets helps find the size of their union: add the sizes of the individual sets, subtract the sizes of the two-set overlaps, and add back the size of the three-set overlap.
Step 1: Determine the number of days in the year
The calls synchronized on December 31, 2016. The ‘next year’ is 2017. Because 2017 is not divisible by 4, it is a regular year with 365 days. We number these days from 1 to 365. Since Day 0 was December 31, calls will happen on days that are exact multiples of 3, 4, and 5.
Step 2: Count the days each grandchild calls individually
To find how many multiples of 3, 4, and 5 exist up to 365, we divide 365 by each interval and take the integer part (rounding down). Let A, B, and C be the sets of days each respective grandchild calls.
Step 3: Count the overlapping calls between pairs of grandchildren
If two grandchildren call on the same day, that day must be a multiple of both their calling intervals. We find the least common multiple (LCM) for each pair and count the multiples up to 365. The LCMs of the pairs are 12, 20, and 15.
Step 4: Count the days when all three grandchildren call
All three grandchildren call on days that are multiples of 3, 4, and 5. The least common multiple of all three numbers is 60. We count the multiples of 60 up to 365.
Step 5: Apply the Inclusion-Exclusion Principle
We combine our counts to find the total number of days with at least one call. We add the individual calls, subtract the two-person overlaps, and add back the three-person overlaps.
Step 6: Find the number of days with no calls
Subtract the days with at least one call from the total days in the year to find the days with no calls.
The answer is 146, which matches choice D. We can check our work with a quick magnitude estimation: since 3, 4, and 5 share no common factors, roughly (2/3) * (3/4) * (4/5) = 40% of the days should have no calls. 40% of 365 is exactly 146.
Periodic Grouping Method
The calling schedule repeats exactly every 60 days (the least common multiple of 3, 4, and 5). We can find the proportion of no-call days in one 60-day block, multiply it by the number of full blocks in a year, and manually check the few leftover days at the end.
Because 3, 4, and 5 share no common factors, divisibility by each is an independent event. The fraction of days not divisible by any of them is the product of the individual non-divisible fractions.
Step 1: Calculate the proportion of days with no calls
Because 3, 4, and 5 are pairwise relatively prime (they share no common factors greater than 1), we can find the fraction of days that are NOT multiples of 3, 4, or 5 by multiplying the independent fractions of days that avoid each interval.
Step 2: Find the number of no-call days in one full cycle
The calling pattern repeats every 60 days, which is the LCM of 3, 4, and 5. We apply the proportion we just found to a 60-day block.
Step 3: Break the year into 60-day blocks
We divide the 365 days of the year by 60 to find how many full blocks fit, and how many leftover days remain at the end of the year.
Step 4: Calculate the no-call days from the full blocks
Multiply the 6 full cycles by the 24 no-call days in each cycle.
Step 5: Check the leftover days manually
The remaining 5 days of the year follow the same pattern as the first 5 days of the cycle (Days 1, 2, 3, 4, and 5). We check each day to see if it is a multiple of 3, 4, or 5: – Day 1: Not a multiple (No call) – Day 2: Not a multiple (No call) – Day 3: Multiple of 3 (Call) – Day 4: Multiple of 4 (Call) – Day 5: Multiple of 5 (Call) There are 2 no-call days in this leftover period.
Step 6: Combine the totals
Add the no-call days from the full blocks to the no-call days from the leftover days.
The answer is 146, which is choice D. Notice that in the leftover 5 days, exactly 2 days had no calls. The ratio 2/5 perfectly matches the overall ratio of the full blocks, which elegantly explains why exactly 2/5 of the entire 365-day year (365 * 2/5 = 146) had no calls.
Similar Problems in Real Exams
The following five AMC 8 problems are closely related to AMC 8 2017 Question 24. These real AMC 8 problems test the same key concept—GCD, LCM & Common-Factor Structure. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.
2017 AMC 8 Problem 12
2016 AMC 8 Problem 20
2013 AMC 8 Problem 10
2013 AMC 8 Problem 16
2012 AMC 8 Problem 15
Adapted Problems to Practice
What Formulas Do You Need for This Problem?
- Inclusion-Exclusion Principle (3 Sets):
|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |C \cap A| + |A \cap B \cap C|
A, B, and C are sets. |A| represents the number of items in set A. Overlaps (intersections) are subtracted to correct for double counting, and the three-set intersection is added back to correct for triple subtraction. Since grandchildren's calling days overlap, simply adding the multiples of 3, 4, and 5 counts overlapping days multiple times. This principle organizes the overlaps perfectly.
- Independent Divisibility Fractions:
If a set of numbers a, b, and c share no common factors greater than 1 (they are pairwise relatively prime), the fraction of integers over a full LCM cycle that are NOT divisible by any of them is (1 - 1/a)(1 - 1/b)(1 - 1/c).
Requires a, b, and c to be pairwise relatively prime. It is exact when evaluated over a full repeating cycle (a multiple of a * b * c). Because 3, 4, and 5 share no common factors, divisibility by one does not affect divisibility by another. We can simply multiply their non-divisible fractions to find the proportion of no-call days.
Key Takeaway: Remember This Strategy for Next Time
A schedule or repeating event starting together on 'Day 0'.
Try: Number the days consecutively (1, 2, 3...) and recognize that the events will always occur on days that are exact multiples of the given interval.
Why: Assigning the synchronized start as 0 perfectly aligns the calendar days with mathematical divisibility rules.
A problem asking for when 'none' of several overlapping events happen.
Try: Use Complement Counting paired with the Inclusion-Exclusion Principle. First find the days when 'at least one' event happens, then subtract from the total.
Why: It is usually much easier to build up the overlapping sets of events using LCMs than to directly count the gaps between the events.
Multiple periodic events whose intervals share no common factors.
Try: Find the cycle length using the Least Common Multiple (LCM), then use the Independent Divisibility Fractions trick to quickly find the proportion of time where nothing happens.
Why: When numbers are relatively prime, divisibility conditions behave like independent probabilities over a full LCM cycle.
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