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18 Feb 2026, 04:59
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If k is a positive integer such that for every odd integer n > 20, n(n - 5)(n + k) is divisible by 24, what is the smallest possible value of k?
(1) 1
(2) 2
(3) 3
(4) 5
(5) 7
(1) 1
(2) 2
(3) 3
(4) 5
(5) 7
18 Feb 2026, 13:21
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Great question — this one trips up a lot of people because it looks like a straightforward divisibility problem but actually requires you to reason carefully about what "for every odd integer n > 20" really means.
Key concept: Divisibility and Number Properties
Step 1: Understand the structure
We need n(n − 5)(n + k) to be divisible by 24 for every odd integer n > 20. Factor 24: 24 = 8 × 3 = 23 × 3. So the product needs to supply both a factor of 8 and a factor of 3.
Step 2: Analyse what n and (n − 5) contribute
Since n is odd, n itself contributes no factors of 2. Now look at (n − 5): odd minus odd = even, so (n − 5) is always even. But how many factors of 2 does it guarantee? If n = 23 → n − 5 = 18 (only one factor of 2). So the pair n and (n − 5) don't reliably cover the factor of 8 — we need (n + k) to help.
Step 3: Handle the factor of 3
Among n, (n − 5), and (n + k), we need at least one divisible by 3 for every odd n. If n ≡ 1 (mod 3), then n − 5 ≡ −4 ≡ 2 (mod 3) — neither n nor (n − 5) is divisible by 3. So (n + k) must be ≡ 0 (mod 3), meaning k ≡ 2 (mod 3).
Step 4: Handle the factor of 8
We need the product divisible by 8. Since n is odd and (n − 5) is always even, we need (n + k) to also be even, meaning k must be odd (odd + odd = even). This gives us at least two factors of 2. For the cases where (n − 5) is divisible by 4, we get more — but the binding constraint is the minimum case.
Step 5: Check answer choices
We need k ≡ 2 (mod 3) AND k to be odd:
- k = 1: 1 mod 3 = 1 ✗
- k = 2: even ✗
- k = 3: 3 mod 3 = 0 ✗
- k = 5: 5 mod 3 = 2 ✓, and 5 is odd ✓
- k = 7: 7 mod 3 = 1 ✗
Answer: D (5)
Quick verification — n = 23: 23 × 18 × 28 = 23 × 504. Since 504 = 24 × 21, this is divisible by 24 ✓
n = 21: 21 × 16 × 26 — 21 provides the factor of 3, 16 provides 24 ✓
Common trap: Students test just one or two values of n and think they're done. The question says "for every odd integer n > 20" — you need the condition to hold universally. A single passing test case doesn't prove sufficiency. The mod analysis approach is what catches failures that lucky numbers hide.
Takeaway: When a divisibility condition must hold for every value in a set, check multiple residue classes mod the prime factors of the divisor — don't rely on testing a single case.
Key concept: Divisibility and Number Properties
Step 1: Understand the structure
We need n(n − 5)(n + k) to be divisible by 24 for every odd integer n > 20. Factor 24: 24 = 8 × 3 = 23 × 3. So the product needs to supply both a factor of 8 and a factor of 3.
Step 2: Analyse what n and (n − 5) contribute
Since n is odd, n itself contributes no factors of 2. Now look at (n − 5): odd minus odd = even, so (n − 5) is always even. But how many factors of 2 does it guarantee? If n = 23 → n − 5 = 18 (only one factor of 2). So the pair n and (n − 5) don't reliably cover the factor of 8 — we need (n + k) to help.
Step 3: Handle the factor of 3
Among n, (n − 5), and (n + k), we need at least one divisible by 3 for every odd n. If n ≡ 1 (mod 3), then n − 5 ≡ −4 ≡ 2 (mod 3) — neither n nor (n − 5) is divisible by 3. So (n + k) must be ≡ 0 (mod 3), meaning k ≡ 2 (mod 3).
Step 4: Handle the factor of 8
We need the product divisible by 8. Since n is odd and (n − 5) is always even, we need (n + k) to also be even, meaning k must be odd (odd + odd = even). This gives us at least two factors of 2. For the cases where (n − 5) is divisible by 4, we get more — but the binding constraint is the minimum case.
Step 5: Check answer choices
We need k ≡ 2 (mod 3) AND k to be odd:
- k = 1: 1 mod 3 = 1 ✗
- k = 2: even ✗
- k = 3: 3 mod 3 = 0 ✗
- k = 5: 5 mod 3 = 2 ✓, and 5 is odd ✓
- k = 7: 7 mod 3 = 1 ✗
Answer: D (5)
Quick verification — n = 23: 23 × 18 × 28 = 23 × 504. Since 504 = 24 × 21, this is divisible by 24 ✓
n = 21: 21 × 16 × 26 — 21 provides the factor of 3, 16 provides 24 ✓
Common trap: Students test just one or two values of n and think they're done. The question says "for every odd integer n > 20" — you need the condition to hold universally. A single passing test case doesn't prove sufficiency. The mod analysis approach is what catches failures that lucky numbers hide.
Takeaway: When a divisibility condition must hold for every value in a set, check multiple residue classes mod the prime factors of the divisor — don't rely on testing a single case.
21 Feb 2026, 07:48
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While handling the factor of 8 why there were no special efforts taken to get three factors of 2. Its explained that "This gives us at least two factors of 2." - But why wasnt efforts taken to find (n+k) such that there is one more factor of 2. And still answer is tallying. May I know why? Edskore
However, Appreciate the answer given!
However, Appreciate the answer given!
