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If m and n are non-zero integers, is m6-m5-m4 divisible by n?   1) m2

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If m and n are non-zero integers, is m6-m5-m4 divisible by n?   1) m2  [#permalink]

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New post 29 Mar 2017, 22:18
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If m and n are non-zero integers, is \(m^6-m^5-m^4\) divisible by n?

1) \(m^2-1=m+n^2\)
2) \(m^4-2m^3+m^2=n^2+11\)

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Re: If m and n are non-zero integers, is m6-m5-m4 divisible by n?   1) m2  [#permalink]

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New post 30 Mar 2017, 10:22
1
ziyuen wrote:
If m and n are non-zero integers, is \(m^6-m^5-m^4\) divisible by n?

1) \(m^2-1=m+n^2\)
2) \(m^4-2m^3+m^2=n^2+11\)


m and n can either be positive or negative.
When m is negative
m^6+m^5-m^4. m4(m2+m-1)
When m is positive
m4(m2-m-1)
1) m2-m-1=n2
Divisible when m is negative but not when m is positive.
2)m2(m2-2m+1)=n2+11
Insuff.

IMO E

Please post the solution
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Re: If m and n are non-zero integers, is m6-m5-m4 divisible by n?   1) m2  [#permalink]

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New post 30 Mar 2017, 15:48
1
ziyuen wrote:
If m and n are non-zero integers, is \(m^6-m^5-m^4\) divisible by n?

1) \(m^2-1=m+n^2\)
2) \(m^4-2m^3+m^2=n^2+11\)


OFFICIAL SOLUTION



If you modify the original condition and the question,\(m^6-m^5-m^4 = nt\)? (t = any non-zero integer), so you get \(m^4(m^2-m-1)=nt\)?. If you look at con 1), from \(m^2-m-1=n^2\), if you substitute it to \(m^4(m^2-m-1)\), you get \(m^4(m^2-m-1)=m^4*n^2=n(m^4n)\), hence yes, it is sufficient. For con 2), it is difficult to approach the question, but if you apply CMT 4 (B: if you get A or B too easily, consider D), it is also sufficient. Therefore, the answer is D. If you actually solve it, for con 2), you get \((m^2)^2-2m(m^2)+m^2=(m^2-m)^2=n^2+11\), \((m^2-m)^2-n^2=(m^2-m-n)(m^2-m+n)=11\),

① When it becomes m^2-m-n=1 and m^2-m+n=11, from m^2-m-n=1, you get m^2-m-1=n, so m^4(m^2-m-1)=m^4*n, hence yes.

② When it becomes m2-m-n=11 and m^2-m+n=1, if you add the two equations, you get 2(m^2-m)=12, m^2-m=6 and n=-5, then m^2-m-1=6-1=5=(-1)(-5)=-n, so from m^4(m^2-m-1)=m^4(-n), it is also yes.

③ When it becomes m^2-m-n=-1 and m^2-m+n=-11, if you add the two equations, from 2(m^2-m)=-12, m^2-m=-6 and n=-5, integer m that satisfies this does not exist, so it is out of scope.

④ When it becomes m^2-m-n=-11 and m^2-m+n=-1, if you add the two equations, from 2(m^2-m)=-12, m^2-m=-6 and n=5, integer m that satisfies this does not exist, so it is out of scope.

Therefore, as ①, ② shown above, it always becomes yes, hence it is sufficient. Therefore, the answer is D.
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Re: If m and n are non-zero integers, is m6-m5-m4 divisible by n?   1) m2  [#permalink]

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New post 31 Mar 2017, 08:02
How to solve this question in 2 minutes ?
Deriving option B takes a lot of time!
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Re: If m and n are non-zero integers, is m6-m5-m4 divisible by n?   1) m2  [#permalink]

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New post 31 Mar 2017, 08:04
askul2389 wrote:
How to solve this question in 2 minutes ?
Deriving option B takes a lot of time!


Dear MathRevolution, Could you help to resolve this?
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Re: If m and n are non-zero integers, is m6-m5-m4 divisible by n?   1) m2  [#permalink]

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New post 23 Jul 2019, 04:33
chetan2u can you solve option 2. I am unable to solve it.
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Re: If m and n are non-zero integers, is m6-m5-m4 divisible by n?   1) m2  [#permalink]

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New post 25 Jul 2019, 19:15
Bunuel chetan2u VeritasKarishma Please help on solving St:2
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Re: If m and n are non-zero integers, is m6-m5-m4 divisible by n?   1) m2  [#permalink]

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New post 25 Jul 2019, 20:05
Awesome! This is really good.

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Re: If m and n are non-zero integers, is m6-m5-m4 divisible by n?   1) m2  [#permalink]

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New post 25 Jul 2019, 22:58
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warrior1991 wrote:
Bunuel chetan2u VeritasKarishma Please help on solving St:2


m and n are non negative integers.

Ques: Is (m^6 - m^5 - m^4) divisible by n?

Stmnt 2:
\(m^2(m^2-2m+1)=n^2+11\)
\(m^2 (m - 1)^2 = n^2 + 11\)

Now note that LHS is a perfect square so RHS must be a perfect square too. n is an integer and its square is n^2.
n^2 + 11 needs to be a perfect square too. Two consecutive perfect squares with a difference of only 11 between them must be 25 and 36 (the difference between consecutive perfect squares keeps increasing)
So n must be 5 so that n^2 + 11 is 36 which is also a perfect square.

\(m^2 (m - 1)^2 = 36\)
\((m(m-1))^2 = 6^2\)
\(m*(m-1) = 6\)
m = 2

We have the values for both m and n so we can find whether the given expression is divisible by n.
Sufficient.
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Re: If m and n are non-zero integers, is m6-m5-m4 divisible by n?   1) m2  [#permalink]

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New post 26 Jul 2019, 08:35
VeritasKarishma wrote:
warrior1991 wrote:
Bunuel chetan2u VeritasKarishma Please help on solving St:2


m and n are non negative integers.

Ques: Is (m^6 - m^5 - m^4) divisible by n?

Stmnt 2:
\(m^2(m^2-2m+1)=n^2+11\)
\(m^2 (m - 1)^2 = n^2 + 11\)

Now note that LHS is a perfect square so RHS must be a perfect square too. n is an integer and its square is n^2.
n^2 + 11 needs to be a perfect square too. Two consecutive perfect squares with a difference of only 11 between them must be 25 and 36 (the difference between consecutive perfect squares keeps increasing)
So n must be 5 so that n^2 + 11 is 36 which is also a perfect square.

\(m^2 (m - 1)^2 = 36\)
\((m(m-1))^2 = 6^2\)
\(m*(m-1) = 6\)
m = 2

We have the values for both m and n so we can find whether the given expression is divisible by n.
Sufficient.


Honestly, I could not even think of a starting point. This explanation is just great.! Thank you! :please :angel:
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Re: If m and n are non-zero integers, is m6-m5-m4 divisible by n?   1) m2   [#permalink] 26 Jul 2019, 08:35
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