# Thread: Help with conjectures (proof)

1. ## Help with conjectures (proof)

Directions : By trying the following examples, and more if necessary, (at least one more) try to make a conjecture (educated guess) inspired by the examples, as indicated. In each case, once you have a conjecture, say how confident you are that your conjecture is correct. (How much are you willing to bet on it?) Then, if you can, provide a completely convincing argument that your conjecture is correct or is incorrect, as appropriate.

Need help here's the question
#2)
a. Examples: copy and complete: 1=1; 1+3=...;1+3+5=...;1+3+5+7=...;
b. Complete to make a conjecture: for n a natural number, 1+3+...+(2n+1)=...

c. Examples: copy and complete: 1=1; 1+8=...;1+8+27=...;1+8+27+64=...;
d. Complete to make a conjecture: for n a natural number, 1+8+...+n^3)=...

e. Say whether each of the following is or is not prime
• 1^2+1+41;
• 2^2+2+41;
• 3^2+3+41;
• 4^2+4+41;

f. make a conjecture: for n a natural number is n^2+n+41 is ...

g.express the even number as a sum of two prime numbers.
• 4=...
• 6=...
• 8=...
• 10=...

h. Complete the conjecture: Every even number ...

MY WORK:

a. 1= 1; 1+3= 4;1+3+5= 9;1+3+5+7= 16;
b. for n a natural number, 1+3+...+(2n+1)= ???

c. 1= 1; 1+8= 9;1+8+27= 36;1+8+27+64= 100...
d. for n a natural number, 1+8+...+n^3)= ???

e.

• 1^2+1+41= 43 is a prime #
• 2^2+2+41= 47 is a prime #
• 3^2+3+41= 53 is a prime #
• 4^2+4+41= 61 is a prime #

f. conjecture: n^2+n+41 will result in a prime number.

g.

• 4= 2 + 2
• 6= 3 + 3
• 8= 5 + 3
• 10= 5 + 5

h. conjecture: Every even number can be the sum of two prime numbers.

2. ## Re: Help with conjectures (proof)

Originally Posted by Jay23456
Directions : By trying the following examples, and more if necessary, (at least one more) try to make a conjecture (educated guess) inspired by the examples, as indicated. In each case, once you have a conjecture, say how confident you are that your conjecture is correct. (How much are you willing to bet on it?) Then, if you can, provide a completely convincing argument that your conjecture is correct or is incorrect, as appropriate.

Need help here's the question
#2)
a. Examples: copy and complete: 1=1; 1+3=...;1+3+5=...;1+3+5+7=...;
b. Complete to make a conjecture: for n a natural number, 1+3+...+(2n+1)=...

c. Examples: copy and complete: 1=1; 1+8=...;1+8+27=...;1+8+27+64=...;
d. Complete to make a conjecture: for n a natural number, 1+8+...+n^3)=...

e. Say whether each of the following is or is not prime
• 1^2+1+41;
• 2^2+2+41;
• 3^2+3+41;
• 4^2+4+41;

f. make a conjecture: for n a natural number is n^2+n+41 is ...

g.express the even number as a sum of two prime numbers.
• 4=...
• 6=...
• 8=...
• 10=...

h. Complete the conjecture: Every even number ...

MY WORK:

a. 1= 1; 1+3= 4;1+3+5= 9;1+3+5+7= 16;
b. for n a natural number, 1+3+...+(2n+1)= ???

c. 1= 1; 1+8= 9;1+8+27= 36;1+8+27+64= 100...
d. for n a natural number, 1+8+...+n^3)= ???

e.

• 1^2+1+41= 43 is a prime #
• 2^2+2+41= 47 is a prime #
• 3^2+3+41= 53 is a prime #
• 4^2+4+41= 61 is a prime #

f. conjecture: n^2+n+41 will result in a prime number.

g.

• 4= 2 + 2
• 6= 3 + 3
• 8= 5 + 3
• 10= 5 + 5

h. conjecture: Every even number can be the sum of two prime numbers.
In a) haven't you seen the pattern 1, 4, 9, 16, 25, 36, ... before?

3. ## Re: Help with conjectures (proof)

I can't recall...

ooooooo wait is it

x^2 ?

Okay here an updated work:

2.
a. 1= 1; 1+3= 4;1+3+5= 9;1+3+5+7= 16; The Next one would be 1 + 3 + 5 + 7 + 9 = 25
b. The sums of each of these sequences are all perfect squares. For n a natural number, 1+3+...+(2n+1) = n^2 .

c. 1= 1; 1+8= 9;1+8+27= 36;1+8+27+64= 100; The next sequence would be 1 + 8 + 27 + 64 + 125 = 225.

d. If you look at all the solutions 1,9,36,100,225, you’ll notice they that are indeed perfect squares there square roots are 1,3,6,10,15 the difference between these roots is 2,3,4,5. A conjecture can be made that: for n a natural number, 1+8+...+n^3) = [(n^2 + n)/2]^2

e.

• 1^2+1+41= 43 is a prime #
• 2^2+2+41= 47 is a prime #
• 3^2+3+41= 53 is a prime #
• 4^2+4+41= 61 is a prime #

f. conjecture: n^2+n+41 will result in a prime number, as long as n is not equal to 41.

g.

• 4 = 2 + 2
• 6 = 3 + 3
• 8 = 5 + 3
• 10 = 5 + 5

h. conjecture: Every even number can be the sum of two prime numbers.

4. ## Re: Help with conjectures (proof)

For f, how did you determine that n must not be 41? Are there other numbers for which the polynomial is not prime?

There is a slight caveat for part h. Can you find out what it is?

Otherwise the rest look like perfectly reasonable conjectures.

5. ## Re: Help with conjectures (proof)

oh good point it should be n must not equal 41k, k being any natural number.

What do you mean by "caveat" should I write it like this

Every even number is the sum of two prime numbers.

Also another question I think it to be much more challenging.

There are three boxes of candy in front of you. One box contains chocolates and mints. One contains mints only, and one contains chocolates only. All three boxes are incorrectly labeled. What is the smallest number of candies you need to remove and sample from the boxes (without looking in the boxes) to be able to correctly label all three boxes. Explain.

Hmm was gonna say 2 from each jar but what if you selected two mints from the mixed jar...then that would be untrue.

6. ## Re: Help with conjectures (proof)

You need to check your answer for part b again. It's not quite right ... example,
We have that 1 + 3 + 5 = 9
According to the problem, 2n + 1 = 5 for the above sum.
If 2n + 1 = 5, then n = 2 ... but 2^2 = 4, not 9 .... so it's not n^2

Hint: If the problem had said 2n - 1, then n^2 would be correct

7. ## Re: Help with conjectures (proof)

ah your right! It should be: 1+3+...+(2n+1) = (n + 1)^2

Originally Posted by Jay23456
There are three boxes of candy in front of you. One box contains chocolates and mints. One contains mints only, and one contains chocolates only. All three boxes are incorrectly labeled. What is the smallest number of candies you need to remove and sample from the boxes (without looking in the boxes) to be able to correctly label all three boxes. Explain.

8. ## Re: Help with conjectures (proof)

For part f, if you try it with 44, you will get $\displaystyle 44^2+44+41 = 2021 = 43\cdot 47$. The point is that for $\displaystyle n > 40$ there are a lot of cases where the polynomial does not give you a prime.

9. ## Re: Help with conjectures (proof)

oh your right 44 does work should i change my conjecture to make it for n < 41 ?

10. ## Re: Help with conjectures (proof)

Originally Posted by Jay23456
I can't recall...

ooooooo wait is it

x^2 ?

Okay here an updated work:

2.
a. 1= 1; 1+3= 4;1+3+5= 9;1+3+5+7= 16; The Next one would be 1 + 3 + 5 + 7 + 9 = 25
b. The sums of each of these sequences are all perfect squares. For n a natural number, 1+3+...+(2n+1) = n^2 .
Squares, yes, but not $\displaystyle n^2$. 5, for example, is 2(2)+ 1 but $\displaystyle 2^2\ne 9= 3^2$. 7= 2(3)+ 1 but $\displaystyle 3^2\ne 16= 4^2$. Can you see what is squared?

c. 1= 1; 1+8= 9;1+8+27= 36;1+8+27+64= 100; The next sequence would be 1 + 8 + 27 + 64 + 125 = 225.

d. If you look at all the solutions 1,9,36,100,225, you’ll notice they that are indeed perfect squares there square roots are 1,3,6,10,15 the difference between these roots is 2,3,4,5. A conjecture can be made that: for n a natural number, 1+8+...+n^3) = [(n^2 + n)/2]^2

e.

• 1^2+1+41= 43 is a prime #
• 2^2+2+41= 47 is a prime #
• 3^2+3+41= 53 is a prime #
• 4^2+4+41= 61 is a prime #

f. conjecture: n^2+n+41 will result in a prime number, as long as n is not equal to 41.

g.

• 4 = 2 + 2
• 6 = 3 + 3
• 8 = 5 + 3
• 10 = 5 + 5

h. conjecture: Every even number can be the sum of two prime numbers.

11. ## Re: Help with conjectures (proof)

The three boxes problem is kind of vague. There's different ways of interpreting the problem.
My first interpretation (and I think this interpretation is most accurate) assumes that you know the number of chocs and mints in the mixed box. If there's x chocs and y mints in the mixed box, then you'd need to get x+1 or y+1 candies, whichever is bigger (or smaller, I'll discuss this in a bit), from each box . The idea is that in the mixed box, you may end up picking either all chocs or all mints, then you'd need to pick one more to definitely get the other type to confirm that that box is the mixed box, and the other two boxes will follow trivially (you can look up "Pigeon Hole Principle" to get a better idea). Now, regarding the bigger/smaller problem. You would choose the bigger of x+1 or y+1 so that you'll ALWAYS be able to label the boxes correctly. If you choose the smaller, then that would be the case if you ended up picking all the chocs or mints (whichever is smaller) plus one, and you'd be able to label the boxes correctly after that. But, if you don't pick all the chocs/mints (whichever is smaller), then you can't label the boxes correctly with 100% certainty.
With that in mind, technically, you don't need to pick any candies to be able to label the boxes correctly, you can do it "accidentally". Again, this is under the assumption that you know how many chocs and mints are in the mixed box.

Another approach, which I'm not sure is correct is the right approach, is to take advantage of the fact that the boxes are labelled incorrectly. Under the assumption that the labels simply need rearranging (i.e. there are only three labels, one for each box), then you would just need to empty out one box to be able to determine what the other two boxes should be labelled. I'll generalize and say we have three boxes X,Y,Z which should be labelled x,y,z, respectively. The boxes are labelled incorrectly, so let's say X is labelled y (I'll write X-y from now on). Then we have either Y-x or Y-z, but if we have Y-x, then the only label left is z and we will have Z-z which contradicts our incorrect labeling situation. So we have X-y, Y-z, and Z-x. To generalize what happened, if Box A is labelled b, then Box B cannot be labelled a (because if you do that, then you'll have Box C labelled c), and thus box B must be labelled c. So, back to X-y, Y-z, and Z-x, if you confirm that X-y, that means the box labelled x cannot be Y, so the box labelled x must be Z, and the last box must be labelled y.
Example, you confirm that the only chocs box is labelled as only mints. This means that the box labelled as only chocs must be the mixed box, and the box labelled as mixed must be the only mints box. It's kinda confusing, you can use a diagram to help understand what's going on . If you want, you can think of the boxes as numbered 1,2, and 3, and the position of the number is its label. So 1,2,3 is the correct positions of the boxes so that they're labelled correctly. 2,3,1 is a valid list where all the boxes are labelled incorrectly, but 2,1,3 is not because 3 is in its right position. Again, this is under the assumption stated before.
If you remove that assumption, and allow for repeated labels, then you'd need to empty out two boxes, which isn't very helpful, so I doubt this case would be an appropriate interpretation.

Ofcourse, there's other practical things you could do, like using densities, etc. But I think one of the two above would be appropriate, just not sure which.

12. ## Re: Help with conjectures (proof)

the solution to the 3 boxes problem is simple:

all 3 boxes are labelled incorrectly. so take one candy (just one!) from the box labelled "mints and chocolates".

if it is a mint, then it is the box of mints. thus the box labelled "chocolates" cannot be chocolates, nor can it be mints, so it must be the mixed candies, and the box labelled "mints" must be the chocolates.

if, however, it is a chocolate, then box labelled "mints" cannot be mints or chocolates, and thus THAT one must be the mixed candies, and the box labelled "chocolates" must be mints.

EDIT: on the caveat on "h":

have you tried to express 2 as the sum of two primes?

h is a very famous conjecture, known as Goldbach's conjecture. no proof is known, but it is widely regarded as true (it has been computer verified for all integers less than 4 x 1018, according to Wikipedia).