2005 JBMO Problems/Problem 4: Difference between revisions
Created page with "Let (a, b, c) = k Then a = pk b = qk c = rk 100a + 10b + c = abc(a + b + c) 100p + 10q + r = pqr(p + q + r)k^3 We see that if any of (a, b, c) is 0, all the terms are 0, an..." |
No edit summary |
||
| Line 2: | Line 2: | ||
Then | Then | ||
a = pk | a = pk | ||
b = qk | b = qk | ||
c = rk | c = rk | ||
100a + 10b + c = abc(a + b + c) | 100a + 10b + c = abc(a + b + c) | ||
100p + 10q + r = pqr(p + q + r)k^3 | 100p + 10q + r = pqr(p + q + r)k^3 | ||
| Line 13: | Line 16: | ||
Let (p, q) = c with p = cs and q = cy | Let (p, q) = c with p = cs and q = cy | ||
Then | Then | ||
c(100s + 10y) + r = (c^2)syr(cs + cu + r)k^3 | |||
c(100s + 10y) + r = (c^2)syr(cs + cu + r)k^3 | |||
We see that c is also a factor of r. | We see that c is also a factor of r. | ||
We get similar results for (p, r) = c and (q, r) = c | We get similar results for (p, r) = c and (q, r) = c | ||
The terms cancel, and thus we get that | The terms cancel, and thus we get that | ||
For some (m, n, t, j) which are respectively factors of (a, b, c) | For some (m, n, t, j) which are respectively factors of (a, b, c) | ||
100m + 10n + t = mnt(m + n + t)j^3 | 100m + 10n + t = mnt(m + n + t)j^3 | ||
With 0 < m, n, t, j < 10 | With 0 < m, n, t, j < 10 | ||
The computation is left to the reader | The computation is left to the reader | ||
Revision as of 10:58, 11 April 2019
Let (a, b, c) = k Then a = pk
b = qk
c = rk
100a + 10b + c = abc(a + b + c)
100p + 10q + r = pqr(p + q + r)k^3
We see that if any of (a, b, c) is 0, all the terms are 0, and such (a, b, c) are nonzero
We will show that we can find (m, n, t) such that m, n, and t are factors of a, b, and c respectively, with (m, n, p) being pairwise relatively prime.
Let (p, q) = c with p = cs and q = cy
Then
c(100s + 10y) + r = (c^2)syr(cs + cu + r)k^3
We see that c is also a factor of r.
We get similar results for (p, r) = c and (q, r) = c
The terms cancel, and thus we get that
For some (m, n, t, j) which are respectively factors of (a, b, c)
100m + 10n + t = mnt(m + n + t)j^3
With 0 < m, n, t, j < 10
The computation is left to the reader
