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If m and n are positive integers, are m and n consecutive

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If m and n are positive integers, are m and n consecutive [#permalink]

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New post 15 Sep 2005, 10:45
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If m and n are positive integers, are m and n consecutive integers?

1)m^2 = n^2 + n + m

2)n^2 = m^2 - 2m + 1


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New post 15 Sep 2005, 10:56
D)

Question Stem, m>0, n>0
1) Suff
Pick a positive value for either m or n=> you get a neg & pos root of the quadratic equation. Question stem tells us it cannot be neg. And the positive value is the consecutive value.

2) Suff
Same explanation.
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New post 15 Sep 2005, 10:59
D. Plugging in numers shows that shows this.

1. m^2=n^2 +n+m, m^2-m=n^2+n, m(m-1)=n(n+1). Plug in numbers. Sufficient

2. Same reasoning. Sufficient
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New post 15 Sep 2005, 11:02
Now that is a good question.

Rewrite equation 1.

(m-1) * m = n * (n+1)

So either m-1 = n or n = -m, but m,n>0 so
m-1 = n sufficient

Rewrite equation 2

n^2 = (m-1)^2

Again since m,n > 0, n=m-1

Both sufficient D
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New post 15 Sep 2005, 11:11
B it is.

i)
m^2 = n^2 + n + m or
m * (m - 1) = n * (n + 1)

lets pick numbers:
m = 6, n = 7

6 * 5 <> 7 * 8 (ans NO)

if m = 7, n = 6
7 * 6 = 6 * 7 (ans YES)

not sufficient

ii)
n^2 = m^2 - 2m + 1
n ^ 2 = (m - 1) ^2

taking all combinations m,n comes to be consecutive integers.

Remember, we need to find out if m & n are consecutive. its not given m < n
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New post 15 Sep 2005, 11:16
duttsit wrote:
B it is.

i)
m^2 = n^2 + n + m or
m * (m - 1) = n * (n + 1)

lets pick numbers:
m = 6, n = 7

6 * 5 <> 7 * 8 (ans NO)

if m = 7, n = 6
7 * 6 = 6 * 7 (ans YES)

not sufficient

ii)
n^2 = m^2 - 2m + 1
n ^ 2 = (m - 1) ^2

taking all combinations m,n comes to be consecutive integers.

Remember, we need to find out if m & n are consecutive. its not given m < n


I don't agree with this approach.

Under 1, m>n since m^2>n^2

The numbers you are picking don't satisfy the original equation.
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New post 15 Sep 2005, 11:16
got B.. just by plugging in numbers as duttsit
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New post 15 Sep 2005, 11:20
m=6, n=7 is not a solution to the equation in 1). So you can't pick these numbers. m=7, n=6 agree with the equation.

I think D still holds. OA please.
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New post 15 Sep 2005, 11:23
richardj wrote:
I don't agree with this approach.

Under 1, m>n since m^2>n^2

The numbers you are picking don't satisfy the original equation.


You got me there.
I thought I was being extra careful. Thanks. going with D.
  [#permalink] 15 Sep 2005, 11:23
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If m and n are positive integers, are m and n consecutive

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