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Bunuel
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If n and m are positive integers, and 3 is not a factor of m, then m 06 May 2021, 07:00
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D
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55% (hard)

Question Stats:

58% (02:03) correct 42% (02:04) wrong based on 185 sessions

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If n and m are positive integers, and 3 is not a factor of m, then m may be which of the following?

A. (n−1)·n·(n+1)
B. (n−3)·(n−1)·(n+1)
C. (n−2)·n·(n+2)
D. (n−1)·n·(n+2)
E. (n−3)·(n+1)·(n+2)
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D
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Bunuel
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If n and m are positive integers, and 3 is not a factor of m, then m 08 May 2021, 05:30
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Bunuel
If n and m are positive integers, and 3 is not a factor of m, then m may be which of the following?

A. (n−1)·n·(n+1)
B. (n−3)·(n−1)·(n+1)
C. (n−2)·n·(n+2)
D. (n−1)·n·(n+2)
E. (n−3)·(n+1)·(n+2)
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m is not a multiple of 3.

A. (n−1)·n·(n+1) --> this is the product of 3 consecutive integers = a multiple of 3. Discard.

B. (n−3)·(n−1)·(n+1) --> n-3 gives the same remainder upon division by 3 as n, so out of n-3 (n), n-1 and n+1 one IS a multiple of 3. Discard.

C. (n−2)·n·(n+2) --> this is the product of 3 consecutive even or odd integers = a multiple of 3.

D. (n−1)·n·(n+2) --> keep.

E. (n−3)·(n+1)·(n+2) --> n-3 gives the same remainder upon division by 3 as n, so out of n-3 (n), n+1 and n+2 one IS a multiple of 3. Discard.

Answer: D.
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If n and m are positive integers, and 3 is not a factor of m, then m 06 May 2021, 07:15
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Bunuel
If n and m are positive integers, and 3 is not a factor of m, then m may be which of the following?

A. (n−1)·n·(n+1)
B. (n−3)·(n−1)·(n+1)
C. (n−2)·n·(n+2)
D. (n−1)·n·(n+2)
E. (n−3)·(n+1)·(n+2)
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Posted from my mobile device

I'm waiting for someone to answer this, i can jot down some value for m making 3 a factor of m..
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If n and m are positive integers, and 3 is not a factor of m, then m 06 May 2021, 08:54
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Here is my approach
If n and m are positive integers, and 3 is not a factor of m, -> m is not divisible by 3-> answer wont be divisible by 3

A. (n−1)·n·(n+1) divisible by 3 because n-1, n, n+1 are 3 consecutive integers so always have 1 of them divisible by 3 then (n-1)n(n+1) divisible by 3
B. (n−3)·(n−1)·(n+1) divisible by 3
n-3 has the same remainder with n when divide by 3
=> (n-3)(n-1)(n+1) has the same remainder with (n-1)n(n+1) when divide by 3
Like choice A-> (n-3)(n-1)(n+1) divisible by 3
C. (n−2)·n·(n+2) divisible by 3
Similar to B, n-2 has the same remainder with n+1 when divide by 3
n+2 has the same remainder with n -1 when divide by 3
=> (n-2)n(n+2) divisible by 3
D. (n−1)·n·(n+2) => seem good. ex: n=5 then m is not divisible by 3 => CORRECT
E. (n−3)·(n+1)·(n+2) divisible by 3
n-3 has the same remainder with n when divide by 3
=> (n−3)·(n+1)·(n+2) has the same remainder with n(n+1)(n+2) when divide by 3
Have: n, n+1, n+2 are 3 consecutive integer then n(n+1)(n+2) is divisible by 3 => (n-3)(n+1)(n+2)


Hence IMO D
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If n and m are positive integers, and 3 is not a factor of m, then m 06 May 2021, 09:25
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ntngocanh19
Here is my approach
If n and m are positive integers, and 3 is not a factor of m, -> m is not divisible by 3-> answer wont be divisible by 3

A. (n−1)·n·(n+1) divisible by 3 because n-1, n, n+1 are 3 consecutive integers so always have 1 of them divisible by 3 then (n-1)n(n+1) divisible by 3
B. (n−3)·(n−1)·(n+1) divisible by 3
n-3 has the same remainder with n when divide by 3
=> (n-3)(n-1)(n+1) has the same remainder with (n-1)n(n+1) when divide by 3
Like choice A-> (n-3)(n-1)(n+1) divisible by 3
C. (n−2)·n·(n+2) divisible by 3
Similar to B, n-2 has the same remainder with n+1 when divide by 3
n+2 has the same remainder with n -1 when divide by 3
=> (n-2)n(n+2) divisible by 3
D. (n−1)·n·(n+2) => seem good. ex: n=5 then m is not divisible by 3 => CORRECT
E. (n−3)·(n+1)·(n+2) divisible by 3
n-3 has the same remainder with n when divide by 3
=> (n−3)·(n+1)·(n+2) has the same remainder with n(n+1)(n+2) when divide by 3
Have: n, n+1, n+2 are 3 consecutive integer then n(n+1)(n+2) is divisible by 3 => (n-3)(n+1)(n+2)


Hence IMO D
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What happens when N = 3 ?

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If n and m are positive integers, and 3 is not a factor of m, then m 06 May 2021, 09:28
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ntngocanh19
Here is my approach
If n and m are positive integers, and 3 is not a factor of m, -> m is not divisible by 3-> answer wont be divisible by 3

A. (n−1)·n·(n+1) divisible by 3 because n-1, n, n+1 are 3 consecutive integers so always have 1 of them divisible by 3 then (n-1)n(n+1) divisible by 3
B. (n−3)·(n−1)·(n+1) divisible by 3
n-3 has the same remainder with n when divide by 3
=> (n-3)(n-1)(n+1) has the same remainder with (n-1)n(n+1) when divide by 3
Like choice A-> (n-3)(n-1)(n+1) divisible by 3
C. (n−2)·n·(n+2) divisible by 3
Similar to B, n-2 has the same remainder with n+1 when divide by 3
n+2 has the same remainder with n -1 when divide by 3
=> (n-2)n(n+2) divisible by 3
D. (n−1)·n·(n+2) => seem good. ex: n=5 then m is not divisible by 3 => CORRECT
E. (n−3)·(n+1)·(n+2) divisible by 3
n-3 has the same remainder with n when divide by 3
=> (n−3)·(n+1)·(n+2) has the same remainder with n(n+1)(n+2) when divide by 3
Have: n, n+1, n+2 are 3 consecutive integer then n(n+1)(n+2) is divisible by 3 => (n-3)(n+1)(n+2)


Hence IMO D

What happens when N = 3 ?

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The question is which maybe m. So when other options always divisible by 3 but only D may not.
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If n and m are positive integers, and 3 is not a factor of m, then m 18 Aug 2025, 11:11
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just select a random postive integer, let's say 2 and then solve through options, you;ll get D as the answer.
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