SOLUTION: Let p(n) and s(n) denote the product and the sum, respectively, of the digits of the integer n. For example, p(23)=6 and s(23)=5. Suppose n is a 2 digit number such that n=p(n) +s(

Algebra ->  Equations -> SOLUTION: Let p(n) and s(n) denote the product and the sum, respectively, of the digits of the integer n. For example, p(23)=6 and s(23)=5. Suppose n is a 2 digit number such that n=p(n) +s(      Log On


   

Question 987739: Let p(n) and s(n) denote the product and the sum, respectively, of the digits of the integer n. For example, p(23)=6 and s(23)=5. Suppose n is a 2 digit number such that n=p(n) +s(n). What could n be?
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
n=p(n) +s(n).

Let the tens digit be t and the units digit be u:

p(n) = tu;  s(n) = t+u; n = 10t+u 

10t+u = tu + t+u

9t-tu = 0

t(9-u) = 0

t=0;  9-u=0
        9=u

The tens digit can't be 0.

So the units digit is 9.

Substituting:

9t-tu=0
9t-t%289%29=0
9t-9t=0
0t=0

Any digit but 0 can be the tens digit.

So there are nine solutions.

19 = 1×9 + 1+9 =  9+10 = 19
29 = 2×9 + 2+9 = 18+11 = 29 
39 = 3×9 + 3+9 = 27+12 = 39
49 = 4×9 + 4+9 = 36+13 = 49
59 = 5×9 + 5+9 = 45+14 = 59
69 = 6×9 + 6+9 = 54+15 = 69
79 = 7×9 + 7+9 = 63+16 = 79
89 = 8×9 + 8+9 = 72+17 = 89
99 = 9×9 + 9+9 = 81+18 = 99

Edwin