AMC 10 · 2006 · #10
Grade 8 arithmeticFor how many real values of x is 120−x an integer?
Pick an answer.
AMC 10 2006 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: The expression $\sqrt{120 - \sqrt{x}}$ is a square root nested inside another square root. We want to count how many real numbers $x$ make the whole expression $\sqrt{120 - \sqrt{x}}$ come out to a whole number (an integer).
Givens: The expression is $\sqrt{120 - \sqrt{x}}$; $x$ ranges over the real numbers; We want the value of the expression to be an integer; Answer choices: (A) 3, (B) 6, (C) 9, (D) 10, (E) 11
Unknowns: How many real values of $x$ make $\sqrt{120 - \sqrt{x}}$ an integer
Understand
Restated: The expression $\sqrt{120 - \sqrt{x}}$ is a square root nested inside another square root. We want to count how many real numbers $x$ make the whole expression $\sqrt{120 - \sqrt{x}}$ come out to a whole number (an integer).
Givens: The expression is $\sqrt{120 - \sqrt{x}}$; $x$ ranges over the real numbers; We want the value of the expression to be an integer; Answer choices: (A) 3, (B) 6, (C) 9, (D) 10, (E) 11
Plan
Primary tool: #4 Introduce a Variable
Secondary: #11 Work Backwards, #2 Make a Systematic List
The phrase 'is an integer' hides a number with no name, so Tool #4 (Introduce a Variable) says: call that integer $n$ and write $\sqrt{120 - \sqrt{x}} = n$. Tool #11 (Work Backwards) then peels the radicals one at a time from the outside in — square to remove the outer root, isolate $\sqrt{x}$, 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.
Execute — Answer: E
8.EE.A.2 Step 1 Name the integer and square once
- Let $n$ be the integer the expression equals, so $\sqrt{120 - \sqrt{x}} = n$.
- Because $n$ is itself a square root, it must be a whole number that is $0$ or larger.
- Squaring both sides removes the outer radical and gives $120 - \sqrt{x} = n^2$.
💡 Giving the mystery integer a name lets you square the equation and strip off one layer of the nesting.
6.EE.B.5 Step 2 Isolate the inner root
- Rearrange to get the inner root alone: $\sqrt{x} = 120 - n^2$.
- Now use the rule that a square root is never negative.
- So the right side must satisfy $120 - n^2 \ge 0$, which means $n^2 \le 120$.
- This single inequality is the whole gate that decides which $n$ are allowed.
💡 Working from the outside in leaves one clean rule: the leftover $120 - n^2$ has to be something a square root can equal, so it can't dip below zero.
6.EE.A.1 Step 3 Count the allowed integers
- List the non-negative integers $n$ with $n^2 \le 120$.
- Since $10^2 = 100 \le 120$ but $11^2 = 121 > 120$, the values are $n = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10$.
- That is $11$ integers in all.
- (Do not forget that $n = 0$ counts — it gives $\sqrt{x} = 120$.)
💡 Squares grow fast, so only $n$ up to $10$ stay under $120$, and $0$ is a legitimate starting point.
8.EE.A.2 Step 4 Match each integer to one real x
- Each allowed $n$ fixes $\sqrt{x} = 120 - n^2$ as one definite non-negative number, and squaring gives exactly one real value $x = (120 - n^2)^2$.
- Different $n$ give different values of $120 - n^2$ (they are $120, 119, 116, \dots, 20$), so they give different $x$.
- Eleven integers produce eleven distinct real $x$, so the count is $\textbf{(E) } 11$.
💡 One value of $n$ pins down $\sqrt{x}$, and one non-negative value of $\sqrt{x}$ pins down a single $x$.
8.EE.A.2 Let $n$ be the integer the expression equals, so $\sqrt{120 - \sqrt{x}} = n$. Be 6.EE.B.5 Rearrange to get the inner root alone: $\sqrt{x} = 120 - n^2$. Now use the rule 6.EE.A.1 List the non-negative integers $n$ with $n^2 \le 120$. Since $10^2 = 100 \le 120 8.EE.A.2 Each allowed $n$ fixes $\sqrt{x} = 120 - n^2$ as one definite non-negative numbe Review
Reasonableness: Check the two endpoints. At $n = 10$: $\sqrt{x} = 120 - 100 = 20$, so $x = 400$; testing, $\sqrt{x} = 20$, $120 - 20 = 100$, and $\sqrt{100} = 10$, a whole number. Good. Push one further to $n = 11$: $\sqrt{x} = 120 - 121 = -1 < 0$, 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.
Alternative: Skip naming $n$ and reason directly: $\sqrt{120 - \sqrt{x}}$ is an integer exactly when the inside $120 - \sqrt{x}$ is a perfect square that a square root can reach — that is, a perfect square from $0$ up to $120$. The perfect squares $\le 120$ are $0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100$, which is $11$ of them. Each one sets $\sqrt{x}$, hence $x$, uniquely, giving the same total of $11$.
CCSS standards used (min grade 8)
8.EE.A.2Use square root and cube root symbols to represent solutions (Reading the nested radical $\sqrt{120 - \sqrt{x}}$, squaring to undo a square root, and recovering $x = (120 - n^2)^2$ from a value of $\sqrt{x}$.)6.EE.B.5Understand solving an equation as finding which values make it true (Turning 'the expression is an integer' into the condition $n^2 \le 120$ and finding which whole numbers $n$ satisfy it.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Squaring integers to compare $n^2$ against $120$ and listing $n = 0$ through $10$.)
⭐ 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.
⭐ 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.
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