AMC 10 · 2013 · #7
Easy mode Grade 4A list has ten numbers in it. Starting from the third number, every number is the sum of the two numbers just before it. So the fifth number equals the third number plus the fourth number, and the same rule holds all the way to the tenth. The seventh number is 42 and the ninth number is 110. What is the fourth number?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A list of ten numbers follows one rule: from the third number on, each number is the sum of the two numbers just before it. The seventh number is 42 and the ninth number is 110. Find the fourth number.
Givens: The rule $S_n = S_{n-2} + S_{n-1}$ holds for every $n$ from 3 to 10.; $S_7 = 42$.; $S_9 = 110$.
Unknowns: The value of $S_4$.
Understand
Restated: A list of ten numbers follows one rule: from the third number on, each number is the sum of the two numbers just before it. The seventh number is 42 and the ninth number is 110. Find the fourth number.
Givens: The rule $S_n = S_{n-2} + S_{n-1}$ holds for every $n$ from 3 to 10.; $S_7 = 42$.; $S_9 = 110$.
Plan
Primary tool: #11 Work Backwards
Secondary: #5 Look for a Pattern, #4 Introduce a Variable
The rule points forward, but the known terms sit above the wanted term, so run the rule in reverse. The rule says that in any block of three consecutive terms, the last is the sum of the first two; addition and subtraction are inverses, so knowing any two terms of such a block forces the third. Starting from the block $(S_7, S_8, S_9)$, which already has two known members, each new block picks up a freshly filled term and hands back one more, walking down the list to $S_4$. Because every filled value is forced, the chain proves $S_4$ can only be one number; a forward run then confirms such a sequence really exists.
Execute — Answer: C
1.OA.B.4 Step 1 Turn the rule into a filling tool
- The rule ties together three terms in a row: $S_{n-2}$, $S_{n-1}$, and $S_n$, with the last equal to the sum of the first two.
- Subtraction undoes addition, so any one of the three can be recovered from the other two.
- In particular the middle term is $S_{n-1} = S_n - S_{n-2}$ and the earliest term is $S_{n-2} = S_n - S_{n-1}$.
- This is what makes it possible to move down the list instead of up.
💡 A block of three terms with two of them known has no freedom left in the third, no matter which one is missing.
4.NBT.B.4 Step 2 Fill the gap between the givens
- Take the block $(S_7, S_8, S_9)$.
- The rule for $n = 9$ says $S_9 = S_7 + S_8$.
- Two of the three terms are given, so the missing middle one is forced: subtract $S_7 = 42$ from $S_9 = 110$.
💡 The gap between the two givens is exactly one term wide, so one subtraction closes it.
4.OA.C.5 Step 3 Step down to $S_6$
- Now $S_7 = 42$ and $S_8 = 68$ are both known, so use the block $(S_6, S_7, S_8)$.
- The rule for $n = 8$ says $S_8 = S_6 + S_7$, and the missing term is the earliest one.
💡 Each newly found term teams up with its neighbour to unlock the term just before them.
4.OA.C.5 Step 4 Step down to $S_5$, then $S_4$
- The same move repeats twice more.
- From the block $(S_5, S_6, S_7)$ with $S_7 = 42$ and $S_6 = 26$, the earliest term is $S_5 = 42 - 26 = 16$.
- Then from the block $(S_4, S_5, S_6)$ with $S_6 = 26$ and $S_5 = 16$, the earliest term is $S_4 = 26 - 16 = 10$.
💡 The backward rule is the same subtraction every time, so the walk down the list is one repeated move.
4.OA.C.5 Step 5 Check the sequence really exists
- The backward chain shows that if such a sequence exists, then $S_4$ has to be 10 — every value along the way was forced.
- It does not by itself show a sequence with $S_7 = 42$ and $S_9 = 110$ exists at all.
- So run the rule forward from the two terms just found and check both givens land.
- Starting at $S_4 = 10$ and $S_5 = 16$: $S_6 = 10 + 16 = 26$, $S_7 = 16 + 26 = 42$, $S_8 = 26 + 42 = 68$, $S_9 = 42 + 68 = 110$.
- Both given values come out right, so the sequence exists and $S_4 = 10$, which is choice (C).
💡 Working backwards says what the answer must be; running forwards says that answer actually happens.
1.OA.B.4 The rule ties together three terms in a row: $S_{n-2}$, $S_{n-1}$, and $S_n$, wi 4.NBT.B.4 Take the block $(S_7, S_8, S_9)$. The rule for $n = 9$ says $S_9 = S_7 + S_8$. T 4.OA.C.5 Now $S_7 = 42$ and $S_8 = 68$ are both known, so use the block $(S_6, S_7, S_8)$ 4.OA.C.5 The same move repeats twice more. From the block $(S_5, S_6, S_7)$ with $S_7 = 4 4.OA.C.5 The backward chain shows that if such a sequence exists, then $S_4$ has to be 10 Review
Reasonableness: A size estimate made without any of the arithmetic above agrees. A sum-of-the-previous-two sequence multiplies by roughly $1.618$ at each step once it settles, so over the three steps from $S_4$ to $S_7$ it grows by about $1.618^3 \approx 4.24$. That predicts $S_4 \approx \frac{42}{4.24} \approx 9.9$, which singles out 10 and rules out 4, 6, and 16 outright while leaving 12 clearly too big. The exact values also behave: pushing the backward rule past $S_4$ gives $S_3 = 16 - 10 = 6$, $S_2 = 10 - 6 = 4$, $S_1 = 6 - 4 = 2$, so the whole list $2, 4, 6, 10, 16, 26, 42, 68, 110, 178$ stays in positive whole numbers with no contradiction anywhere.
Alternative: Introduce variables instead of walking backwards. Let $S_1 = a$ and $S_2 = b$, then expand forward: $S_3 = a + b$, $S_4 = a + 2b$, $S_5 = 2a + 3b$, $S_6 = 3a + 5b$, $S_7 = 5a + 8b$, $S_8 = 8a + 13b$, $S_9 = 13a + 21b$ — the coefficients are Fibonacci numbers. The givens become the system $5a + 8b = 42$ and $13a + 21b = 110$. Its determinant is $5 \cdot 21 - 8 \cdot 13 = 105 - 104 = 1$, which is not zero, so the system has exactly one solution; this is the real reason two scattered givens are enough to pin the whole sequence down. Solving gives $a = 2$ and $b = 4$, hence $S_4 = a + 2b = 2 + 8 = 10$, matching (C).
CCSS standards used (min grade 4)
1.OA.B.4Understand subtraction as an unknown-addend problem (Rewriting $S_n = S_{n-2} + S_{n-1}$ as a subtraction so a missing term in any block of three can be recovered.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Carrying out the subtractions $110 - 42$, $68 - 42$, $42 - 26$, and $26 - 16$.)4.OA.C.5Generate a number or shape pattern following a given rule (Applying the sum-of-the-previous-two rule repeatedly, both backwards down to $S_4$ and forwards to verify $S_7$ and $S_9$.)
⭐ When three terms in a row add up in a fixed way, knowing any two of them forces the third — so you can walk backwards down the list just as easily as forwards.
⭐ When three terms in a row add up in a fixed way, knowing any two of them forces the third — so you can walk backwards down the list just as easily as forwards.
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