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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