When I started it, I had titled it as "Love and Machines" and later changed the title to "my thought store". I later moved to LiveJournal at http://phoe6.livejournal.com
Two completely *ineffective* perl scripts for doing the above is here. Do u hav any improvements ?????? The scripts consumed 45% CPU each....Imagine the extent of ineffectiveness :)
A. Difference of cubes of two numbers which is a square
Hey hey hey ...A big mistake in my script...which I just now noticed...Oh GOd...teh anwer is simple 6 and 10... The problem in the first script is the assigning of the variable $a=20....I just changed it to 100...So that we can compare the diferenc of the cubes to upto square of 100.....
The difference between cubes of 6 an 10 leads to 784 whose square root is 48. In my script i compared only till 20 :(
So All the previous results die here... The Answer is 6 and 10...And hey I havent seen the answer as yet...for my 2 secs of fame that is :)
manually took up numbers to get the first thing, surprisingly landed up at 6,10 but I guess that moment I overlooked/ lost sight and went ahead :) to write futile C program. :( sometime later checked your comments..only to jump back and check the paper where I had calculations.. verified with sqrt(784) ... oh, dont to loose track.
Your approach to the problem is structured and good. I went right away for the brute force. Only after sometime sat to check if there are some algebraic ways of solving the problem. Say something like using the formulas of (a power 2)- (b power 2) = (a-b)*(a+b) (a power 3)-(b power 3) = (a-b)*(a power 2 + ab + b power 2)
Continued thinking in this way only to see my head turn a lil more bald. huh
Anyhow there surely has to be some way of solving this problem other than brute force.
Out of Context http://math.ucr.edu/home/baez/dublin/index.html#preskill
Back to context: I dint believe the problem had a smaller answer. Part of the reason for this thought was my working on another problem which have a sol for only very high numbers.
Problem no.5 on http://olympiads.win.tue.nl/imo/imo97/imo97d2.html
Looks like an easy prob. But this is pegging the mind for a lot lot time.
Using Brute Force and a comp program, the olympiad problem is solved as (1,1) (16,2) (27,3) for (a,b). Cant identify any pattern and dunno as yet how to solve it by hand.
The funny part is after 1 after many numbers it doesnt match...but matches only at 199981 and obviously matches from there on for the next 9 numbers until 199990.
19 comments:
The only clue I got till now is that atleast One of the two numbers cannot be below 1000.
Provided the question isnt a riddle ^_^)
The results of first 100 numbers are here
A. Difference of cubes of two numbers which is a square
Number 7
Number 8
Difference = 169
B. Difference of squares of two numbers which is a cube.
Number = 1
Number = 3
Difference = 8
Number = 3
Number = 6
Number = 27
3
15
216
6
10
64
6
42
1728
8
24
512
10
15
125
11
43
1728
13
14
27
15
17
64
15
21
216
15
35
1000
15
60
3375
21
28
343
24
48
1728
25
29
216
27
81
5832
28
36
512
35
63
2744
36
45
729
39
57
1728
45
55
1000
46
62
1728
48
80
4096
49
76
3375
53
55
216
55
66
1331
55
80
3375
62
63
125
62
66
512
63
99
5832
66
78
1728
78
91
2197
Two completely *ineffective* perl scripts for doing the above is here. Do u hav any improvements ??????
The scripts consumed 45% CPU each....Imagine the extent of ineffectiveness :)
A. Difference of cubes of two numbers which is a square
$temp=100;
$k=1;
while ($k<$temp)
{
$i=1;
while ($i<$temp){
$j=$i*$i*$i - $k*$k*$k;
if ($j>0) {
$a=1;
while($a < 20) {
$squire=$a*$a;
if ($j == $squire){
print("$k\n");
print("$i\n");
print ("$j\n");
print("\n\n");
}
$a=$a+1;
}
}
$i=$i+1;
}
$k=$k+1;
}
B. Difference of squares of two numbers which is a cube
$temp=100;
$k=1;
while ($k<$temp)
{
$i=1;
while ($i<$temp){
$j=$i*$i - $k*$k;
if ($j>0) {
$a=1;
while($a < 20) {
$cube=$a*$a*$a;
if ($j == $cube){
print("$k\n");
print("$i\n");
print ("$j\n");
print("\n\n");
}
$a=$a+1;
}
}
$i=$i+1;
}
$k=$k+1;
}
The scripts started consuming 45% cpu when wanted to check for numbers upto 10000. Otherwise it was quite normal.
!riddle
http://www.dansmath.com/probofwk/probs12.html
Problem No. 118 b)
havent yet seen the answer....lemme try exhausting my tries before checkinn the answer
http://www.dansmath.com/probofwk/probs12.html
Problem No. 118 b)
havent yet seen the answer....lemme try exhausting my tries before checkinn the answer
hey, atleast one of the answers will be grater than 10,000. What do u say ???
correction :::
atleast One of the numbers is grater than 5000.
ten thousand am not sure//
Frankly speaking,me too working on this.this should not be a difficult problem as it seems.
Hey hey hey ...A big mistake in my script...which I just now noticed...Oh GOd...teh anwer is simple 6 and 10...
The problem in the first script is the assigning of the variable $a=20....I just changed it to 100...So that we can compare the diferenc of the cubes to upto square of 100.....
The difference between cubes of 6 an 10 leads to 784 whose square root is 48. In my script i compared only till 20 :(
So All the previous results die here...
The Answer is 6 and 10...And hey I havent seen the answer as yet...for my 2 secs of fame that is :)
Well, you win!
Mine went like this:
took a notebook and wrote:
1 1 1
2 4 8
3 - 9 - 27
4 - 16 - 64
5 - 25 - 125
6 - 36 - 216
7 - 49 - 273
8 - 64 - 512
9 - 81 - 729
10 - 100 - 1000
manually took up numbers to get the first thing, surprisingly landed up at 6,10 but I guess that moment I overlooked/ lost sight and went ahead :) to write futile C program. :(
sometime later checked your comments..only to jump back and check the paper where I had calculations.. verified with sqrt(784) ...
oh, dont to loose track.
Your approach to the problem is structured and good.
I went right away for the brute force. Only after sometime sat to check if there are some algebraic ways of solving the problem. Say something like using the formulas of
(a power 2)- (b power 2) = (a-b)*(a+b)
(a power 3)-(b power 3) = (a-b)*(a power 2 + ab + b power 2)
Continued thinking in this way only to see my head turn a lil more bald. huh
Anyhow there surely has to be some way of solving this problem other than brute force.
Out of Context http://math.ucr.edu/home/baez/dublin/index.html#preskill
Back to context:
I dint believe the problem had a smaller answer. Part of the reason for this thought was my working on another problem which have a sol for only very high numbers.
Problem no.5 on
http://olympiads.win.tue.nl/imo/imo97/imo97d2.html
Looks like an easy prob. But this is pegging the mind for a lot lot time.
There is this problem which consumed hours on paper and tried upto number 1000 and lil beyond. Yet unable to find answer
problem No.17 on
http://www.google.com/googleblog/GLAT4.gif
Using Brute Force and a comp program, the olympiad problem is solved as (1,1) (16,2) (27,3) for (a,b). Cant identify any pattern and dunno as yet how to solve it by hand.
>>problem No.17 on
>>http://www.google.com/googleblog/GLAT4.gif
199981Woah! :))
Yeah the answer is 199981. Cool man.
The funny part is after 1 after many numbers it doesnt match...but matches only at 199981 and obviously matches from there on for the next 9 numbers until 199990.
then 200000, 200001 also match.
The result upto 100 lacs is
1
199981
199982
199983
199984
199985
199986
199987
199988
199989
199990
200000
200001
1599981
1599982
1599983
1599984
1599985
1599986
1599987
1599988
1599989
1599990
2600000
2600001
What pattern to make here... huh :)
One for the mind
Hey the problem no.1 on
http://www.google.com/googleblog/GLAT2.gif
has been solved by hand. Two hours and 15 mins though :(
Answer is
in the alphabetical order of the nine different letters in the problem
4,5,3,1,0,6,8,9,7
A few logical conclusions followed by two trial and err steps.
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