Join the Unexplained Mysteries community today! It's free and setting up an account only takes a moment.

# Mathematics Brain Teasers. ## Recommended Posts  18 hours ago, Ozymandias said:

That's a cool one.

Here's my one.

Find the sets of  whole numbers whose sum is equal to their total when multiplied?

• Replies 426
• Created

#### Top Posters In This Topic

• 166

• 144

• 42

• 25

#### Top Posters In This Topic

• Ozymandias 166 posts

• danydandan 144 posts

• ant0n 42 posts

• Golden Duck 25 posts

#### Posted Images  4 hours ago, danydandan said:

Find the sets of  whole numbers whose sum is equal to their total when multiplied?

Assuming I have understood the question correctly, I can only find one such set:

{1,2,3}  where 1 + 2+ 3 = 6

and 1 x 2 x 3 = 6

Edited by Ozymandias
##### Share on other sites  15 hours ago, Ozymandias said:

Assuming I have understood the question correctly, I can only find one such set:

{1,2,3}  where 1 + 2+ 3 = 6

and 1 x 2 x 3 = 6

Yeap that is one of them;

2+2=4, 2x2=4 is another.

##### Share on other sites  1 hour ago, danydandan said:

Yeap that is one of them;

2+2=4, 2x2=4 is another.

Only half right then! Should have gotten that other one. You'll have to set another problem.

Edited by Ozymandias
##### Share on other sites  59 minutes ago, Ozymandias said:

Only half right then! Should have gotten that other one. You'll have to set another problem.

I only have one more, I heard it on a podcast I was listening to.

Using only addition and the number 8. Show me an equation that equals to 1000.

##### Share on other sites  10 hours ago, danydandan said:

I only have one more, I heard it on a podcast I was listening to.

Using only addition and the number 8. Show me an equation that equals to 1000.

8+8+8+8+8+8+....... = 1000  (where you add 125 number 8s in a row together - I'm not writing out 125 number 8s, but you get the idea).

Or, if you like, 888+88+8+8+8 = 1000

Edited by Ozymandias
##### Share on other sites  6 hours ago, Ozymandias said:

8+8+8+8+8+8+....... = 1000  (where you add 125 number 8s in a row together - I'm not writing out 125 number 8s, but you get the idea).

Or, if you like, 888+88+8+8+8 = 1000

That's one solution,

888+88+8+8+8=1000. Is the one I had.

##### Share on other sites  John is twice as old as Bill. Fred is 15 years younger than Bill. If John was ten years older he would be four as old as Fred is now. What is the ages of the three men.

##### Share on other sites  8 minutes ago, Ozymandias said:

John is twice as old as Bill. Fred is 15 years younger than Bill. If John was ten years older he would be four as old as Fred is now. What is the ages of the three men.

J = 2B

F = B - 15

J + 10 = 4F

Therefore Fred is 20, John is 70 and Bill is 35.

I'll provide a new teaser later. I've gotta run.

##### Share on other sites  Let ABCDEFGH be a cube.

Let (A; vectorAB, vectorAD, vectorAE) be the coordinate system.

Let P be the plan which equation is x + 1/2.y + 1/3.z - 1 = 0.

-> Please draw the section of the cube by the plan P AND explain your reasoning. ##### Share on other sites  I have been very busy yesterday and this am, but here is my solution:-

This is a three-dimensional vector algebra question and, although it is simple enough, I would think it is undoubtedly beyond most peoples' capability. The general equation of a plain in a 3D space is given as ax + by + cz = d where the intercept points of the section are given by (d/a, 0, 0), (0, d/b, 0) and (0, 0, d/c). Our plain section P is given by x + (1/2)y + (1/3)z - 1 = 0

ax + by + cz = d

x + (1/2)y + (1/3)z - 1 = 0

x + (1/2)y + (1/3)z = 1     so a = 1, b = 1/2, c = 1/3 and d = 1

Intercept on X-axis where y = z = 0: (d/a, 0 , 0) = (1/1, 0, 0) = (1, 0, 0)

Intercept on Y-axis where x = z = 0: (0, d/b , 0) = (0, 1/(1/2), 0) = (0, 2, 0)

Intercept on Z-axis where x = y = 0: (0, 0 , d/c) = (0, 0, 1/(1/3)) = (0, 0, 3)

These three points dilineate the plain triangular section P (see diagram above)

Edited by Ozymandias
##### Share on other sites  1 hour ago, Ozymandias said:

I have been very busy yesterday and this am, but here is my solution:-

This is a three-dimensional vector algebra question and, although it is simple enough, I would think it is undoubtedly beyond most peoples' capability. The general equation of a plain in a 3D space is given as ax + by + cz = d where the intercept points of the section are given by (d/a, 0, 0), (0, d/b, 0) and (0, 0, d/c). Our plain section P is given by x + (1/2)y + (1/3)z - 1 = 0

ax + by + cz = d

x + (1/2)y + (1/3)z - 1 = 0

x + (1/2)y + (1/3)z = 1     so a = 1, b = 1/2, c = 1/3 and d = 1

Intercept on X-axis where y = z = 0: (d/a, 0 , 0) = (1/1, 0, 0) = (1, 0, 0)

Intercept on Y-axis where x = z = 0: (0, d/b , 0) = (0, 1/(1/2), 0) = (0, 2, 0)

Intercept on Z-axis where x = y = 0: (0, 0 , d/c) = (0, 0, 1/(1/3)) = (0, 0, 3)

These three points dilineate the plain triangular section P (see diagram above)

Your cube is misplaced in that coordinate system: the length of each of its sides is 1 unit.

##### Share on other sites  23 minutes ago, ant0n said:

Your cube is misplaced in that coordinate system: the length of each of its sides is 1 unit.

On your original diagram you did not say that your cube had sides representing orthogonal unit vectors. There was no scaling given. You only indicated three directions.  I have not the time now to redraw the diagram from scratch showing the correct triangular section through your cube.

Edited by Ozymandias
##### Share on other sites  6 minutes ago, Ozymandias said:

On your original diagram you did not say that your cube had sides representing orthogonal unit vectors. There was no scaling given. You only indicated three directions.  I have not the time now to redraw the diagram from scratch showing the correct triangular section through your cube.

Calm down.

That exercise is a literal translation of an official exercise for the oldest pupils in high school in France. As shown, the coordinate system is an orthonormal one, the magnitude of vectors AB, AD & AE being 1 unit.

##### Share on other sites  54 minutes ago, Ozymandias said:

On your original diagram you did not say that your cube had sides representing orthogonal unit vectors. There was no scaling given. You only indicated three directions.  I have not the time now to redraw the diagram from scratch showing the correct triangular section through your cube.

For your information, here's the original exercise ("bac" = "A Levels"): Edited by ant0n
##### Share on other sites  OK, Anton. I accept what you say although the original problem did not state that the vectors were orthonormal whereas their orthogonality is obvious. Anyway, I have redrawn the visual solution and added some explanatory details underneath the diagram:- When you rescale the cube and the axes, the section through the cube is no longer triangular but is now represented by the red-shaded plain abcd in the diagram above. By vector geometry the parallelogram sides ab is parallel to dc and ad is parallel to bc. The co-ordinates of points a, b c, and d are easily found by inspection using known points, ratio and/or vector translation. For example, the rule that translates point a to d will be the same that translates b to c; that is, for the x co-ordinate subtract 1/3, for the y co-ord no change, and for the z co-ord add 1. So b(1/2, 1, 0) goes to c(1/6, 1, 1)

a = ( 1, 0, 0)          b = (1/2, 1, 0)          c = (1/6, 1, 1)          d = (2/3, 0, 1)

You can show the same result using caculus but I think the vector/geometric explanation is easier to understand.

##### Share on other sites  3 hours ago, Ozymandias said:

OK, Anton. I accept what you say although the original problem did not state that the vectors were orthonormal whereas their orthogonality is obvious.

You know, when I was translating that exercise, I was about to add 'orthonormal'. I was very close to doing it but I finally assumed the original exercise was accurate enough... Next time, I won't hesitate to add details that are not specified in the original exercise.

Edited by ant0n
##### Share on other sites  3 hours ago, Ozymandias said:

OK, Anton. I accept what you say although the original problem did not state that the vectors were orthonormal whereas their orthogonality is obvious. Anyway, I have redrawn the visual solution and added some explanatory details underneath the diagram:- When you rescale the cube and the axes, the section through the cube is no longer triangular but is now represented by the red-shaded plain abcd in the diagram above. By vector geometry the parallelogram sides ab is parallel to dc and ad is parallel to bc. The co-ordinates of points a, b c, and d are easily found by inspection using known points, ratio and/or vector translation. For example, the rule that translates point a to d will be the same that translates b to c; that is, for the x co-ordinate subtract 1/3, for the y co-ord no change, and for the z co-ord add 1. So b(1/2, 1, 0) goes to c(1/6, 1, 1)

a = ( 1, 0, 0)          b = (1/2, 1, 0)          c = (1/6, 1, 1)          d = (2/3, 0, 1)

You can show the same result using caculus but I think the vector/geometric explanation is easier to understand.

Well done, congratulations! Your turn ##### Share on other sites  14 hours ago, Ozymandias said:

Co-ordinates of the points of the parallelogram:   a = ( 1, 0, 0)    b = (1/2, 1, 0)    c = (1/6, 1, 1)    d = (2/3, 0, 1)

My new problem is based on the last problem.

Problem: The section through the cube represented by the parallelogram abcd cuts the volume of the cube into two unequal parts. Using geometry and mensuration calculate the volume of the smaller of these two parts as a percentage of the whole cube.

Edited by Ozymandias
##### Share on other sites  No takers? Will I put up the solution?

##### Share on other sites  2 hours ago, Ozymandias said:

No takers? Will I put up the solution?

Wait a moment, I'm going to try to find that solution.

##### Share on other sites  On 9/16/2020 at 11:23 AM, Ozymandias said:

My new problem is based on the last problem.

Problem: The section through the cube represented by the parallelogram abcd cuts the volume of the cube into two unequal parts. Using geometry and mensuration calculate the volume of the smaller of these two parts as a percentage of the whole cube.

Volume of the bigger half of the sectioned cube, EFcdabIO:

- area of EFcd = (dE + cF)/2.EF

d(2/3; 0; 1), E(0; 0; 1) -> dE = ((2/3)² + 0² +  0² )^0.5 = 2/3

c(1/6; 1; 1), F(0; 1; 1) -> cF = 1/6

U = area(EFcd) = (2/3 + 1/6)/2 = 5/12 square unit

- area of abIO = (aO + bI)/2.OI

aO = 1

b(1/2; 1; 0), I(0; 1; 0) -> bI = 1/2

V = area(abIO) = 3/4 square unit

Volume(EFcdabIO) = (U + V)/2 x FI = (5/12 + 3/4)/2 = 7/12 cubic unit

therefore the volume of the smaller half is (1 - 7/12) = 5/12 cubic unit = approximately 41.67% of the volume of that cube.

Edited by ant0n
##### Share on other sites  16 minutes ago, ant0n said:

Volume of the bigger half of the sectioned cube, EFcdabIO:

- area of EFcd = (dE + cF)/2.EF

d(2/3; 0; 1), E(0; 0; 1) -> dE = ((2/3)² + 0² +  0² )^0.5 = 2/3

c(1/6; 1; 1), F(0; 1; 1) -> cF = 1/6

U = area(EFcd) = (2/3 + 1/6)/2 = 5/12 square unit

- area of abIO = (aO + bI)/2.OI

aO = 1

b(1/2; 1; 0), I(0; 1; 0) -> bI = 1/2

V = area(abIO) = 3/4 square unit

Volume(EFcdabIO) = (U + V)/2 x FI = (5/12 + 3/4)/2 = 7/12 cubic unit

therefore the volume of the smaller half is (1 - 7/12) = 5/12 cubic unit = approximately 41.67% of the volume of that cube.

That is correct, Anton. My solution took a slightly different route. Refer to diagram below:- Point Co-ordinates:    a(1, 0, 0)      b(1/2, 1, 0)      c(1/6, 1, 1)      d(2/3, 0, 1)

Right-angles: agb = eht = bfc = aed = 90º

The parallelogram abcd represents the plain along which the cube is cut into two parts. We are interested in the volume of the smaller part nearest to us as we look at the diagram. This in turn is made up of two parts: a regular prism agbfeh whose base area agb (shaded blue) is the same as its top area ehf,  and a wedge or oblique prism abcdef whose end area bfc (shaded red) is the same as its other end area aed.

Volume of the Regular Prism agbfeh = Cross-sectional area x length

= Blue-shaded Area agb x Height gh

= 1/2 (base ag)(perp. height gb) x gh

=  1/2 (1)(1/2)  x  1

=  1/4 unit^3

Volume of the Oblique Prism abcdef = Red-shaded Area bfc  x Perp. Height (green line)

= 1/2 (base fb)(perp. height fc) x (green line)

=  1/2 (1)(1/2 -1/6)  x  1            x         1

=      1/2 (1)(1/3)  x  1               x         1

=   1/6 unit^3

Total Volume = 1/4 + 1/6 units^3      =   5/12 unit^3

Percentage of Whole Cube                  = [(5/12)/1] x 100             =          41.666%

• 1
##### Share on other sites  Great Express this complex number z = (1 - i.sqrt(3))^11 in the form z = |z|.e^(arg(z)). Edited by ant0n
##### Share on other sites  3 hours ago, ant0n said:

Volume of the bigger half of the sectioned cube, EFcdabIO:

- area of EFcd = (dE + cF)/2.EF

d(2/3; 0; 1), E(0; 0; 1) -> dE = ((2/3)² + 0² +  0² )^0.5 = 2/3

c(1/6; 1; 1), F(0; 1; 1) -> cF = 1/6

U = area(EFcd) = (2/3 + 1/6)/2 = 5/12 square unit

- area of abIO = (aO + bI)/2.OI

aO = 1

b(1/2; 1; 0), I(0; 1; 0) -> bI = 1/2

V = area(abIO) = 3/4 square unit

Volume(EFcdabIO) = (U + V)/2 x FI = (5/12 + 3/4)/2 = 7/12 cubic unit

therefore the volume of the smaller half is (1 - 7/12) = 5/12 cubic unit = approximately 41.67% of the volume of that cube.

Sorry, I forgot to display the cube with all the letters I involved: Edited by ant0n