Dan McQuillan
@normalsubgroup.bsky.social
Mathematician, mathematics educator and mathematics enthusiast. He/him.
Solution (2 of 2). The other key point is, with d the number of digits in x (choosing x from 1, 16, 166, 1666, etc) the right sides are:
(x)10^{2d}+(1/2)10^{2d}+(2x+1). But 6x+4=10^d. So the right sides are
(x+0.5)(6x+4)^2+(2x+1).
(x)10^{2d}+(1/2)10^{2d}+(2x+1). But 6x+4=10^d. So the right sides are
(x+0.5)(6x+4)^2+(2x+1).
December 1, 2024 at 12:50 PM
Solution (2 of 2). The other key point is, with d the number of digits in x (choosing x from 1, 16, 166, 1666, etc) the right sides are:
(x)10^{2d}+(1/2)10^{2d}+(2x+1). But 6x+4=10^d. So the right sides are
(x+0.5)(6x+4)^2+(2x+1).
(x)10^{2d}+(1/2)10^{2d}+(2x+1). But 6x+4=10^d. So the right sides are
(x+0.5)(6x+4)^2+(2x+1).
Solution. (Part 1 of 2). A key point is that the left sides of the equations have the form:
X^3+(2x+3)^3+(3x+2)^3. Also, for the x we are using (x=1, 16, 166, etc), 6x+4 is a power of 10, and therefore, so is (6x+4)^2.
X^3+(2x+3)^3+(3x+2)^3. Also, for the x we are using (x=1, 16, 166, etc), 6x+4 is a power of 10, and therefore, so is (6x+4)^2.
December 1, 2024 at 12:35 PM
Solution. (Part 1 of 2). A key point is that the left sides of the equations have the form:
X^3+(2x+3)^3+(3x+2)^3. Also, for the x we are using (x=1, 16, 166, etc), 6x+4 is a power of 10, and therefore, so is (6x+4)^2.
X^3+(2x+3)^3+(3x+2)^3. Also, for the x we are using (x=1, 16, 166, etc), 6x+4 is a power of 10, and therefore, so is (6x+4)^2.
Testing alt text to make sure I can do it. I saw this on another social media site. If it looks like alt text works for me, I’ll post the answer in 2 replies. A picture in each reply. The other site does not have an alt text option that I know about.
December 1, 2024 at 12:22 PM
Testing alt text to make sure I can do it. I saw this on another social media site. If it looks like alt text works for me, I’ll post the answer in 2 replies. A picture in each reply. The other site does not have an alt text option that I know about.