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 Getting the Numbers With Product And Quotient (Posted on 2007-09-17)
(A) Analytically determine all possible two digit positive decimal integers n, that satisfy this equation:

n = p + 4c

where p is the product of the digits of n, while c is the result when the units digit is divided by the tens digit.

(B) For what positive integer bases N< 10 do valid solutions to the above equation exist?

 Submitted by K Sengupta Rating: 1.5000 (2 votes) Solution: (Hide) (A) 39 is the only possible two digit integer. (B) Valid solutions exist for N = 3, 4, 6, 8, 9 EXPLANATION: 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 (i) 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..(ii) 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: base     n 10       39 3         22 4         11 6         24 8         12 9         26 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.

 Subject Author Date re: non-analytic (spoiler) K Sengupta 2008-08-11 12:53:55 re: Solution - Part (A) + Part(B) Ady TZIDON 2007-09-18 04:32:13 Solution - Part (A) hoodat 2007-09-17 17:14:25 Analytical solution Ady TZIDON 2007-09-17 12:56:41 re: non-analytic (spoiler) Ady TZIDON 2007-09-17 12:48:34 non-analytic (spoiler) Charlie 2007-09-17 11:53:36 i got 2 bases- - WHO'S on the 3rd?? Ady TZIDON 2007-09-17 11:30:00

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