27505 is a product to two 4k+1 primes and so it can be expressed as a sum of two squares in two different way viz. 64^2+153^2 and 84^2+143^2.
27505 is a composite numbers containing the digit 5 at least once whose prime factors each contain the digit 5 as well so that, overall, the digit 5 occurs five times.
There are 27 such numbers in the range up to 40000. The following algorithm will generated them (permalink).
L=[]
d=5
upper=40000
for n in [1..upper]:
if is_prime(n)==0:
D=n.digits()
total=D.count(d)
if d in D:
F=list(factor(n))
OK=1
for f in F:
if d not in f[0].digits():
OK=0
break
else:
total+=(f[0].digits()).count(d)
if OK==1 and total==d:
print(n,"=",factor(n),"with total of",total)
L.append(n)
print()
print(L)
print("There are",len(L),"numbers in the range up to", upper)
OUTPUT
5255 = 5 * 1051 with total of 5
5755 = 5 * 1151 with total of 5
7555 = 5 * 1511 with total of 5
12755 = 5 * 2551 with total of 5
15635 = 5 * 53 * 59 with total of 5
17555 = 5 * 3511 with total of 5
25055 = 5 * 5011 with total of 5
25295 = 5 * 5059 with total of 5
25505 = 5 * 5101 with total of 5
25535 = 5 * 5107 with total of 5
25565 = 5 * 5113 with total of 5
25595 = 5 * 5119 with total of 5
25765 = 5 * 5153 with total of 5
25855 = 5 * 5171 with total of 5
26755 = 5 * 5351 with total of 5
27505 = 5 * 5501 with total of 5
27515 = 5 * 5503 with total of 5
27535 = 5 * 5507 with total of 5
27595 = 5 * 5519 with total of 5
27655 = 5 * 5531 with total of 5
27785 = 5 * 5557 with total of 5
27955 = 5 * 5591 with total of 5
28255 = 5 * 5651 with total of 5
28555 = 5 * 5711 with total of 5
29255 = 5 * 5851 with total of 5
32755 = 5 * 6551 with total of 5
35755 = 5 * 7151 with total of 5
[5255, 5755, 7555, 12755, 15635, 17555, 25055, 25295, 25505, 25535, 25565, 25595, 25765, 25855, 26755, 27505, 27515, 27535, 27595, 27655, 27785, 27955, 28255, 28555, 29255, 32755, 35755]
There are 27 numbers in the range up to 40000
27505 is a happy and Duffinian number.
27505 can be rendered as a digit equation as follows:
-2 + 7 - 5 = 0 * 5
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