AMC 10 · 2017 · #1
Easy mode Grade 5Work out the value of this stacked expression:
(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)
Every layer does the same thing: take the number inside it, double it, then add 1. Start at the innermost part, 2+1, and work your way outward one layer at a time. What number does the whole expression equal?
Pick an answer.
AMC 10 2017 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: Evaluate the nested expression $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$.
Givens: The expression $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$; Each layer of parentheses says: take what is inside, double it, then add $1$; Answer choices: (A) $70$, (B) $97$, (C) $127$, (D) $159$, (E) $729$
Unknowns: The single number the whole expression equals
Understand
Restated: Evaluate the nested expression $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$.
Givens: The expression $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$; Each layer of parentheses says: take what is inside, double it, then add $1$; Answer choices: (A) $70$, (B) $97$, (C) $127$, (D) $159$, (E) $729$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #5 Look for a Pattern, #3 Eliminate Possibilities
The expression is one big calculation built from many identical small ones: every layer is just "double the inside, then add $1$." Tool #7 (Identify Subproblems) lets us peel the parentheses one at a time, innermost first, turning a scary tower into six easy steps. Tool #5 (Look for a Pattern) reveals that each running total is one less than a power of $2$, and Tool #3 (Eliminate Possibilities) uses that pattern to lock onto the one choice that fits.
Execute — Answer: C
5.OA.A.1 Step 1 Read the nesting inside-out
- The parentheses are stacked inside one another, so order of operations says start at the very center, $(2+1)$, and move outward one shell at a time.
- Each shell does the same job to whatever number sits inside it: multiply that number by $2$, then add $1$.
💡 Nested parentheses are like nested boxes — you can only open the next one after you finish the one inside it.
3.NBT.A.2 Step 2 Double-and-add, first three layers
- Begin at the center and apply "double, then add $1$" three times.
- The innermost value is $2+1=3$.
- Feed it outward: $2 \times 3 + 1 = 7$, then $2 \times 7 + 1 = 15$.
💡 Each layer only needs the single number the layer below already produced, so you never juggle the whole tower at once.
4.NBT.B.5 Step 3 Finish the remaining layers
- Keep applying the same rule to the last three shells: $2 \times 15 + 1 = 31$, then $2 \times 31 + 1 = 63$, then $2 \times 63 + 1 = 127$.
- The outermost shell is the whole expression, so the expression equals $127$.
💡 Doubling a number is the easiest multiplication there is — just add the number to itself — so each outward step stays quick even as the numbers grow.
4.OA.C.5 Step 4 Pattern check picks the answer
- Look at the running totals: $3, 7, 15, 31, 63, 127$.
- Each one is exactly one less than a power of $2$: $4{-}1, 8{-}1, 16{-}1, 32{-}1, 64{-}1, 128{-}1$.
- So the final value is $2^7 - 1 = 127$.
- Among the choices, only $127$ is one less than a power of $2$, which rules out $70, 97, 159,$ and $729$.
- The answer is (C).
💡 Doubling and adding $1$ each time keeps every total just short of the next power of $2$, so the pattern names the answer instantly.
5.OA.A.1 The parentheses are stacked inside one another, so order of operations says star 3.NBT.A.2 Begin at the center and apply "double, then add $1$" three times. The innermost 4.NBT.B.5 Keep applying the same rule to the last three shells: $2 \times 15 + 1 = 31$, th 4.OA.C.5 Look at the running totals: $3, 7, 15, 31, 63, 127$. Each one is exactly one les Review
Reasonableness: Two independent routes agree on $127$: direct inside-out computation ($3 \to 7 \to 15 \to 31 \to 63 \to 127$) and the pattern $2^7 - 1$. The result sits between the choices $97$ and $159$, which is sensible for an expression whose biggest step is $2 \times 63 + 1$, and it matches (C).
Alternative: Distribute everything instead of evaluating step by step. Multiplying out, the expression becomes $2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1 + 2^0 = 64+32+16+8+4+2+1 = 127$, the sum of all powers of $2$ up to $2^6$ — again (C).
CCSS standards used (min grade 5)
5.OA.A.1Use parentheses, brackets, or braces in numerical expressions, and evaluate expressions with these symbols (Reading the nested parentheses correctly — evaluating the innermost group first and working outward.)3.NBT.A.2Fluently add and subtract within 1000 (Adding $1$ at each layer of the inside-out computation.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Doubling the running total (multiplying by $2$) at each layer, including $2 \times 63 = 126$.)4.OA.C.5Generate a number or shape pattern that follows a given rule (Spotting that each running total is one less than a power of $2$ and using that to confirm and select the answer.)
⭐ When parentheses are stacked inside each other, open them from the inside out — one easy "double and add one" step at a time.
⭐ When parentheses are stacked inside each other, open them from the inside out — one easy "double and add one" step at a time.
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