Can You Solve These Math and Logic Challenges?

  • Thread starter kant
  • Start date
In summary: Last three digits of 3^4798?By the pattern of the last three digits of powers of 3, it repeats every four powers. So, divide 4798 by 4, and the remainder is 2. So the last three digits are the same as 3^2, which is 9.
  • #1
kant
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1) Using a compass, and a straight edge( with no marks). Construct 4 points such that when you connect them, they would form a square.

2) Find all prime p such that 17p+1 is a perfect square.

3) You are in a island with three kinds of people.
Normal: might or might not lie.
knight: Always do the right thing, and tell the turth.
lier: That is obvious.

Suppose you encounter two person walking side-by-side in this island( label them A, and B). You go up to Person A, armed with a single question to ask A, you corructly deduce the identity of Person B. What is that unique question?

4) Suppose you have a 6 digit number. Suppose you switch the first 3 digit of that number to the last three digits of that number; the new number would be six times the original number. Obviously, my question is what is that number?

5) (x^2) + ( y^2) + 1 = (z^2) prove infinite many integer solution.

6) Last three digits of 3^4798?

7) You have a u shape tude. The open ends of the tude have the same cross sectional area, and is pointing up. Water is pored into the tude until the water level is M above the ground. Now, if i pore some oil into the right side of the tude, obviously, the water level on the left side of tude would rise by some number h, because of the inclusion of the oil. What is the height of the oil-air interface on the right side of the tude? what is the height of the water-oil interface on the right side of the tude? Intruduce more stuff if you need to.
 
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  • #2
kant said:
6) Last three digits of 3^4798?
som

889

eom
 
  • #3
3) Assuming that you don't know whether A or B are knights, knaves, or jesters, this can't be solved with a single yes or no question, since there are at least three possibilities for B, but only two possible answers.
 
  • #4
kant said:
1) Using a compass, and a straight edge( with no marks). Construct 4 points such that when you connect them, they would form a square.


1) Adjust compass to length L
2) Draw point A
3) Draw point B at L away from A. Draw line AB with straight edge.
4) Using the compass, sketch a circle around points A and B of radius L
5) The two circles intersect at points C and D. Draw CD with the straight edge, which bisects AB at point E.
6) Adjust the compass to the length of AE (length M)
7) Draw point F along CD, at M away from point E.
8) Draw circles of radius M around points A and F.
9) Circles intersect at two points, E and G.
10) Square is now formed by AEFG.


kant said:
2) Find all prime p such that 17p+1 is a perfect square.


Not sure why it seems to be the ONLY one, but I get 19


kant said:
3) You are in a island with three kinds of people.
Normal: might or might not lie.
knight: Always do the right thing, and tell the turth.
lier: That is obvious.

Suppose you encounter two person walking side-by-side in this island( label them A, and B). You go up to Person A, armed with a single question to ask A, you corructly deduce the identity of Person B. What is that unique question?

Are we assured that person A and person B are different? I can't seem to come up with a solution that isn't spoiled by the possibility of both people being 'normal', since, within the scope of only one question, they could emulate either a truth teller OR a liar, with no means of distinction, and no further questions to anyone else.

kant said:
4) Suppose you have a 6 digit number. Suppose you switch the first 3 digit of that number to the last three digits of that number; the new number would be six times the original number. Obviously, my question is what is that number?


142857


kant said:
5) (x^2) + ( y^2) + 1 = (z^2) prove infinite many integer solution.

Hmm...

kant said:
6) Last three digits of 3^4798?


Well, the digits repeat, so you can figure out based on that: 889


kant said:
7) You have a u shape tude. The open ends of the tude have the same cross sectional area, and is pointing up. Water is pored into the tude until the water level is M above the ground. Now, if i pore some oil into the right side of the tude, obviously, the water level on the left side of tude would rise by some number h, because of the inclusion of the oil. What is the height of the oil-air interface on the right side of the tude? what is the height of the water-oil interface on the right side of the tude? Intruduce more stuff if you need to.


Height of oil/water: M-h
Height of oil/air: (h*mass of water)/(mass of oil)
(think that's right)


DaveE
 
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  • #5
kant said:
5) (x^2) + ( y^2) + 1 = (z^2) prove infinite many integer solution.

Acha!


if x is a power of 2, then using:

y = 2^(2*(lg x)-1)

(lg being log base 2) which yields:

z = 2^(2*(lg x)-1)+1

So, for example:

2^2 + 2^2 + 1 = 3^2
4^2 + 8^2 + 1 = 9^2
8^2 + 32^2 + 1 = 33^2
16^2 + 128^2 + 1 = 129^2
32^2 + 512^2 + 1 = 513^2

And since there are an infinite number of powers of 2, there are obviously an infinite number of integer solutions.



DaveE
 

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