Kvant Math Problem 1072

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Problem

Factor the number $989 \cdot 1001 \cdot 1007 +320$ into prime factors.

S. V. Fomin

Leningrad City Mathematical Olympiad (1987)

Exploration

The expression $989 \cdot 1001 \cdot 1007 + 320$ appears to involve three numbers spaced by six units: $989$, $1001$, $1007$. Observing the spacing suggests attempting a transformation using a central term, $1001$, and writing the others in terms of $1001$. Let $n = 1001$; then $989 = n - 12$ and $1007 = n + 6$. The product becomes $(n-12)n(n+6) + 320$. Expanding this may produce a factorable cubic or suggest grouping.

Another approach is to check small integer factors first. Since $n$ is odd and $320$ is divisible by $2^6 \cdot 5$, factoring out small primes like $2$, $3$, $5$, $7$ may help. Observing that $1001 = 7 \cdot 11 \cdot 13$, perhaps this decomposition plays a role. Testing simple divisibility by $7$, $11$, and $13$ seems promising.

The crucial step is spotting a factorization that absorbs the constant term $320$. Expanding $(n-12)n(n+6)$ gives $n^3 -6 n^2 -72 n$, which plus $320$ gives $n^3 - 6 n^2 - 72 n + 320$. Factoring this cubic carefully will likely reveal linear factors connected to small primes.

Problem Understanding

The problem asks to factor completely into primes the number $989 \cdot 1001 \cdot 1007 + 320$. This is a Type A problem: determine all prime factors. The difficulty lies in handling the three-term product and constant efficiently, without blindly expanding a huge number. The expected insight is to use the near-arithmetic progression structure of $989$, $1001$, $1007$ and exploit the factorization of $1001 = 7 \cdot 11 \cdot 13$.

The answer intuitively should involve small primes, especially those dividing $1001$, and possibly powers of $2$ and $5$ to account for $320$.

Proof Architecture

Lemma 1: Writing the product in terms of the central number $n = 1001$ gives $(n-12)n(n+6) + 320 = n^3 -6 n^2 -72 n + 320$. This follows from direct substitution and expansion.

Lemma 2: The cubic $n^3 -6 n^2 -72 n + 320$ is divisible by $n-10$. This is verified by the remainder theorem: substituting $n = 10$ yields zero.

Lemma 3: The remaining quadratic $n^2 + 4 n - 32$ factors further over integers as $(n+8)(n-4)$. Direct factorization checks: $(n+8)(n-4) = n^2 + 4 n -32$.

Lemma 4: Substituting back $n = 1001$ gives three integer factors: $991$, $1009$, $1012$. Each must be factored into primes to complete the solution.

The hardest step is recognizing the linear factor $n-10$, which absorbs the constant term $320$ and aligns the cubic for integer factorization.

Solution

Let $n = 1001$. Then

$$989 \cdot 1001 \cdot 1007 + 320 = (n-12)n(n+6) + 320.$$

Expanding, we have

$$(n-12)n(n+6) = n(n^2 + 6 n -12 n -72) = n(n^2 -6 n -72) = n^3 -6 n^2 -72 n.$$

Adding $320$ gives

$$n^3 -6 n^2 -72 n + 320.$$

We attempt to factor this cubic. Possible small integer factors are suggested by the divisors of $320$. Testing $n = 10$, we compute

$$10^3 - 6 \cdot 10^2 - 72 \cdot 10 + 320 = 1000 - 600 -720 + 320 = 0,$$

so $n - 10$ is a factor. Performing polynomial division of $n^3 -6 n^2 -72 n + 320$ by $n-10$ gives the quotient

$$n^2 + 4 n - 32.$$

Factoring the quadratic yields integers $n^2 + 4 n -32 = (n+8)(n-4)$. Hence the cubic factors as

$$(n-10)(n+8)(n-4).$$

Substituting back $n = 1001$ produces

$$1001 -10 = 991, \quad 1001 +8 = 1009, \quad 1001 -4 = 997.$$

All three numbers, $991$, $1009$, and $997$, are prime. Therefore, the prime factorization of the original number is

$$\boxed{991 \cdot 997 \cdot 1009}.$$

This completes the proof.

∎

Verification of Key Steps

The crucial check is that $n = 10$ gives a zero remainder. Evaluating $10^3 - 6 \cdot 10^2 - 72 \cdot 10 + 320$ explicitly step by step gives $1000 - 600 -720 + 320 = 0$. Any error in arithmetic would break the factorization. Next, factoring the quadratic $n^2 + 4 n -32$ requires solving for integer roots. The candidate factor pairs of $-32$ are $(1, -32)$, $(2, -16)$, $(4, -8)$, $(8, -4)$, etc. Only $8$ and $-4$ sum to $4$, confirming $(n+8)(n-4)$. Finally, verifying primality of $991$, $997$, and $1009$ requires checking divisibility by primes up to approximately $32$, confirming no smaller prime factors exist.

Alternative Approaches

Another approach begins by directly factoring each number near $1000$ as $1000 + k$ and expanding $(1000 - 11)(1000 +1)(1000 +7) + 320$, collecting powers of $1000$, and attempting to factor the resulting cubic in $1000$. This produces a similar cubic with smaller coefficients, but requires more arithmetic care and offers no advantage over centering at $1001$, which aligns with the factorization of $1001$ itself. Using $n = 1001$ immediately exploits the known factorization $7 \cdot 11 \cdot 13$ to guide factor choices, making the main approach preferable.