ASF Spoilers Thread

[bellenuit]

Q5: 35 cm^2

For those who are having difficulty working out the solution without resorting to computing interim fractions or using algebra, this is how Number 3 is worked out.



You can easily calculate the missing rectangle in the top left hand corner (delineated in red) as its sides are obviously 1 X 5. It has a width 1 because its width is the difference between the widths of the rectangle immediately to its right (4) and that below it (5). It has a height of 5 because it has the same height as the rectangle beside it, which must be 20 divided by 4.

If you then add the areas of this new rectangle with those of the rectangle to its left and the rectangle below it, you get a new combination rectangle of 5 + 20 + 14 or 39cm2. This combination rectangle is exactly half the area of the large rectangle on the right (78cms) and as it shares the same height, its width then must be half that of the bigger rectangle. But we know the combination rectangle has a width of 5, therefore the bigger one must have a width of 10cm, which is the answer we are looking for.

According to the source article, the reason these puzzles are to be solved without use of fractions or algebra, basically with just the knowledge that the area of a rectangle equals its width by height, is that these are problems set for Japanese school kids who haven't yet got to the stage where they have learned about algebra or fractions.

 

[skc]

Q4 = 37 cm^2

Got everything else except this one. A hint please??

(Hopefully I can solve it before a response!)

According to the source article, the reason these puzzles are to be solved without use of fractions or algebra, basically with just the knowledge that the area of a rectangle equals its width by height, is that these are problems set for Japanese school kids who haven't yet got to the stage where they have learned about algebra or fractions.

Thanks Bellenuit. I enjoyed them. It was a good workout for the brain.

 

[cynic]

I used formulae for #4.

L1 = ?
L2 = 4

W1 = ?
W2 = ?
W3 = ?

W2 + w3 = 10
4 * (w1 + w2) = 29
L1 * w3 = 43
L1 * w1 = 21

L1 * (w3-w1) = 22

Etc

 

[skc]

I used formulae for #4.

I worked it out with algebra too. But supposedly you can work it out without such... Anyone else?

 

[bellenuit]

I worked it out with algebra too. But supposedly you can work it out without such... Anyone else?

Official answer to Number 4 ....



The left rectangle with a blue border covers the rectangle with area 21 and a section of the rectangle below. Imagine placing an identical rectangle to the right of the grey rectangle. The shaded blue area must have area 21, and the dotted blue area must be z.

The area x is 43 - 21 = 22

The area of the rectangle with the red border is 4 x 10 = 40. But it is also 29 + y, since a + z = 29. Therefore y = 11

So x is equal to double y. Which means that any rectangle above the horizontal line must have twice the area of the rectangle beneath the line that it shares a side with.

The green rectangle must have area double the rectangle underneath it, which has area 29. So the green rectangle has area 2 x 29 = 58

The missing value is 58 - 21 = 37cm2

An answer hasn't been provided for Number 5, other than saying the readers should work it out for themselves. I suspect it is too complex to explain the solution by a non-algebraic method, so have left it as a teaser for the readers to do who can then post their answers in the blog.

 

[cynic]

I worked it out with algebra too. But supposedly you can work it out without such... Anyone else?

Apologies.

The only thing that occurs to me outside of direct formulae is a brute force approach that presumes that the uppermost rectangles height are a whole multiple of the lower.

 

[pixel]

For those who are having difficulty working out the solution without resorting to computing interim fractions or using algebra, this is how Number 3 is worked out.

View attachment 63718

You can easily calculate the missing rectangle in the top left hand corner (delineated in red) as its sides are obviously 1 X 5. It has a width 1 because its width is the difference between the widths of the rectangle immediately to its right (4) and that below it (5). It has a height of 5 because it has the same height as the rectangle beside it, which must be 20 divided by 4.

If you then add the areas of this new rectangle with those of the rectangle to its left and the rectangle below it, you get a new combination rectangle of 5 + 20 + 14 or 39cm2. This combination rectangle is exactly half the area of the large rectangle on the right (78cms) and as it shares the same height, its width then must be half that of the bigger rectangle. But we know the combination rectangle has a width of 5, therefore the bigger one must have a width of 10cm, which is the answer we are looking for.

According to the source article, the reason these puzzles are to be solved without use of fractions or algebra, basically with just the knowledge that the area of a rectangle equals its width by height, is that these are problems set for Japanese school kids who haven't yet got to the stage where they have learned about algebra or fractions.

I didn't bother with the rectangle top left. For #3 we don't need a calculator:

The height of the 20sq has to be 20 / 4 = 5
The height of the 14sq has to be 14 / 5 = 2.8
add the two together, then divide 78 by the result --> 10

 

[skc]

Official answer to Number 4 ....

Very nice. I tried to do something with the rectangle on the bottom right, but I didn't think of mirroring the thing...

An answer hasn't been provided for Number 5, other than saying the readers should work it out for themselves. I suspect it is too complex to explain the solution by a non-algebraic method, so have left it as a teaser for the readers to do who can then post their answers in the blog.

#5 was actually easier than #4. It required you to "work around the board" to get to the dimensions of the rectangle in question... but it all involved just one method and didn't need any lateral thinking like #4.

It fact it was easy where, anytime you face a calculation involving fractions, you know you are on the wrong track.

 

[cynic]

Given that algebra is essentially a methodology for expressing and determining unknown quantities from known, I find it amusing that all solutions thus far are essentially based upon the same underlying problem solving principle, irrespective of whether they were algebraically expressed.

 

[bellenuit]

I didn't bother with the rectangle top left. For #3 we don't need a calculator:

The height of the 20sq has to be 20 / 4 = 5
The height of the 14sq has to be 14 / 5 = 2.8
add the two together, then divide 78 by the result --> 10

Yes, but remember the challenge was to solve the problem without using fractions (2.8 being an implied fraction). I didn't explicitly state that as a rule, as I wasn't quite sure if that was what was intended, but I did add...

For Number 1 I stated....

Note: For all puzzles, all answers are whole numbers (no fractions). The instructions are a bit ambiguous, but I also understand that to solve the puzzles you do not need to deal with fractions.

And for Number 3 I added...

Note: I can confirm that having now solved Number 3, that 1, 2 and 3 can be solved without resorting to using fractions in interim results. I assume the same applies to 4 and 5.

Nevertheless, congrats on your correct answers.

 

[keithj]

A fairly easy one.

View attachment 63788

You have a board marked into a grid of 20 X 20 squares. You also have rectangular dominoes that are two grid squares in size, like the one shown above. Clearly you could cover the total board surface with 200 of these dominoes if they are laid flat, each covering two squares ((20 * 20) /2).

However, two diagonally opposite corner squares are cut from the board as shown above. Is it possible to place 199 dominoes flat down on the board so that they completely cover the remaining surface of the board?

If yes, prove it. If no, prove it..

NO.

Assume the board is a chess board with alternating black & white squares. A domino must cover one black square & one white square.

The board has 400 squares - 200 black & 200 white. Two black ones are already covered. There are only 198 vacant black ones, but 200 white ones. Therefore there is NO arrangement of dominos that can cover the whole board.

 

[cynic]

NO.

Assume the board is a chess board with alternating black & white squares. A domino must cover one black square & one white square.

The board has 400 squares - 200 black & 200 white. Two black ones are already covered. There are only 198 vacant black ones, but 200 white ones. Therefore there is NO arrangement of dominos that can cover the whole board.

I did very much suspect that No was the answer but couldn't figure out a way to prove or disprove (apart from quasi brute force).

 

[bellenuit]

NO.

Assume the board is a chess board with alternating black & white squares. A domino must cover one black square & one white square.

The board has 400 squares - 200 black & 200 white. Two black ones are already covered. There are only 198 vacant black ones, but 200 white ones. Therefore there is NO arrangement of dominos that can cover the whole board.

Yep, good on you, that is the smart way to solve it. If the squares are alternate colours like a chess board, then diagonally opposite corners are the same colour so long as the number of rows and columns are even, which is the case here. So depending on whether the corner squares removed are white or black, there will be 2 less of that colour than the other. And as you said, since a domino always cover two squares of opposite colours, then each time a domino is placed, there will always be the same number of each colour now covered. So the task is impossible if there are not the same number of squares of each colour.

 

[skc]

NO.

Assume the board is a chess board with alternating black & white squares. A domino must cover one black square & one white square.

The board has 400 squares - 200 black & 200 white. Two black ones are already covered. There are only 198 vacant black ones, but 200 white ones. Therefore there is NO arrangement of dominos that can cover the whole board.

Thanks. I like the simplicity of the solution.

 

[cynic]

Correction to my previous answer. PI/16 is the probability that the coin doesn't overlap adjacent squares. Needless to say 1-(PI/16) is the probability that it does.

 

[cynic]

Correction to my previous answer. PI/16 is the probability that the coin doesn't overlap adjacent squares. Needless to say 1-(PI/16) is the probability that it does.

Reasoning is that the centre of the coin will land in a precise place within or on the boundaries of one square. There is circle of diameter of 1cm at the centre of every square within which the coins centre could land without the entire coin overlapping any adjacent squares.

 

[cynic]

Correction to my previous answer. PI/16 is the probability that the coin doesn't overlap adjacent squares. Needless to say 1-(PI/16) is the probability that it does.

Reasoning is that the centre of the coin will land in a precise place within or on the boundaries of one square. There is circle of diameter of 1cm at the centre of every square within which the coins centre could land without the entire coin overlapping any adjacent squares.

I just noticed that I probably should have used a 1cm square instead of a circle making the probabilities 1/4 and 3/4.

 

[pixel]

I just noticed that I probably should have used a 1cm square instead of a circle making the probabilities 1/4 and 3/4.

I came to the same conclusion: one throw in four.
Rationale: The centre of the coin will need to be half the coin diameter inside either border. Assuming that "touching" the edges satisfies the condition, that makes the permitted locale for the coin centre exactly one square centimeter.

 

[cynic]

I came to the same conclusion: one throw in four.
Rationale: The centre of the coin will need to be half the coin diameter inside either border. Assuming that "touching" the edges satisfies the condition, that makes the permitted locale for the coin centre exactly one square centimeter.

Great pixels look alike, and cynics rarely differ.

Agree that there is a 3/4 probability that the coin will land straddling two or more squares.

 

[bellenuit]

Yes, 3/4 is correct for the reasoning given above. If the coin centre lands anywhere within the imaginary 1cm square that is central to each chessboard square then it will not straddle a border. Outside that it will. Since that square is 1/4 the size of the chessboard square, then overall 1/4 of the total board area is safe from straddling with 3/4 not. Congrats.