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Originally Posted by April madnessWhat about those who haven't learned at school?
They will need to guess the answers.
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Originally Posted by snookersfunnice trip, was that a cruise by any chance? any good pics? Put them in the gallery. How is Gibraltar? Never been there...
Gibraltar was nice. Played with the apes on the Rock. There are also some good tunnels and caves, and tax-free shopping opportunities.
I'll have a look and see what I can do with my many photos...
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Originally Posted by davis_greatest(An extra rule, which I never mentioned, is that none of my questions should require any knowledge of maths beyond being able to count and multiply and what is learnt at school.)
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Originally Posted by snookersfunHas to be somewhat in the lines:
we are looking for a number 1000.....-1= 999..... that can be constructed from the smallest numbers of squares possible.
Lagrange's four-square theorem states, that every "sufficiently large" integer is a sum of no more than 4 positive squares, which reduces to three only, if the number is not of the form 4^k(8l+7). As #s 9999.... are not devidable by 4, this reduces to (8l+7) and thus all possible #s 999999.... fall exactly into this category.
Thus a minimum of 4 different colours necessary.
This (when you add on the white to get 5 colours) is all absolutely true. As you say, any number that is 999999.... (all nines, and at least 3 of them) is of the form 8l+7, i.e. it leaves a remainder of 7 when divided by 8.
However, I am going to be a little mean and not allow you simply to quote Lagrange's four-square theorem, which not everyone will have heard of! (An extra rule, which I never mentioned, is that none of my questions should require any knowledge of maths beyond being able to count and multiply and what is learnt at school.)
For the point, you must explain please WHY any number that leaves a remainder of 7 when divided by 8 can never be written as the sum of three (or fewer) squares.
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nice trip, was that a cruise by any chance? any good pics? Put them in the gallery. How is Gibraltar? Never been there...
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Congratulations davis_greatest!
Originally Posted by snookersfunhow about 5? France, Belgium, Germany, Switzerland, Austria?
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Vidas solves the tiling problem
Originally Posted by Vidastiling(question 25)...
147x147, total 21609 squares.
Let's paint bathroom floor, using 4 colors(red, blue, green, white)
rbrbrbrbrbrb...r
wgwgwgwgwgwg...w
rbrbrbrbrbrb...r
wgwgwgwgwgwg...w
.....
rbrbrbrbrbrb...r
red/blue/green/white squares:
red = 74*74 = 5476
blue = 73*74 = 5402
white = 73*74 = 5402
green = 73*73 = 5329.
any 4-inch piece covers red,blue,white, and green squares.(a - amount of 4-pcs.)
3-inch piece can be of 4 kinds: b(covers red, white, blue), c(red, green, blue), d(red,white, green), e(white, green, blue).
a+b+c+d=5476
a+b+d+e=5402
a+b+c+e=5402 so d=c
a+c+d+e=5329.
-------------------
as d=c,
a+b+2c=5476
a+b+c+e=5402
a+2c+e=5329.
(1)-(2): c-e=74,-> c>=74. -> a<=5329-2*74 =5181, 4a<20724
So b+c+d+e >=(21609-20724)/3=295. total amount of 3-inch pieces exceeds 295.
295=147*2+1, so WT will use at least one colour more than twice.
Nicely done, Vidas - although with your use of 4 colours you could have made things a bit easier for yourself by taking the following approach:
(i) Suppose there are A tiles of 4 square inches and B tiles of 3 square inches.
Then 4A + 3B = 147^2 = 21609 .......(1)
(ii) With your colouring, no tile can cover more than one red square. As there are 5476 red squares, there must be at least 5476 tiles.
So A+B >=5476
or, multiplying both sides by 4, we get
4A + 4B >= 21904 ............. (2)
(iii) Now subtract (1) from (2) to get:
B >= 21904-21609 = 295
i.e. there are at least 295 tiles of 3 square inches.
Then, as you noted, 295 is greater than 2x147, so at least one colour must be used for a third time.
But good answer. Well done!
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Originally Posted by davis_greatestHow many and which European countries did I see on my holiday?
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ok here is basic not correct answer: 5
31x31 blue
6x6 green
1x1 black
1x1 red
1x1 white
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Originally Posted by snookersfunHas to be somewhat in the lines:
we are looking for a number 1000.....-1= 999..... that can be constructed from the smallest numbers of squares possible.
Lagrange's four-square theorem states, that every "sufficiently large" integer is a sum of no more than 4 positive squares, which reduces to three only, if the number is not of the form 4^k(8l+7). As #s 9999.... are not devidable by 4, this reduces to (8l+7) and thus all possible #s 999999.... fall exactly into this category.
Thus a minimum of 4 different colours necessary.
Leave a comment:
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Has to be somewhat in the lines:
we are looking for a number 1000.....-1= 999..... that can be constructed from the smallest numbers of squares possible.
Lagrange's four-square theorem states, that every "sufficiently large" integer is a sum of no more than 4 positive squares, which reduces to three only, if the number is not of the form 4^k(8l+7). As #s 9999.... are not devidable by 4, this reduces to (8l+7) and thus all possible #s 999999.... fall exactly into this category.
Thus a minimum of 4 different colours necessary.
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Originally Posted by VidasHonestly, i doubt four colours(including white) is possible here... let alone three...
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Honestly, i doubt four colours(including white) is possible here... let alone three...
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smallest number of colours not balls then, thought so.
This has to be less, than what I gave. Actually, I think I gave a rather high number of colors (as I learned, that every number can be represented by four squares). I am trying to find a proof (and the number) for three colors right now
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