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Getting the Numbers With Product And Quotient (Posted on 2007-09-17) Difficulty: 3 of 5
(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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: non-analytic (spoiler)K Sengupta2008-08-11 12:53:55
Solutionre: Solution - Part (A) + Part(B)Ady TZIDON2007-09-18 04:32:13
SolutionSolution - Part (A)hoodat2007-09-17 17:14:25
Analytical solutionAdy TZIDON2007-09-17 12:56:41
re: non-analytic (spoiler)Ady TZIDON2007-09-17 12:48:34
Solutionnon-analytic (spoiler)Charlie2007-09-17 11:53:36
Some Thoughtsi got 2 bases- - WHO'S on the 3rd??Ady TZIDON2007-09-17 11:30:00
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