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1. n mod 122 = z
n mod 141 = z
n mod 179 = z.
Find two values of z and n that satisfy the equation above.
2. If x + y = xy, and x and y > 1, y=
a) x/(x-1), b)(x+1)/x, c)(x-1)/x, d)x/(x+1).
Manipulating the equation, I could not get one of the anwsers shown above.
3. 10,10,15,18,27 are numbers of a set N.
Find the probability when 2 randomly chosen numbers are in the ratio of 2:3. The 2 tens in the list are DISTINCT, i.e, 1st 10 and 15 and 2nd 10 and 15 are different.
I got 3/20 for this.
4. Adam could not remember his 3 digit combo to his combo lock at school. He remembered that the 3 digits were 1,3, and 4, but not the order. Nor did he remember if he turned it counter clockwise or clockwise after the first number. He did remember that he, after the 2nd number, turned it in the opposite direction than he did after the first number. How many different ways does Adam have to try to be sure he opens his lock?12?
5. There are 3 people Anne Beth and Carles. They all put their hats on top of their books. Anne, Beth and Carles grab a hat without looking(i.e, randomly.). What is the probability that noone chooses their own hat? . WITHOUT REPLACEMENT. Is this 5/6?
6.An classroom with 10 rows of seats has 10 seats in the front row, 11 seats in the 2nd row...students are permitted to sit in any row, but any student cannot sit next to another student. What is the MAXIMUM number of students that can be seated in the class.
Last edited by Mathegocart (2016-07-30 13:58:25)
The integral of hope is reality.
May bobbym have a wonderful time in the pearly gates of heaven.
He will be sorely missed.
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Hi;
1. n mod 122 = z
n mod 141 = z
n mod 179 = z.
Find two values of z and n that satisfy the equation above.
2. If x + y = xy, and x and y > 1, y=
a) x/(x-1), b)(x+1)/x, c)(x-1)/x, d)x/(x+1).
Manipulating the equation, I could not get one of the anwsers shown above.
3. 10,10,15,18,27 are numbers of a set N.
Find the probability when 2 randomly chosen numbers are in the ratio of 2:3. The 2 tens in the list are DISTINCT, i.e, 1st 10 and 15 and 2nd 10 and 15 are different.
I got 3/20 for this.
4.
4 seems to have been lost in the Bermuda Triangle.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi;
1. n mod 122 = z
n mod 141 = z
n mod 179 = z.
Find two values of z and n that satisfy the equation above.
2. If x + y = xy, and x and y > 1, y=
a) x/(x-1), b)(x+1)/x, c)(x-1)/x, d)x/(x+1).
Manipulating the equation, I could not get one of the anwsers shown above.
3. 10,10,15,18,27 are numbers of a set N.
Find the probability when 2 randomly chosen numbers are in the ratio of 2:3. The 2 tens in the list are DISTINCT, i.e, 1st 10 and 15 and 2nd 10 and 15 are different.
I got 3/20 for this.
4.4 seems to have been lost in the Bermuda Triangle.
For number 3, with.
The integral of hope is reality.
May bobbym have a wonderful time in the pearly gates of heaven.
He will be sorely missed.
Offline
Hi;
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
Hi;
But 3/20 would be without replacement, right?
The integral of hope is reality.
May bobbym have a wonderful time in the pearly gates of heaven.
He will be sorely missed.
Offline
Yes, without replacement there are 5 x 4 total choices. With replacement there are 5 x 5 total choices.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
Yes, without replacement there are 5 x 4 total choices. With replacement there are 5 x 5 total choices.
Bobbym, could you explain 1 ? Did you use Chinese Remainder theorem on no 1?
Last edited by Mathegocart (2016-07-30 14:17:41)
The integral of hope is reality.
May bobbym have a wonderful time in the pearly gates of heaven.
He will be sorely missed.
Offline
I have thought about CRT on it but settled on something faster. There are also other solutions.
Never use a cannon to kill a mosquito.
Use a laser!
https://www.youtube.com/watch?v=BKm8FolQ7jw
The easiest way seemed to be just multiply the three mods.
This leaves a remainder of 0 when divided by all three mods so z = 0.
Now just add one to get another solution
this will obviously now leave a remainder of 1 when divided by all three mods so z = 1
Of course, every solution must be backed up by another different solution. I ran it off by computer too.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
Last edited by thickhead (2016-07-30 17:12:31)
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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Hi;
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
Offline
Hi;
I tried number 5 again and I found out my mistake. Doh!
The integral of hope is reality.
May bobbym have a wonderful time in the pearly gates of heaven.
He will be sorely missed.
Offline
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