Given that x is a positive integer, determine all possible values of a positive integer constant a such that the product of the digits in the base ten representation of x is equal to x² – ax - 22.

*** For an extra challenge, solve this puzzle without using a computer program.

let p(x) be the product of the digits of x (base 10)
then we have
x^2-ax-22=p(x)
ax=x^2-p(x)-22
a=(x^2-p(x)-22)/x
a=x-(p(x)+22)/x

thus, for a to be an integer
we need for x to divide p(x)+22
now if x has n digits then p(x) is at most 9n
and x is at least 10^(n-1)
thus we need 9n+22>10^(n-1)
if n=1 we have 9+22=31>10^0=1
if n=2 we have 18+22=40>10^1=10
if n=3 we have 27+22=59>10^2=100 which fails
thus x can only have 1 or 2 digits

if x has 1 digit:
p(x)=x and we have
a=x-(x+22)/x=x-1-(22/x)
thus x needs to be a 1 digit divisor of 22, which gives
us x=1, thus a=1-1-22=-22, but this is not positive
thus no 1 digit x's work

if x has 2 digits:
let x=10m+n
then p(x)=mn
then we have
a=10m+n-(mn+22)/(10m+n)
thus we need
mn+22>=10m+n
mn-10m>=n-22
(m-10)n>=n-22
m-10>=(n-22)/n
m>=10-(22-n)/n
thus for
n=1 m>=-11
n=2 m>=0
n=3 m>=4
n=4 m>=6
n=5 m>=7
n=6 m>=8
n=7 m>=8
n=8 m>=9
n=9 m>=9
checking all of these yields
x=12 a=10
x=34 a=33
x=45 a=46
x=57 a=56
x=78 a=77

thus all of the possible values for a are
(10,33,46,56,77)

Posted by Daniel on 2011-10-05 11:09:58