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 !!!
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.
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.
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.
| infinisa wrote: |
| 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. |
Wow you have a computer like brain or what. That was right on the money. Great job.
Hi metalfreek
Thanks, I think I have a "maths brain". Any more maths problems would be highly appreciated
| metalfreek wrote: |
| Wow you have a computer like brain or what. That was right on the money. Great job. |
Thanks, I think I have a "maths brain". Any more maths problems would be highly appreciated
| metalfreek wrote: | ||
Wow you have a computer like brain or what. That was right on the money. Great job. |
Wow; I am impressed too. wow....
I will post some more maths problem soon. So wait for it.
Ya..pretty simple....the concept ehind it is simple..
But post more probs of this/Math sort...waitin...
But post more probs of this/Math sort...waitin...
| metalfreek wrote: |
| I will post some more maths problem soon. So wait for it. |
We're looking forward to more maths problems - just a reminder
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.
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.
| metalfreek wrote: |
| 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. |
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.
| metalfreek wrote: |
| 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. |
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...
Now let's screw everyone and ask how many rectangles are in the chess board. )
)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
| Quote: |
| 961=31^2, upside down is 196=14^2 - another solution! |
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.
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.
| metalfreek wrote: |
| 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. |
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!?
| Code: |
| Number of Hens = X
Number of Pigs = Z X*4=Y Z*2=U => Y + U = V (number of feet) (1) => Y/2 + U = W (number of eyes) |
Now we know that:
| Code: |
| V=86
W=60 |
So....
| Code: |
| Y+U=86
Y/2+U=60 (Put them together gives): 1/2Y = 52 52+U=86 (1) U=34 |
Which gives us:
| Code: |
| X*4=52
X=13 and Z*2=34 Z=17 |
I also wrote a little program for a TI-84+ (also compatible with TI-83 and TI-83+)
So I would have been first to post if I hadn't wrote this program.
| Code: |
| ClrHome
Input "Number of eyes? ",A Input "Number of feet? ",B B-A->C C*2->C C/4->D B-C->E E/2->E ClrHome Disp "----------------","4 feeted Animal:",D,"","2 feeted Animal:",E,"---------------- Pause Stop |
Adri
| adri wrote: |
| I also wrote a little program for a TI-84+ (also compatible with TI-83 and TI-83+)
So I would have been first to post if I hadn't wrote this program. :) |
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.
| guissmo wrote: |
| Now let's screw everyone and ask how many rectangles are in the chess board. ) |
Hi guissmo
I think this problem deserves a thread of its own!
I've posted my answer here
Ok its time for answer.
Total Number of hens = 17
Total number of pigs = 13
cheers to all who got correct answer.
Total Number of hens = 17
Total number of pigs = 13
cheers to all who got correct answer.
I started a topic with some questions as well: http://www.frihost.com/forums/vt-108646.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. 
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. 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.
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.
| metalfreek wrote: |
| 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. |
Well... You can do this with two coins. You can have a half-dollar and another half-dollar. The difference between them is zero, which equals the difference of their squares obviously since they have the same value.
| metalfreek wrote: |
| 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. |
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
| Afaceinthematrix wrote: | ||
Well... You can do this with two coins. You can have a half-dollar and another half-dollar. The difference between them is zero, which equals the difference of their squares obviously since they have the same value. |
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
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.
| Quote: |
| Well... You can do this with two coins. You can have a half-dollar and another half-dollar. The difference between them is zero, which equals the difference of their squares obviously since they have the same value. |
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?
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?
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.
+
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.
| adri wrote: |
| 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. |
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 + (n-1)d ... (1)
Sum of first n terms S(n) = (n/2) (2a + (n-1)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
| infinisa wrote: |
|
Sum of first n terms S(n) = (n/2) (2a + (n-1)d) ... (2) |
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+(n-1)*v (v=d in your example)
Un=2+(n-1)*4
Un=2+4n-4
Un=-2+4n is the formula to calculate any number. (So u(50) would be 198 I think...)
Adri
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?
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?
| metalfreek wrote: |
| 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? |
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
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.
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.
After a while... (~10minutes)
9+9-9+(9/9)=10
((9^9)/(9^9))+9=10
Adri
9+9-9+(9/9)=10
((9^9)/(9^9))+9=10
Adri
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.
2 w x y z 47
Find the values of w,x,y and z. Also these numbers are in Arithmetic series.
| metalfreek wrote: |
| 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. |
2, 11, 20, 29, 38, 47...
| infinisa wrote: | ||
Hi guissmo I think this problem deserves a thread of its own! I've posted my answer here |
Hi guissmo
Have you looked at my solution? Is it OK?
| Code: |
| #!/usr/bin/env perl
# aseq [ start terms end [ start terms end [ ... ] ] ] # prints out t - 1 terms between start and end use strict; use warnings; while (@ARGV >= 3) { my ($s,$t,$e) = (shift,shift,shift); my $diff = $e - $s; my $step = $diff / $t; while (($s += $step) < $e) { print "$s "; } print "\n"; } |
11, 20, 29, 38
Related topics
 |