Two natural numbers, p and q, do not end in 0. The products of any pair, p and q, is a power of 10 (that is 10, 100, 1000, 10000 etc). If p q, the last digit of p-q cannot be
a-1
b-3
c-5
d-7
e-9

Question

Two natural numbers, p and q, do not end in 0. The products of any pair, p and q, is a power of 10 (that is 10, 100, 1000, 10000 etc). If p> q, the last digit of p-q cannot be
a-1
b-3
c-5
d-7
e-9

anonymous · Answer

Is it C?

anonymous · Answer

Note that, in order to have a power of ten as $N$, we must have a prime factorization of:
$$
N=2^a5^b
$$But, since none of our number end in zero, the only possible factorization is:
$$
pq=N
$$Where:
$$
p=2^a
$$And:
$$
q=5^b
$$With $a,b>0$, without loss of generality.

So, then, we must have:
$$
|p-q|=|2^a-5^b|
$$

Note that any power of 5 must end in 5, while powers of two end in all even digits (excluding 0).

You could end here and say that, since we have to have an even number subtracted from 5, then 5 cannot be a number hence C is correct, but, to expand:

The previous means that our possible numbers are:
$$
\{5-2,5-4,5-6,5-8\}=\{3,1,-1,-3\}=\{1,3,7,9\}
$$Which excludes 5.