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# If M = (x^2 - 7x + 11) and N = (x^2 - 13x + 42), what is the number of

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If M = (x^2 - 7x + 11) and N = (x^2 - 13x + 42), what is the number of  [#permalink]

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01 Feb 2020, 18:13
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GMATBusters’ Quant Quiz Question -3

If $$M = (x^2-7x+11)$$ and $$N = (x^2-13x+42)$$, what is the number of roots of the equation $$M^N = 1$$?

A. 3
B. 4
C. 5
D. 6
E. 7

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Re: If M = (x^2 - 7x + 11) and N = (x^2 - 13x + 42), what is the number of  [#permalink]

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01 Feb 2020, 18:13
The Official Solution is as follows:

Attachment:

WhatsApp Image 2020-02-03 at 6.49.55 PM (2).jpeg [ 62.15 KiB | Viewed 339 times ]

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Re: If M = (x^2 - 7x + 11) and N = (x^2 - 13x + 42), what is the number of  [#permalink]

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Updated on: 03 Feb 2020, 10:30
Factorizing
N=(x-6)(x-7)
for M^N=1, we have the following possibilities:
1) N=0 and M is different than 0
N=0, then (x-6)(x-7)=0
here x=6 or x=7
2) M=1 v M=-1
x^2-7x+11=1
x^2-7x+10=0
(x-2)(x-5)=0
here x=2 or x=5
^2-7x+11=-1 and N is even
x^2-7x+12=0
(x-3)(x-4)=0
here x=3 or x=4. In both cases N is even
Number of roots=6

Originally posted by AndreV on 01 Feb 2020, 19:01.
Last edited by AndreV on 03 Feb 2020, 10:30, edited 1 time in total.
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Re: If M = (x^2 - 7x + 11) and N = (x^2 - 13x + 42), what is the number of  [#permalink]

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01 Feb 2020, 21:11
M^N = 1 => M^N = 1^0 => N=0

Now N= (x^2-13x+42)=0 will derive to x = 6 or 7

that means M^N = 1 => (x^2-7x+11)^6 =1 or (x^2-7x+11)^7 =1

So the number of roots should be 7
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Re: If M = (x^2 - 7x + 11) and N = (x^2 - 13x + 42), what is the number of  [#permalink]

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01 Feb 2020, 21:33
sol;
we can see from the question that multiple roots are possible.
M^N=1 the equation can be satisfied by M=1,-1 or N=0 or integer

case-1:
lets consider if M=1,

x^2-7x+11= 01
x^2-7x =10=0
(x-2)(x-5)=0
x= 2, 5

N= 0 >>>>X^2-13x+42= 0
(x-6)(x-7)= 0,
x= 6,7
so roots = 2,5,6,7
case-2:
now consider other condition
M= -1, then N=even,
x^2-7x+11= -1
x^2-7x+12= 0
(x-3)(x-4)=0
x= 3,4
hence no. of total roots = 6
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Re: If M = (x^2 - 7x + 11) and N = (x^2 - 13x + 42), what is the number of  [#permalink]

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02 Feb 2020, 00:43
not sure
given
x^2-7x+11^( x^2-13x+11) = 1
we get
x^2-7x+11 ^ ( (x-6)*(x-7) = 1
total possible roots would be 4
IMO B
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Re: If M = (x^2 - 7x + 11) and N = (x^2 - 13x + 42), what is the number of  [#permalink]

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02 Feb 2020, 05:48
Ans is B
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Re: If M = (x^2 - 7x + 11) and N = (x^2 - 13x + 42), what is the number of  [#permalink]

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Updated on: 04 Feb 2020, 10:37
Given: M = (x^2−7x+11) and N = (x^2−13x+42)
Asked: What is the number of roots of the equation M^N= 1?
A. 3
B. 4
C. 5
D. 6
E. 7

When N=0; M^N=1
N=0 when x=6 or x=7; 2 roots

When M=1; M^N=1
x^2-7x+11 = 1; x^2 -7x+10 = 0; x = 2 or x=5; 2 roots

When M=-1; M^N=1 when N is even
x^2-7x+11 = -1; x^2 -7x+12= 0; x = 4 or x=3
When x=4; N=6; 1 root
When x=3; N= 12; 1 root

There are total 6 roots of the equation.

IMO D
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Originally posted by Kinshook on 02 Feb 2020, 07:14.
Last edited by Kinshook on 04 Feb 2020, 10:37, edited 2 times in total.
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Re: If M = (x^2 - 7x + 11) and N = (x^2 - 13x + 42), what is the number of  [#permalink]

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03 Feb 2020, 09:49
Here we can divide solution into three parts.
Anything raised to 0 is 1 so equating equation n to zero gives us two roots.
2nd part= 1 to the power of anything will give us 1 so equating equation M to 1 we get two solutions for M
3rd part = -1 to any even power of anything will give us 1 so equating equation M to -1 we get two solutions for M.
So total 6 solutions.
Re: If M = (x^2 - 7x + 11) and N = (x^2 - 13x + 42), what is the number of   [#permalink] 03 Feb 2020, 09:49