AMC 10 · 2006 · #10

Grade 8 algebracounting
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
perfect-squaressystematic-enumerationinterval-arithmetic identify-subproblems ↑ Prerequisites: perfect-squares
📏 Medium solution 💡 2 insights
Problem
The expression √(120 - √(x)) is a square root nested inside another square root, where x is a real number and every square root means the non-negative principal root. How many real values of x make √(120 - √(x)) an integer?

Pick an answer.

(A)
3
(B)
6
(C)
9
(D)
10
(E)
11

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The phrase 'is an integer' hides a number with no name, so Tool #4 (Introduce a Variable) says: call that integer n and write √(120 - √(x)) = n. Tool #11 (Work Backwards) then peels the radicals one at a time from the outside in — square to remove the outer root, isolate √(x), and read off the condition on n. The question asks 'how many', so Tool #2 (Make a Systematic List) counts the whole numbers n that survive the condition. Naming the integer first is what turns a scary nested radical into a plain counting problem.

1STEP 1

Name the integer and square once

Let n be the integer, so √(120 - √(x)) = n with n at least 0; squaring once gives 120 - √(x) = n².

√(120 - √(x)) = n → 120 - √(x) = n², n = 0, 1, 2, …
2STEP 2

Isolate the inner root

Isolating the inner root gives √(x) = 120 - n², and a square root is never negative, so n² ≤ 120.

√(x) = 120 - n² ≥ 0 → n² ≤ 120
3STEP 3

Count the allowed integers

Only n = 0 through 10 fit, since 10² = 100 is under 120 but 11² = 121 is not — that is 11 integers, zero included.

n ∈ {0, 1, 2, …, 10} → 11 values of n
4STEP 4

Match each integer to one real x

Each n fixes x = (120 - n²)² as one distinct real value, so there are 11 real values of x — choice (E).

x = (120 - n²)² → 11 distinct real x → (E)
Answer
11
Check the two endpoints. At n = 10: √(x) = 120 - 100 = 20, so x = 400; testing, √(x) = 20, 120 - 20 = 100, and √(100) = 10, a whole number. Good. Push one further to n = 11: √(x) = 120 - 121 = -1, which is below zero and impossible, so 11 is correctly excluded. Both the top (n = 10) and the bottom (n = 0) are valid, and everything between is too, so 0 through 10 is 11 values. The tempting wrong answer 10 (choice D) comes from forgetting that n = 0 counts.
💡Key takeaway

Give the hidden integer a name, square to peel off one root, and let the rule 'a square root is never negative' cap how many values fit — here n = 0 through 10 gives 11, and forgetting that zero counts is the trap.

  • Name the integer and square once
  • Isolate the inner root
  • Count the allowed integers
  • Match each integer to one real x

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