AMC 10 · 2005 · #24
Grade 8 number-theoryFor each positive integer m>1, let P(m) denote the greatest prime factor of m. For how many positive integers n is it true that both P(n)=n and P(n+48)=n+48?
Pick an answer.
AMC 10 2005 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: For a number $m>1$, $P(m)$ is its greatest prime factor. Count the positive integers $n$ for which the greatest prime factor of $n$ equals $\sqrt{n}$, and at the same time the greatest prime factor of $n+48$ equals $\sqrt{n+48}$.
Givens: $P(m)$ means the greatest prime factor of $m$; The condition $P(n)=\sqrt{n}$ must hold; The condition $P(n+48)=\sqrt{n+48}$ must hold at the same time; Answer choices: (A) $0$, (B) $1$, (C) $3$, (D) $4$, (E) $5$
Unknowns: How many positive integers $n$ satisfy both conditions at once
Understand
Restated: For a number $m>1$, $P(m)$ is its greatest prime factor. Count the positive integers $n$ for which the greatest prime factor of $n$ equals $\sqrt{n}$, and at the same time the greatest prime factor of $n+48$ equals $\sqrt{n+48}$.
Givens: $P(m)$ means the greatest prime factor of $m$; The condition $P(n)=\sqrt{n}$ must hold; The condition $P(n+48)=\sqrt{n+48}$ must hold at the same time; Answer choices: (A) $0$, (B) $1$, (C) $3$, (D) $4$, (E) $5$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
The condition $P(n)=\sqrt{n}$ is a disguise. Unpacking it (Tool #4, Introduce a Variable) shows $n$ must be the square of a prime: write $n=p^2$ and $n+48=q^2$ with $p,q$ prime. Subtracting turns the whole problem into one clean equation $q^2-p^2=48$. Factoring the difference of squares gives $(q-p)(q+p)=48$, so the primes are pinned down by the factor pairs of $48$. Tool #2 (Make a Systematic List) enumerates those pairs, and Tool #3 (Eliminate Possibilities) throws out the pairs whose $p$ or $q$ is not prime, leaving the exact count.
Execute — Answer: B
8.EE.A.2 Step 1 Decode P(n)=√n
- A greatest prime factor is a whole number, so $\sqrt{n}$ must be a whole number — meaning $n$ is a perfect square.
- Say $\sqrt{n}=p$, so $n=p^2$.
- The greatest prime factor of $p^2$ is $p$ only when $p$ itself is prime; if $p$ were composite it would split into smaller primes and its greatest prime factor would be less than $p=\sqrt{n}$.
- So the condition $P(n)=\sqrt{n}$ holds exactly when $n$ is the square of a prime.
💡 $P(m)=\sqrt{m}$ is just a code for “$m$ is a prime squared.”
6.EE.B.6 Step 2 Apply it to both numbers
- The same decoding works for $n+48$: it must also be the square of a prime, say $n+48=q^2$ with $q$ prime.
- Now both conditions are captured by two clean equations.
- Subtracting the first from the second removes $n$ and leaves a relation between the two primes.
💡 Naming both square roots lets the $+48$ gap become a single equation.
6.EE.A.3 Step 3 Factor the difference of squares
- A difference of two squares factors as $q^2-p^2=(q-p)(q+p)$.
- So $(q-p)(q+p)=48$.
- The two whole-number factors $q-p$ and $q+p$ must multiply to $48$, which lets us hunt through the factor pairs of $48$ instead of guessing primes blindly.
💡 Turning a subtraction of squares into a product opens the door to factor-pair reasoning.
6.NS.B.4 Step 4 Both factors must be even
- The two factors $q-p$ and $q+p$ add up to $2q$, an even number, so they are both even or both odd.
- Two odd numbers multiply to an odd product, but $48$ is even, so they cannot both be odd — they must both be even.
- The even factor pairs of $48$ (smaller times larger) are $2\times24$, $4\times12$, and $6\times8$.
💡 Same-parity factors of an even number are forced to both be even.
6.NS.B.4 Step 5 Solve each pair and keep only primes
- For a pair with $q-p=d$ and $q+p=s$, adding and halving gives $q=\tfrac{s+d}{2}$ and $p=\tfrac{s-d}{2}$.
- Pair $(2,24)$ gives $p=11$, $q=13$ — both prime, so $n=11^2=121$ works and $n+48=169=13^2$.
- Pair $(4,12)$ gives $p=4$, $q=8$ — not prime.
- Pair $(6,8)$ gives $p=1$, $q=7$ — but $1$ is not prime.
- So exactly one value, $n=121$, survives, and the count of valid $n$ is $1$, which is choice (B).
💡 Each factor pair proposes a $(p,q)$; keep it only if both are genuinely prime.
8.EE.A.2 A greatest prime factor is a whole number, so $\sqrt{n}$ must be a whole number 6.EE.B.6 The same decoding works for $n+48$: it must also be the square of a prime, say $ 6.EE.A.3 A difference of two squares factors as $q^2-p^2=(q-p)(q+p)$. So $(q-p)(q+p)=48$. 6.NS.B.4 The two factors $q-p$ and $q+p$ add up to $2q$, an even number, so they are both 6.NS.B.4 For a pair with $q-p=d$ and $q+p=s$, adding and halving gives $q=\tfrac{s+d}{2}$ Review
Reasonableness: Check the survivor directly: $n=121=11^2$, whose greatest prime factor is $11=\sqrt{121}$, and $n+48=169=13^2$, whose greatest prime factor is $13=\sqrt{169}$. Both conditions hold, so $121$ is genuinely valid. The two rejected pairs fail on primality, so exactly one $n$ works — matching choice (B) $=1$.
Alternative: Search directly over prime squares. The gap between consecutive squares is $(k+1)^2-k^2=2k+1$, which exceeds $48$ once $k\ge 24$, so $\sqrt{n}=p$ can only be a prime below $24$: $2,3,5,7,11,13,17,19,23$. Test $p^2+48$ for each: $4+48=52$, $9+48=57$, $25+48=73$, $49+48=97$, $121+48=169=13^2$, $169+48=217$, $289+48=337$, $361+48=409$, $529+48=577$. Only $p=11$ lands on a prime square ($13^2$), confirming a single solution and answer (B).
CCSS standards used (min grade 8)
8.EE.A.2Use square root symbols to represent and evaluate square roots of perfect squares (Recognizing that $P(n)=\sqrt{n}$ forces $n$ to be a perfect square whose root is prime, i.e. $n=p^2$.)6.EE.B.6Use variables to represent numbers and write expressions when solving a problem (Naming $n=p^2$ and $n+48=q^2$ so the two conditions become the single equation $q^2-p^2=48$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring the difference of squares $q^2-p^2$ into the product $(q-p)(q+p)$.)6.NS.B.4Find factors of whole numbers and reason about prime factorization (Listing the even factor pairs of $48$ and checking which give prime values of $p$ and $q$.)
⭐ “Greatest prime factor equals the square root” secretly means “prime squared,” so set $q^2-p^2=48$, factor it as $(q-p)(q+p)$, and only $p=11,q=13$ gives two primes.
⭐ “Greatest prime factor equals the square root” secretly means “prime squared,” so set $q^2-p^2=48$, factor it as $(q-p)(q+p)$, and only $p=11,q=13$ gives two primes.
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