AMC 10 · 2021 · #1

Grade 6 arithmetic
exponentsorder-of-operationsmental-arithmetic identify-subproblemspattern-recognition ↑ Prerequisites: exponents
📏 Short solution 💡 1 insight
Problem
Compute a number whose first term is a single power of two with a sum as its exponent, minus a sum of three separate powers of two. Find the value.

Pick an answer.

(A)
0
(B)
50
(C)
52
(D)
54
(E)
57
How to solve
Strategy Identify Subproblems

The expression is one subtraction between two independent chunks, so Tool #7 (Identify Subproblems) splits it: evaluate 2¹⁺²⁺³ on its own, evaluate 2¹+2²+2³ on its own, then subtract. Inside the first chunk the exponent has its own small subproblem, 1+2+3, that must be finished before the power is taken. Tool #5 (Look for a Pattern) supplies the doubling ladder 2, 4, 8, 16, 32, 64, which produces every power of 2 needed without multiplying from scratch. The whole difficulty of the problem is the placement of the exponent, so keeping the two chunks strictly separate is what protects against the trap.

1STEP 1

Split the expression into two chunks

Split into two chunks.

2¹⁺²⁺³-(2¹+2²+2³) = 2⁶ - (2¹+2²+2³)
2STEP 2

Evaluate the single power

Evaluate the single power with the added exponent.

2⁶ = 2 · 2 · 2 · 2 · 2 · 2 = 64
3STEP 3

Evaluate the sum in parentheses

Evaluate the sum in parentheses.

2¹+2²+2³ = 2+4+8 = 14
4STEP 4

Subtract the two chunk values

Subtracting gives 50.

2⁶ - (2¹+2²+2³) = 64 - 14 = 50
Answer
50
Recompute the first chunk a different way to be sure. Adding exponents on a common base corresponds to multiplying the powers, so 2¹⁺²⁺³=2¹ · 2² · 2³ = 2 · 4 · 8 = 64, agreeing with the doubling ladder. Two quick sanity filters back up the final number: 64 is much larger than 14, so the difference must be positive, which rules out (A) 0; and 64 and 14 are both even, so their difference must be even, which rules out (E) 57. The exact arithmetic 64-14=50 then lands on (B).
💡Key takeaway

A sum sitting in the exponent is not the same as a sum of powers: finish the exponent first, because adding up there multiplies the powers while adding down here only adds them.

  • Split the expression into two chunks
  • Evaluate the single power
  • Evaluate the sum in parentheses
  • Subtract the two chunk values