Some multi-choice questions...

Toast · 2007-01-22 21:04:32

A woman has three school age children. The product of her age and the ages of her three children is 16555. The difference between the ages of her eldest and youngest children is:

(A) 4 (B) 5 (C) 6 (D) 7 (E) 11

Not much of a clue... something to do with factors maybe
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Consider all the numbers, bigger than 8, which when divided by 2, by 3, by 4, by 5, by 6, by 7 and by 8, each time give a remainder of 1. Find the sum of the two smallest such numbers.

(A) 842 (B) 2522 (C) 3362 (D) 912 (E) 2532

Not much of a clue...
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A cube is given. How many planes are there which pass through at least three of its vertices?

(A) 8 (B) 12 (C) 16 (D) 20 (E) 36

What is this question even asking???

Dross · 2007-01-22 21:22:07

For the first question...

We know that 16555 is the product of four integers (the woman's and children's ages). Note that you can split 16555 into it's prime factors in the following way:

16555 = 5 x 3311
= 5 x 11 x 301
= 5 x 11 x 7 x 43

16555 = 5 x 7 x 11 x 43

Now, there happens to be a theorem that says that any integer can be represented uniquely by a product of prime numbers. That means that there is no other way to multiply any number of prime numbers together to get 16555 other than the way we've just found. So those four numbers above are the only ones that could have been multiplied together to get 16555, which means that the difference in age between the eldest and youngest child is 6 years.

Was that all clear?

Toast · 2007-01-22 21:28:09

Yeah, very clear, thanks Dross XD

Dross · 2007-01-22 21:51:28

For the second question, I'm not sure but I think a start might be that we're looking for the numbers n such that:

2|(n-1)
3|(n-1)
.
.
8|(n-1)

(where a|b means "a divides b")

To get the smallest such number, notice that if 8|(n-1), then 2|(n-1) since 2 is a factor of 8 - similarly with 3 and 4. Then (5 x 6 x 7 x 8 + 1) gives us the smallest number with the required property. Not sure how to get the next smallest number, though.

For the third question, imagine a cube sat on a set of 3-dimensional coordinate axes. Now draw a plane (as in 2-dimensional flat surface) which intersects that cube - move around until it intersects with three vertices (corners of the cube). There are a few ways you could do this:

1) have the plane line up with a face of the cube (8 faces = 8 ways)
2) have the plane go through two opposite edges of the cube (12 edges = 6 ways)

there are other ways, gotta go do some "work" now though - hope this helps!

mathsyperson · 2007-01-22 22:59:29

I agree with most of Dross's work for the second question so far, but he's repeated a factor of 2 in the 6 and the 8. All you actually need is 8x3x5x7 to make a number that's divisible by everything from 2 to 8, and then you add one to get a number where the remainder is one all the time.

Also, the next number with the needed property is almost certainly 2x(8x3x5x7)+1. I don't see how any other number would work.
Adding the two numbers together will give 3x(8x3x5x7)+2 = 2522.
And as that's one of the options given, it's quite likely that that's the right answer.

---

For 3), there's another way of looking at it that you might find easier. You can define a plane by 3 points. That is, if you chose three vertices on your cube, then you could always draw a plane that goes through them all. So therefore, you could just choose combinations of vertices and count how many combinations you have at the end.

The problem with that method is that there's lots of repetition. There are quite a few planes that go through 4 vertices instead of just 3, and so you'd need to consider that to make sure you're not counting things twice. But who knows, you might find this way easier. It's another option, anyway.

Toast · 2007-01-23 03:27:18

Hmmm, how did you come up with the conclusion that 8*3*5*7 will give the lowest multiple of 1 through 8?

mathsyperson · 2007-01-23 04:05:53

Every number is a multiple of 1, so that doesn't matter.
Then we need our number to be a multiple of 2, so it needs to have a factor of 2.
In addition, it needs to be a multiple of 3, so it needs to have factors of 2*3.
It also needs to be a multiple of 4, but 4=2*2, so if we include 4 then we don't need the 2 anymore because that's already covered. So now we have 4*3.

To make it a multiple of 5, we need a factor of 5 in there: 4*3*5.
The 6 is already covered because 6=2*3, and those two are already in.

We need to have a factor of 7 for it to be divisible by 7: 4*3*5*7.
And using similar reasoning to what we did for 4, we can replace the 4 with an 8 to include 8 but also still include 4.

Therefore, we have 8*3*5*7. And then to get a remainder of 1 all the time, we just do (8*3*5*7)+1.

Sorry if that's badly explained, but I'm in a bit of a hurry because I've got an exam in about half an hour. Integration and Differential Equations. Should be great fun.

Toast · 2007-01-23 05:55:35

Thanks mathsy and dross. Questions 1-2 make sense and I'll try gnawing through question 3.

Oh, and I hope you did well in your exam mathsy, next time if you've got an exam coming up you should attend to it first instead of worrying about my question.

mathsyperson · 2007-01-23 07:05:14

Don't worry, I'd spent quite a while revising before I posted that. I answered your question while I was having a break.

I think I did alright. The exam has two sections in it: one that has lots of little questions worth a few marks each, and then another one that has 3 big questions and you need to answer two of them.

Most of the first section was fine, but one of the big questions I chose to do took me ages. Not really because it was difficult, but because I kept making stupid mistakes in it. So then by the time I'd finally finished that, I only had 5 minutes left to do the other big question. I did as much of it as I could, but it was only half-finished when my time was up.

Still, I'm content with how I did. I think I'll get around 70% or something, which is perfectly reasonable. It could be better, but it could also be a lot worse so it's fine.

Anyway, enough about me. Let's look at this cube again.

---

I think the best way to solve this is to combine my and Dross's methods.

8C3 = 56, so according to my method, there are 56 planes that go through 3 vertices on the cube. Unfortunately, some of those planes are the same as each other, because there are some planes that go through 4 vertices. Each plane that goes through 4 vertices is counted 4 times, when it should only be counted once.

Thankfully, Dross has made a list of such planes.
He started off by making the planes coincide with faces of the cube, so that's 6 (not 8 ) planes there. Then he's made them intersect opposite edges, which makes another 6. So we have 12 planes that go through 4 vertices.

Each of these has been counted 4 times by my method, when it should only have been counted once. So to correct this, we do 56 - 12*3 = 20. Now there are 20 planes.

It's possible that there may have been some more 4-vertex planes that Dross missed, but if that was the case, then we'd need to take away another 3 for each one there was. However, the only given option that you can make by taking multiples of 3 away from 20 is 8.
There can't only be 8 planes there, because Dross has counted 12 of them.

Therefore, the only option that can possibly be right is 20.

dr_shiv_88 · 2007-05-12 04:33:43

i actually got 36, for the third question. (12 that pass through 4 vertices's and 24 that pass through 3). Are there answers to these questions anywhere? Anyway, i'm not so good at explaining so just email me if you don't get this.

you get the first 6 which are the actual 6 faces of the cube.

you get 6 more if you get planes passing from one side of the cube to the extreme other side of the cube

and 24 is if you draw a line from one vertex to the extreme opposite vertex on the cube and create a plane passing through a third vertex. As there are four combinations for extreme opposite vertex's and 6 other vertex's for each combination to pass through, that makes 8 x 3.

see i told you i can't explain very well lol.

mathsyperson · 2007-05-12 10:58:20

I just drew a few diagrams, and found out that any plane that contains 2 extreme opposite vertices has to have either 2 or 4 vertices in total, which means that they've either been counted before or are uninteresting.

killea · 2021-03-01 08:44:41

The answer to the cube question is 20.
6 + 8 + 2*6/2 = 6+ 8+ 6 =20
The cube has 6 surfaces, each surface is a plane, this is the first 6 in the equation.
The cube has 8 vertices, each vertice faces a triangle plane, there are 8 triangle plane(8 in the equation)
The cube has 6 surfaces, we see each surface has two crossed planes if we look at the surface ⊥ly ( looks like an X ), however, actually we counted the planes twice, and that's why we need to divide it by 2.

Math Is Fun Forum

#1 2007-01-22 21:04:32

Some multi-choice questions...

#2 2007-01-22 21:22:07

Re: Some multi-choice questions...

#3 2007-01-22 21:28:09

Re: Some multi-choice questions...

#4 2007-01-22 21:51:28

Re: Some multi-choice questions...

#5 2007-01-22 22:59:29

Re: Some multi-choice questions...

#6 2007-01-23 03:27:18

Re: Some multi-choice questions...

#7 2007-01-23 04:05:53

Re: Some multi-choice questions...

#8 2007-01-23 05:55:35

Re: Some multi-choice questions...

#9 2007-01-23 07:05:14

Re: Some multi-choice questions...

#10 2007-05-12 04:33:43

Re: Some multi-choice questions...

#11 2007-05-12 10:58:20

Re: Some multi-choice questions...

#12 2021-03-01 08:44:41

Re: Some multi-choice questions...

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