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 Big areas (Posted on 2013-09-17)
How many Pythagorean triangles are there whose area is expressed by 9 different digits?

Provide the values for the smallest and the biggest of them.

 No Solution Yet Submitted by Ady TZIDON No Rating

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 re: No Subject -- solution continued | Comment 3 of 16 |
(In reply to No Subject by Charlie)

`  22836   82175   85289    938274150     132     475     493  18623  101580  103273    945862170      11      60      61  21952   86436   89180    948721536      16      63      65   3936  484120  484136    952748160     492   60515   60517   3128  611520  611528    956417280     391   76440   76441  20930   91392   93758    956417280    1495    6528    6697  24240   78982   82618    957261840     120     391     409  33500   58011   66989    971684250   33500   58011   66989   4992  389360  389392    971842560     312   24335   24337  16140  120512  121588    972531840      15     112     113  25824   75320   79624    972531840      12      35      37  43040   45192   62408    972531840      20      21      29  17874  108832  110290    972631584    8937   54416   55145  28510   68424   74126    975384120       5      12      13  38256   51008   63760    975681024       3       4       5  21217   92544   94945    981753024   21217   92544   94945  43300   45465   62785    984317250      20      21      29  25980   75775   80105    984317250      12      35      37`

DECLARE FUNCTION gcd# (a#, b#)
DEFDBL A-Z
CLS
OPEN "bigareas.txt" FOR OUTPUT AS #2
doing = 1
WHILE doing
n = n + 1
m = n + 1
IF m * m * m * n - m * n * n * n > 1000000000# THEN doing = 0
WHILE m * m * m * n - m * n * n * n < 1000000000#
IF gcd(m, n) = 1 THEN
a = m * m - n * n: b = 2 * m * n: c = m * m + n * n
IF b < a THEN SWAP b, a
area0 = a * b / 2
k0 = INT(SQR(100000000# / area0))
FOR k = k0 TO -INT(-SQR(10) * k0) + 2
area1 = k * k * area0
s\$ = LTRIM\$(STR\$(area1))
IF LEN(s\$) = 9 THEN
good = 1
FOR i = 1 TO 8
IF INSTR(MID\$(s\$, i + 1), MID\$(s\$, i, 1)) THEN good = 0: EXIT FOR
NEXT
IF good THEN
PRINT USING "####### ####### #######    #########"; k * a; k * b; k * c; area1
PRINT #2, USING "####### ####### #######    #########"; k * a; k * b; k * c; area1;
PRINT #2, USING "########"; a; b; c
ct = ct + 1
END IF
END IF
NEXT
END IF
m = m + 2
WEND
WEND
PRINT ct

CLOSE 2

FUNCTION gcd (a, b)
x = a: y = b
DO
q = INT(x / y): r = x - y * q
IF r = 0 THEN gcd = y: EXIT FUNCTION
x = y: y = r
LOOP
END FUNCTION

 Posted by Charlie on 2013-09-17 18:17:29

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