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13 messages · Page 1 of 1 (latest)

left raptor Jan 19, 2024, 2:04 PM

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Prove that 8n^3-12n^2+4n is divisible by 24 for any integers

empty solarBOT Jan 19, 2024, 2:04 PM

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help

vital zinc Jan 19, 2024, 4:03 PM

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lets take 4n common

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$4n(2n^2 - 3n + 1)$

hybrid finchBOT Jan 19, 2024, 4:03 PM

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ᴰᴱᴬᴰ ★ ᴋɪʟʟᴇʀ ⁰⁴

vital zinc Jan 19, 2024, 4:03 PM

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$4n (2n-1) (n-1)$

hybrid finchBOT Jan 19, 2024, 4:03 PM

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ᴰᴱᴬᴰ ★ ᴋɪʟʟᴇʀ ⁰⁴

vital zinc Jan 19, 2024, 4:03 PM

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so definitely its divisible by 4

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so whats left is $n (n-1) (2n-1)$ to get divisible by 6

hybrid finchBOT Jan 19, 2024, 4:04 PM

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ᴰᴱᴬᴰ ★ ᴋɪʟʟᴇʀ ⁰⁴

vital zinc Jan 19, 2024, 4:05 PM

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so if you put n = 3k, 3k+1, 3k+2

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then this eqn is divisble by 6