A certain positive integer N is twice the product of its digits.
Find N.
For N=3 digits, 100a +10b +c = 2(a + b + c)
or 98a + 8b = c which requires a=0. A larger value of N is similarly ruled out.
So 10a + b = 2a + 2b, 8a = b.
Then a=1, b=8, and N=18 = 2(1+8).

Posted by xdog
on 20191029 12:01:05 