AMC 10 · 2021 · #3

Grade 7 algebra
fraction-arithmeticlinear-equations-one-varorder-of-operationsformula-substitution work-backwardsidentify-subproblemseasier-related-problem ↑ Prerequisites: fraction-arithmeticlinear-equations-one-var
📏 Medium solution 💡 2 insights
Problem
A number sits at the bottom of a stacked fraction. The whole stack equals one hundred forty-four over fifty-three. Find that number.

Pick an answer.

(A)
$\frac34$
(B)
$\frac78$
(C)
$\frac{14}{15}$
(D)
$\frac{37}{38}$
(E)
$\frac{52}{53}$
How to solve
Strategy Work Backwards

The value of the whole stack is given and the starting number x is hidden at the very bottom, which is exactly the setup for Tool #11 (Work Backwards): undo the operations in reverse order. Reading the left side from the outside in, the operations are add 2, take a reciprocal, add 1, take a reciprocal, add 2, take a reciprocal (of the 2 on top), add 3. Tool #7 (Identify Subproblems) makes each layer its own tiny task, and after each undo the equation that remains is a shorter stack of the same shape — Tool #9 (Solve an Easier Related Problem) in action. No expanding of the giant fraction is needed.

1STEP 1

Undo the outer plus two

Undo the outer addition first.

1{1+1/(2+2/(3+x))}=144/53-2=38/53
2STEP 2

Flip to undo the reciprocal

Flip to undo the reciprocal.

1+1/(2+2/(3+x))=53/38
3STEP 3

Subtract one, then flip again

Subtract, then flip again.

1/(2+2/(3+x))=15/38 → 2+2/(3+x)=38/15
4STEP 4

Peel down to three plus x

Peel down to the bottom layer.

2/(3+x)=8/15 → 3+x=(2 · 15)/8=15/4
5STEP 5

Solve for x

Solving gives three quarters.

x=15/4-3=15/4-12/4=3/4 → (A)
Answer
3/4
Rebuild the stack forward with x=3/4: the bottom is 3+3/4=15/4, so 2/(15/4)=8/15 and 2+8/15=38/15. Then 1/(38/15)=15/38 and 1+15/38=53/38. Then 1/(53/38)=38/53 and 2+38/53=144/53, exactly the target. A size check agrees too: 144/53 is a little over 2.7, and the left side is 2 plus a fraction that is less than 1, so the total had to land between 2 and 3.
💡Key takeaway

A tall stacked fraction unwraps from the outside in: subtract whatever was added, flip whatever sits under the 1, and repeat until only x is left.

  • Undo the outer plus two
  • Flip to undo the reciprocal
  • Subtract one, then flip again
  • Peel down to three plus x
  • Solve for x