metalfreek
Hay guys can you count total number of squares of all possible length on a chess board? Guess work would be fine if you don't want to give stress to your brain.

Just think for a while !!!metalfreek
Hay guys can you count total number of squares of all possible length on a chess board? Guess work would be fine if you don't want to give stress to your brain.
infinisa
Hi metalfreek
That would be: 1 8x8 square 2x2=4 7x7 squares 3x3=9 6x6 squares 4x4=16 5x5 squares 5x5=25 4x4 squares 6x6=36 3x3 squares 7x7=49 2x2 squares 8x8=64 1x1 squares Total = 204, I think. tchaunt
Wow. I'm going to like.......take a sharpie and go all over a cheap chess board to test this. xD It could take a while though.
metalfreek
Wow you have a computer like brain or what. That was right on the money. Great job. infinisa
Hi metalfreek
Thanks, I think I have a "maths brain". Any more maths problems would be highly appreciated tony
Wow; I am impressed too. wow.... metalfreek
I will post some more maths problem soon. So wait for it.
nanunath
Ya..pretty simple....the concept ehind it is simple..
But post more probs of this/Math sort...waitin... infinisa
We're looking forward to more maths problems  just a reminder metalfreek
Here is another one. Very easy. Just think for second and you will get an answer. No mathematical calculation is required.
From number 1 to 500 find a number which is perfect square and if the number is upside down the number will still be a perfect square. Afaceinthematrix
100  depending on how you write the number "1." Most people's "1's" are symmetrical. However, some people actually write the horizontal line at the bottom and the diagonal line at the top. infinisa
Thanks for the problem! 1 is obviously a solution, but there are probably some more... What about 100? Upside down gives 001. Does this count? The digits we can use are only 0, 1, 6, 8, 9 Well, the squares I can form from these digits are: 9=3^2, upside down is 6  not a square 16=4^2, upside down is 91  not a square 81=9^2, upside down is 18  not a square 169=13^2, upside down is 691  not a square 900=30^2, upside down is 006  not a square 961=31^2, upside down is 196=14^2  another solution! Right now, can't see any more solutions... guissmo
Now let's screw everyone and ask how many rectangles are in the chess board. )
metalfreek
The answer is not 001. You can consider 1 as symmetrical number and the answer has its first digit 1. So the right answer is here
metalfreek
Ok here is another one. For this you might want to do some calculation in paper. Take a paper and pen and than try to solve this one.
In a farm house there are some hens and pigs. I am not going to tell you how many hens and pigs are there in the farm. You have to find it out. The hint is that there are altogether 60 eyes and 86 feet. So, give me the total number of hens and pigs in the farm. Afaceinthematrix
Hens = Y Pigs = X Each hen has two legs and two eyes. Each pig has four legs and two eyes. 2Y + 4X = 86 2Y + 2X = 60  2X = 60  2Y 2Y + 2(60  2Y) = 86 2Y  4Y = 34 2Y = 34 Y = 17 There are 17 Hens, which produces 34 eyes and 34 legs. Therefore, there must be 13 pigs, which will produce the additional 26 eyes and 52 legs. Just to check... 34 + 26 = 60 and 34 + 52 = 86. Am I right!? adri
Now we know that:
So....
Which gives us:
I also wrote a little program for a TI84+ (also compatible with TI83 and TI83+) So I would have been first to post if I hadn't wrote this program.
Adri Afaceinthematrix
Ahhh... So you may not have been the first contributor but you were the best contributor. I just did a simple system of equations. You actually wrote a program for it! I might bust out my calculator and play with the program. I don't usually use programs (I actually don't like to use calculators  I like to try and do things without them) but that looks fun. infinisa
Hi guissmo I think this problem deserves a thread of its own! I've posted my answer here metalfreek
Ok its time for answer.
Total Number of hens = 17 Total number of pigs = 13 cheers to all who got correct answer. adri
I started a topic with some questions as well: http://www.frihost.com/forums/vt108646.html
Just trying to make this forum more fun. So feel free to try to solve these too... PS: metalfreek, keep posting your questions too. I like them. metalfreek
Here is another question.
I have some coins of $1 (I mean it adds up to $1), if you divide my coins into two parts the difference between the numbers will be the same as the difference between their squares. All you have to do is find how many coins I have? Very Easy. Afaceinthematrix
Well... You can do this with two coins. You can have a halfdollar and another halfdollar. The difference between them is zero, which equals the difference of their squares obviously since they have the same value. infinisa
Hi metalfreek I'm not sure that I understand your question. Are you saying that you have a certain number of coins whose total value is $1, divide them into two groups of coins, with say m in the first group and n in the second group, in such a way that m  n = m^2  n^2? If I understand this right: We can suppose that m ≥ n, so the equation becomes m  n = m^2  n^2 m  n = (m + n)x(m  n) So (m  n)x(m + n  1) = 0 So m=n or m + n = 1 Assuming you require both m ≥ 1 and n ≥ 1 this eliminates m + n = 1 leaving m=n So you have 2n coins whose total value is $1. The US coins with value less than $1 are: $0.01, $0.05, $0.10, $0.25, $0.50 So you could have any even number of coins totalling $1: Possibilities with all the same coins: 2 x $0.50 4 x $0.25 10 x $0.10 20 x $0.05 100 x $0.01 Possibilities with 1 x $0.50, 1 x $0.25 (=> 1 x $0.10): 1 x $0.50, 1 x $0.25, 1 x $0.10, 15 x $0.01 (18 coins) 1 x $0.50, 1 x $0.25, 1 x $0.10, 1 x $0.05, 10 x $0.01 (14 coins) 1 x $0.50, 1 x $0.25, 1 x $0.10, 2 x $0.05, 5 x $0.01 (10 coins) Possibilities with 1 x $0.50, 0 x $0.25 (=> 1 x $0.10): 1 x $0.50, 1 x $0.10, 40 x $0.01 (42 coins) 1 x $0.50, 1 x $0.10, 1 x $0.05, 35 x $0.01 (38 coins) 1 x $0.50, 1 x $0.10, 2 x $0.05, 30 x $0.01 (34 coins) 1 x $0.50, 1 x $0.10, 3 x $0.05, 25 x $0.01 (30 coins) 1 x $0.50, 1 x $0.10, 4 x $0.05, 20 x $0.01 (26 coins) 1 x $0.50, 1 x $0.10, 5 x $0.05, 15 x $0.01 (22 coins) 1 x $0.50, 1 x $0.10, 6 x $0.05, 10 x $0.01 (18 coins) 1 x $0.50, 1 x $0.10, 7 x $0.05, 5 x $0.01 (14 coins) 1 x $0.50, 1 x $0.10, 8 x $0.05 (10 coins) Possibilities with 0 x $0.50, 1 x $0.25 (=> even number (0 to 6) of $0.10): Possibilities with 0 x $0.50, 1 x $0.25, 0 x $0.10: 1 x $0.25, 75 x $0.01 (76 coins) 1 x $0.25, 1 x $0.05, 70 x $0.01 (72 coins) 1 x $0.25, 2 x $0.05, 65 x $0.01 (68 coins) 1 x $0.25, 3 x $0.05, 60 x $0.01 (64 coins) and so on (with 60, 56, 52, 48, 44, 40, 36, 32, 28, 24, 20 coins) down to: 1 x $0.25, 15 x $0.05, (16 coins) Possibilities with 0 x $0.50, 1 x $0.25, 2 x $0.10: 1 x $0.25, 2 x $0.10, 55 x $0.01 (58 coins) 1 x $0.25, 2 x $0.10, 1 x $0.05, 50 x $0.01 (54 coins) 1 x $0.25, 2 x $0.10, 2 x $0.05, 45 x $0.01 (50 coins) and so on (with 46, 42, 38, 34, 30, 26, 22, 18 coins) down to: 1 x $0.25, 2 x $0.10, 11 x $0.05 (14 coins) Possibilities with 0 x $0.50, 1 x $0.25, 4 x $0.10: 1 x $0.25, 4 x $0.10, 35 x $0.01 (40 coins) 1 x $0.25, 4 x $0.10, 1 x $0.05, 30 x $0.01 (36 coins) 1 x $0.25, 4 x $0.10, 2 x $0.05, 25 x $0.01 (32 coins) and so on (with 28, 24, 20, 16 coins) down to: 1 x $0.25, 4 x $0.10, 7 x $0.05 (12 coins) Possibilities with 0 x $0.50, 1 x $0.25, 6 x $0.10: 1 x $0.25, 6 x $0.10, 15 x $0.01 (22 coins) 1 x $0.25, 6 x $0.10, 1 x $0.05, 10 x $0.01 (18 coins) 1 x $0.25, 6 x $0.10, 2 x $0.05, 5 x $0.01 (14 coins) 1 x $0.25, 6 x $0.10, 3 x $0.05 (10 coins) Possibilities with 0 x $0.50, 0 x $0.25 => (even number (0 to of $0.10): Possibilities with 0 x $0.50, 0 x $0.25, 2 x $0.10: 2 x $0.10, 80 x $0.01 (82 coins) 2 x $0.10, 1 x $0.05, 75 x $0.01 (78 coins) 2 x $0.10, 2 x $0.05, 70 x $0.01 (74 coins) and so on (with 70, 66, 62, 58, 54, 50, 46, 42, 38, 34, 30, 26, 22 coins) down to: 2 x $0.10, 16 x $0.05 (18 coins) Possibilities with 0 x $0.50, 0 x $0.25, 4 x $0.10: 4 x $0.10, 60 x $0.01 (64 coins) 4 x $0.10, 1 x $0.05, 55 x $0.01 (60 coins) 4 x $0.10, 2 x $0.05, 50 x $0.01 (56 coins) and so on (with 52, 48, 44, 40, 36, 32, 28, 24, 20 coins) down to: 4 x $0.10, 12 x $0.05 (16 coins) Possibilities with 0 x $0.50, 0 x $0.25, 6 x $0.10: 6 x $0.10, 40 x $0.01 (46 coins) 6 x $0.10, 1 x $0.05, 35 x $0.01 (42 coins) 6 x $0.10, 2 x $0.05, 30 x $0.01 (38 coins) and so on (with 34, 30, 26, 22, 18 coins) down to: 6 x $0.10, 8 x $0.05 (14 coins) Possibilities with 0 x $0.50, 0 x $0.25, 8 x $0.10: 8 x $0.10, 20 x $0.01 (28 coins) 8 x $0.10, 1 x $0.05, 15 x $0.01 (24 coins) 8 x $0.10, 2 x $0.05, 10 x $0.01 (20 coins) 8 x $0.10, 3 x $0.05, 5 x $0.01 (16 coins) 8 x $0.10, 4 x $0.05 (12 coins) On the other hand, I may have completely misunderstood your question infinisa
Hi metalfreek Are we supposed to find A solution or ALL solutions to your problem? The question is certainly very easy if we just have to find a single solution, but as I interpreted the problem as having to find all possible solutions I didn't find it very easy at all metalfreek
I think I have to clarify the question. You have some coins. The difference between the values of the coins will be the same as the difference between their squares. Someone have just given the right answer here.
metalfreek
I am firing question after question. Here is another one.
One day I and my friend went to eat ice cream. He began to eat icecream one after the other, after finishing he ordered more. When he finished eating he said to me "The icecream I ate was 50th in the last five days. Each day I ate 4 more than on the previous day." All you have to find is how many icecreams he will eat tomorrow? adri
x (first day)
+ x+4 (4 more than the previous day) + x+4+4 (for more than the previous day  third day) + x+4+4+4 (fourth day) + x+4+4+4+4 (fifth day) so we get: (x)+(x+4)+(x++(x+12)+(x+16)=50 5x+40=50 x=2 Fill this in: 2 (day 1) + 6 (day 2) + 10 (day 3) + 14 (day 4) + 18 (day 5) = 50 So 18 + 4 = 22 He ate 22 icecreams the sixth day. infinisa
Hi adri Your solution is correct, but here is a more general approach: The number of icecreams eaten each day is an arithmetic sequence, so if: a = first term and d = common difference (in this case 4), we have: n th term u(n) = a + (n1)d ... (1) Sum of first n terms S(n) = (n/2) (2a + (n1)d) ... (2) Applying (2) with n = 5, d = 4 and S(n) = 50, we get: 50 = (5/2) (2a + 4x4) so 50 = 5a + 40 so 5a = 10 and a = 2 Now applying (1) with n = 6, a = 2, d = 4, we get: u(6) = 2 + 5x4 so u(6) = 22  the number of icecreams eaten on the 6th day This solution may seem more complex at first, but if you were told the number of icecreams eaten in the first 50 days, I think you'd find this method easier adri
That was the formula that I had forgotten Well anyway if you know your u1, you just need to fill it in in this formula: Un=u1+(n1)*v (v=d in your example) Un=2+(n1)*4 Un=2+4n4 Un=2+4n is the formula to calculate any number. (So u(50) would be 198 I think...) Adri metalfreek
another question
My mother is twice as old as my sister and my father is 24 years older than me. At the time of my sisters birth I was 5 years old. My sister is now 25. What is the difference in the age of my parents? infinisa
M = mother's age F = father's age S = sister's age I = my age My mother is twice as old as my sister: M = 2S ...(1) My father is 24 years older than me: F = I + 24 ...(2) At the time of my sisters birth I was 5 years old: I = S + 5 ...(3) My sister is now 25: S = 25 ...(4) What is the difference in the age of my parents? (4) in (3) gives I = 25 + 5 = 30 In (2) gives F = 30 + 24 = 54 In (1) gives M = 2 x 25 = 50 So the difference in the age of my parents is 54  50 = 4. Thanks, metalfreek, keep them coming metalfreek
Another question.
Can you write 10 by using only five nines? Allowed operation Plus, Minus, Divide, Multiply and power. example: 9+(99/99)=10 Give me another two ways. adri
After a while... (~10minutes)
9+99+(9/9)=10 ((9^9)/(9^9))+9=10 Adri metalfreek
Some simple calculation is required to solve this.
2 w x y z 47 Find the values of w,x,y and z. Also these numbers are in Arithmetic series. Bikerman
2, 11, 20, 29, 38, 47... infinisa
Hi guissmo Have you looked at my solution? Is it OK? sreservoir
11, 20, 29, 38 metalfreek
Looks like I have ignored this thread for a long long time. But I am back again. So, let's begin our brain stretching journey.
22 is the right answer for ice cream problem. Here i another question for you guys "John had ten billion dollars. If he wants to give a $500 bill every minute then how long would it take to give all of his money?" adri
Doesn't look to difficult if I'm not misreading... 10.000.000.000 $ / (500$ / min) = (10.000.000.000$) * (1 min / 500$) = 20 000 000 minutes = 333 333 hours = 13 888 days = 15, 43 years. adri saberlivre
If metalfreek allow me I would like to post a question:
find the remainder when 2^1990 is divided by 1990. Sorry for my English metalfreek
Ok another question from me. Carefully read this equalities and find the answer.
Lets see how many people can solve this 52=11 65=21 83=17 62=? chatrack
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