(A) 39 is the only possible two digit integer.
(B) Valid solutions exist for N = 3, 4, 6, 8, 9
Let the units digit and the tens digit are respectively denoted by x and y. then, by the given conditions:
Nx + y = xy + 4c
If y=0, then it follows that c=0/x =0, giving: x=0. this is a contradiction, since 0/0 is undefined.
If x=0, then we have x=0, which is a contradiction for the reasons explained above.
Now, y= cx (given), and so:
Nx+cx = cx2 + 4c
Or, cx2 - (N+c)x + 4c = 0
The discriminant D = (N+c) 2 - 16c2 = (N-3c)(N+5c) ≥ 0
If both N-3c and N+5c are ≤ 0. then c ≤ -N/5, which is a contradiction, since both c and N are positive.
Accordingly, (N-3c) ≥ 0 and (N+5c) ≥ 0, so that: c ≤ N/3
From (ii), we observe that:
For N=2, we have c ≤ 2/3, so that c=0, which contradicts that both x and y are nonzero.
For N ε (3,4,5), we have: c= 1
For N ε (6,7,8), we have: c= 1,2
For N ε (9,10), we have: c= 1,2,3
For the original question, we have: N=10. Substituting (N,c) = (10, 1), (10,2) in turn in (i) does not yield any integer solution in x.
For (N,c) = (10,3), we have: x=3 as the integer root, so that :y=9, giving n=39
For N=3, substituting c=1 in (i), we obtain: x=2, so that: y=2, giving: n= 22
For N=4, substituting (N,c) = (4,1) in (i), we have: x = 1,4 as the roots of (i) . Since x > c, we have: x=1, giving: y=1, so that: n=11.
For N=5, we observe that, there does not exist any integer root for (i), whenever (N, c) = (5, 1)
For, (N,c) = (6,2), we obtain: x=2, so that: (y,n) = (4, 24)
For (N,c) = (7,1), (7,2) there does not exist any integer root for (i)
(N,c) = (8,1) do not yield any integer root of (i), while (s,c) = (8,2) gives: x=4, so that: (y,n) = (2, 12).
(N,c) = (9,1) and (9,2) do not yield integer roots to (i), while for (N,c) = (9,3), we have: x=2, so that: (y,n) = (6,26)
Summarizing, we have:
An extension to the problem:
Using computer assisted techniques here, Charlie has shown that the minimum positive integer base N such that the given equation possesses at least two solutions, occur at N=12, when n = 13, or 28.