a,b and c are integers and they are not equal to 1 or zero
a does not equal b does not equal c
2006-11-04 08:43:42 · 5 answers · asked by yourdaddy0 2 in Science & Mathematics ➔ Mathematics
does have a solution
2006-11-04 09:14:23 · update #1
I believe you mean "abc" to be a 3 digit number! So, the problem is to find digits a, b and c such that
100a + 10b + c = a + b^c
that is, b^c + digit = three digit number (not beginning with 1)
The only possibilities for b^c are
256, 512; 243, 729; 256; 625; 216; 343; 512; 729
After trying these, find that there is only one answer corresponding to 736:
a = 7, b = 3 and c = 7
2006-11-04 13:03:14 · answer #1 · answered by p_ne_np 3 · 1⤊ 0⤋
You can subtract a from both sides, then divide by bc-1 to get the equation Hermione gave you:
a = b^c/(bc-1)
But wait, bc-1 is relatively prime to b, and hence relatively prime to b^c. So the only way bc-1 can divide b^c is if bc-1 = +/-1.
But if bc-1 = -1, then bc=0, and one of them is zero.
If bc-1 = 1, then bc = 2, and hence one of them has to be one or c has to be negative (c=-2 or c=-1.) If c is negative, b^c is not an integer (hence abc would not be an integer) unless b = +/-1. So b=-1, c=-2.
abc = 2a = a+(-1)^(-2) = a+1
So a=1.
So there are no solutions.
2006-11-04 09:03:42 · answer #2 · answered by thomasoa 5 · 0⤊ 1⤋
HuYaYa!i knew that!i got 1 ques tin,wats that supposed to mean?
2006-11-04 08:45:23 · answer #3 · answered by M!stakenMe 3 · 0⤊ 2⤋
Okay...what's YOUR question about this equation?
2006-11-04 08:50:51 · answer #4 · answered by Anonymous · 0⤊ 2⤋
a=(b^c)/(bc-1)
logb=loga/(logac-c)
c=i dont know
2006-11-04 08:47:52 · answer #5 · answered by Anonymous · 0⤊ 0⤋