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 5-Digit Squares (Posted on 2009-12-06)

 No Solution Yet Submitted by brianjn No Rating

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 Computer solution | Comment 1 of 5
Well, somehow I managed to remember the site after 2 years of inactivity, and decided to start wasting some homework time here, again.

First off, I decided to find the maximum value of A by determining when +(A+1)² was as close to 100,000 as possible without going over. That gets us a range of A's to check from 100 to 223.

We check B over the range A+1 to √(100000-A²).

To check all possible values for A, B, and C, I used the following code in Python.

def five_digit_squares():
cdu=check_digits_used
for val_a in range(100,224):
for val_b in range(val_a+1,int((100000-val_a**2)**.5)):
works=1
val_c=int((val_a**2+val_b**2+0.00000001)**.5)
if val_c**2==(val_a**2+val_b**2) and val_c<=316:
works=cdu(val_a**2)*cdu(val_b**2)*cdu(val_c**2)
if works==1:
used=[]
str_digits=str(val_a**2)+str(val_b**2)+str(val_c**2)
for digit in range(0,15):
if used.count(str_digits[digit])==0:
works=works+1
used=used+[str_digits[digit]]
if works==11:
print(val_a,val_b,val_c)

This ends up getting us a solution of 136 273 305. So, A is 136, B is 273, and C is 305.

136²=1 8 4 9 6
273²=7 4 5 2 9
305²=9 3 0 2 5

Edited on December 6, 2009, 9:39 pm
 Posted by Justin on 2009-12-06 21:34:31

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