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#1 2024-03-14 20:44:01

tony123
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Registered: 2007-08-03
Posts: 229

5 problems

1.Positive integers m and n satisfy


Find

2.Find the positive integer n such that

3.A trapezoid has side lengths

and
in some order. Find its area

4.A rectangle with integer side lengths has the property that its area minus 5 times its perimeter equals

Find the minimum possible perimeter of this rectangle.

5.For real numbers a, b, and c, the roots of the polynomial

form an
arithmetic progression. Find

Last edited by tony123 (2024-03-14 21:20:40)

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#2 2024-03-14 22:45:03

Bob
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Registered: 2010-06-20
Posts: 10,610

Re: 5 problems

Thanks Tony,

More (hopefully) later.

Bob

ps. Still stuck with http://www.mathisfunforum.com/viewtopic.php?pid=432520#p432520

Maybe I need a little hint please.


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2024-03-26 21:25:46

Bob
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Registered: 2010-06-20
Posts: 10,610

Re: 5 problems

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#4 2024-05-08 21:27:26

Keep_Relentless
Member
From: Queensland, Australia
Registered: 2024-05-05
Posts: 61

Re: 5 problems

For Q4, is it the area minus (5 times the perimeter),

As in A - 5P = 2023

Or is it (the area minus 5) times the perimeter,

As in (A - 5) * P = 2023

I'm guessing it's the first one??

Last edited by Keep_Relentless (2024-05-08 23:03:09)


"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

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#5 2024-05-08 22:36:31

Keep_Relentless
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From: Queensland, Australia
Registered: 2024-05-05
Posts: 61

Re: 5 problems

Last edited by Keep_Relentless (2024-05-08 23:02:32)


"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

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#6 2024-05-08 23:00:39

Keep_Relentless
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From: Queensland, Australia
Registered: 2024-05-05
Posts: 61

Re: 5 problems

Oh I forgot the sides are integers.

Last edited by Keep_Relentless (2024-05-08 23:32:24)


"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

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#7 2024-05-08 23:31:08

Keep_Relentless
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From: Queensland, Australia
Registered: 2024-05-05
Posts: 61

Re: 5 problems

I think for Q3 you use Brahmagupta's formula


"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

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#8 2024-05-09 00:33:02

Keep_Relentless
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From: Queensland, Australia
Registered: 2024-05-05
Posts: 61

Re: 5 problems

Regarding Q5, help is needed on finding roots of quintic equations and what's needed to see the answer, thanks smile

Last edited by Keep_Relentless (2024-05-09 00:33:29)


"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

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#9 2024-05-12 18:57:47

Keep_Relentless
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From: Queensland, Australia
Registered: 2024-05-05
Posts: 61

Re: 5 problems

Can anyone show how to do Q5??


"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

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#10 2024-05-12 19:36:48

Bob
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Registered: 2010-06-20
Posts: 10,610

Re: 5 problems

I called the roots a, a+d, a+2d, a+3d and a+4d and expanded that quintic. There were things about the resulting expression that suggested a + b  + c = 9, but I'm not confident that's the correct answer so this one is still on my 'to-do list'.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#11 2024-05-14 00:18:14

KerimF
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From: Aleppo-Syria
Registered: 2018-08-10
Posts: 238

Re: 5 problems

tony123 wrote:

5.For real numbers a, b, and c, the roots of the polynomial

form an
arithmetic progression. Find

This exercise has a solution if the right side of the equation is 32 instead of 320.


Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
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#12 2024-05-14 00:42:50

KerimF
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From: Aleppo-Syria
Registered: 2018-08-10
Posts: 238

Re: 5 problems

Well, I also found the solution in case the right side is 320.
a+b+c=49

Kerim


Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
But, by opposing it, such a human becomes no more of this world.

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#13 2024-05-14 00:44:53

Keep_Relentless
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From: Queensland, Australia
Registered: 2024-05-05
Posts: 61

Re: 5 problems

Interesting, any chance you can explain how you worked it out? We all had a go and didn't get there.

K_R


"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

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#14 2024-05-14 01:24:36

KerimF
Member
From: Aleppo-Syria
Registered: 2018-08-10
Posts: 238

Re: 5 problems

Keep_Relentless wrote:

Can anyone show how to do Q5??

(x-A)*(x-B)*(x-C)*(x-D)*(x-E) = 0

x^5 - (x^4)*(A + B + C + D + E) + (x^3)*(A*B + A*C + A*D + B*C + B*D + C*D + A*E + B*E + C*E + D*E) - (x^2)*(A*B*C + A*B*D + A*C*D + B*C*D + A*B*E + A*C*E + A*D*E + B*C*E + B*D*E + C*D*E) + x*(A*B*C*D + A*B*C*E + A*B*D*E + A*C*D*E + B*C*D*E] - A*B*C*D*E = 0
or
x^5 - (x^4)*(A + B + C + D + E) + (x^3)*(A*B + A*C + A*D + B*C + B*D + C*D + A*E + B*E + C*E + D*E) - (x^2)*(A*B*C + A*B*D + A*C*D + B*C*D + A*B*E + A*C*E + A*D*E + B*C*E + B*D*E + C*D*E) + x*(A*B*C*D + A*B*C*E + A*B*D*E + A*C*D*E + B*C*D*E] = A*B*C*D*E

We have:
x^5 - 10x^4 + ax^3 + bx^2 + cx = 320

By comparing, we get
A + B + C + D + E = 10
A*B + A*C + A*D + B*C + B*D + C*D + A*E + B*E + C*E + D*E = a
A*B*C + A*B*D + A*C*D + B*C*D + A*B*E + A*C*E + A*D*E + B*C*E + B*D*E + C*D*E = b
A*B*C*D + A*B*C*E + A*B*D*E + A*C*D*E + B*C*D*E = c
A*B*C*D*E = 320

A + B + C + D + E = 10

In general, Sum = [2*m + (n-1)*r]*n/2
where
m is the first term
r is the common difference
n the number of terms

[2*A + (5-1)*r]*5/2 = 10
A + 2*r = 2
r = (2 - A)/2

For A = -4 , r = 3
Therefore
A = -4
B = -1
C =  2
D = 5
E = 8

Their sum is 10
And their product is 320

a = A*B + A*C + A*D + B*C + B*D + C*D + A*E + B*E + C*E + D*E = -5
b = A*B*C + A*B*D + A*C*D + B*C*D + A*B*E + A*C*E + A*D*E + B*C*E + B*D*E + C*D*E = 190
c = A*B*C*D + A*B*C*E + A*B*D*E + A*C*D*E + B*C*D*E = -136

Therefore:
a + b + c = 49

Kerim

I did a mistake (sorry, a mistyping, not about the asnswer); will someone correct it? smile

Last edited by KerimF (2024-05-14 01:41:52)


Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
But, by opposing it, such a human becomes no more of this world.

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#15 2024-05-14 08:50:47

Europe2048
Member
Registered: 2024-01-03
Posts: 38

Re: 5 problems

KerimF, use the hide tag if you don't want to straight-up reveal the answer.

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