AMC 10 · 2002 · #7
Grade 5 arithmeticLet n be a positive integer such that 21+31+71+n1 is an integer. Which of the following statements is not true:
Pick an answer.
AMC 10 2002 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: A positive whole number $n$ is chosen so that $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{n}$ comes out to a whole number. Among five statements about $n$, find the one that is not true.
Givens: $n$ is a positive integer; $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{n}$ equals a whole number; Statements: (A) $2$ divides $n$, (B) $3$ divides $n$, (C) $6$ divides $n$, (D) $7$ divides $n$, (E) $n>84$
Unknowns: Which one of the five statements is false
Understand
Restated: A positive whole number $n$ is chosen so that $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{n}$ comes out to a whole number. Among five statements about $n$, find the one that is not true.
Givens: $n$ is a positive integer; $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{n}$ equals a whole number; Statements: (A) $2$ divides $n$, (B) $3$ divides $n$, (C) $6$ divides $n$, (D) $7$ divides $n$, (E) $n>84$
Plan
Primary tool: #14 Extreme Principle
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
You cannot judge the five statements until you know $n$, so the real work is finding $n$. First (Tool #7, Identify Subproblems) add the three known fractions to get one fixed number, $\frac{41}{42}$. Then the key move is Tool #14 (Extreme Principle): because $\frac{1}{n}$ is small and positive, the running total is squeezed between two limits and can only land on the whole number $1$ — that boundary argument forces $n=42$. With $n$ known, Tool #3 (Eliminate Possibilities) tests each statement and keeps the one that fails.
Execute — Answer: E
5.NF.A.1 Step 1 Add the three known fractions
- First combine the parts that do not involve $n$.
- The denominators $2$, $3$, and $7$ have least common denominator $42$, so rewrite each fraction over $42$: $\frac{1}{2}=\frac{21}{42}$, $\frac{1}{3}=\frac{14}{42}$, $\frac{1}{7}=\frac{6}{42}$.
- Adding the numerators gives $\frac{21+14+6}{42}=\frac{41}{42}$.
💡 Putting the three fixed fractions over one denominator collapses them into a single number to work with.
4.NF.A.2 Step 2 Trap the total between two limits
- The full sum is now $\frac{41}{42}+\frac{1}{n}$.
- Notice $\frac{41}{42}$ is just below $1$.
- Since $n$ is a positive integer, $\frac{1}{n}$ is greater than $0$ and at most $1$, so the total is greater than $\frac{41}{42}$ but less than $\frac{41}{42}+1$, which is under $2$.
- The only whole number caught strictly between $\frac{41}{42}$ and $2$ is $1$, so the sum must equal exactly $1$.
💡 A tiny positive push added to something just under $1$ can only reach the very next whole number, $1$.
5.NF.A.1 Step 3 Solve for n
- If the total is $1$, then $\frac{1}{n}$ is whatever is missing from $\frac{41}{42}$ to reach $1$: $\frac{1}{n}=1-\frac{41}{42}=\frac{1}{42}$.
- Two unit fractions are equal only when their denominators match, so $n=42$.
💡 Whatever gap is left below $1$ has to be exactly the last fraction, and $\frac{1}{42}$ names it.
4.OA.B.4 Step 4 Test each statement against n = 42
- Now check the statements with $n=42$.
- Since $42=2\times3\times7$, it is divisible by $2$ (A true), by $3$ (B true), by $6=2\times3$ (C true), and by $7$ (D true).
- Statement (E) claims $n>84$, but $42$ is not greater than $84$, so (E) is false.
- The only statement that is not true is (E).
💡 Break $42$ into its prime factors and every small-divisor claim is easy to confirm, leaving only the size claim to fail.
5.NF.A.1 First combine the parts that do not involve $n$. The denominators $2$, $3$, and 4.NF.A.2 The full sum is now $\frac{41}{42}+\frac{1}{n}$. Notice $\frac{41}{42}$ is just 5.NF.A.1 If the total is $1$, then $\frac{1}{n}$ is whatever is missing from $\frac{41}{4 4.OA.B.4 Now check the statements with $n=42$. Since $42=2\times3\times7$, it is divisibl Review
Reasonableness: Check the found value directly: $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{42}=\frac{21+14+6+1}{42}=\frac{42}{42}=1$, a whole number, so $n=42$ is correct. Four of the statements (A, B, C, D) are then true and exactly one, (E), is false — matching the question's promise that a single statement fails. The trap answer would be to pick a divisibility statement, but $42$ really is a multiple of $2$, $3$, $6$, and $7$; only the size comparison $n>84$ breaks.
Alternative: Skip the bounding argument and work with a common denominator throughout: for the sum to be a whole number, $\frac{21+14+6}{42}+\frac{1}{n}=\frac{41}{42}+\frac{1}{n}$ must be an integer, so $\frac{1}{n}$ must supply the missing $\frac{1}{42}$ (any larger jump would overshoot past $1$ before reaching $2$). That again gives $n=42$, and the same statement check singles out (E).
CCSS standards used (min grade 5)
5.NF.A.1Add and subtract fractions with unlike denominators (Adding $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}$ over the common denominator $42$ to get $\frac{41}{42}$, and subtracting $1-\frac{41}{42}=\frac{1}{42}$ to solve for $n$.)4.NF.A.2Compare two fractions with different numerators and different denominators (Comparing $\frac{41}{42}$ to $1$ and bounding $\frac{1}{n}$ to trap the total between $\frac{41}{42}$ and $2$, forcing the sum to equal $1$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Factoring $42=2\times3\times7$ and confirming it is a multiple of $2$, $3$, $6$, and $7$ to test statements (A)-(D).)
⭐ Add the fractions you can, and if what's left is a tiny positive piece, the total can only climb to the next whole number — here that pins $n=42$, so the false claim is $n>84$.
⭐ Add the fractions you can, and if what's left is a tiny positive piece, the total can only climb to the next whole number — here that pins $n=42$, so the false claim is $n>84$.
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