The product of the digits of a two digit number is 1/3rd of the number itself. Assuming the tens digit is
and units digits then how many such numbers are possible?
Answers
A different (and admittedly more complicated) method of tackling this problem is by solving a system of linear congruences using the Chinese Remainder Theorem.
Suppose X is the two-digit number we’re looking for. From the question we know that X is a multiple of 7, and is 3 more than a multiple of 10. We can define a relation between two integers (say a and b) as being “similar” if they have the same remainder when divided by some integer n. We’ll denote this relation as
a ~ b (mod n).
Notice how both X and 0 have the remainder zero when divided by 7. Then
X ~ 0 (mod 7).
X divided by 10 gives a remainder of 3, and 3 divided by 10 has itself as a remainder (the latter may be a foreign concept. In grade school, 3 ÷ 10 is usually expressed as a fraction or decimal instead of 0 remainder 3). Then
X ~ 3 (mod 10).
Solving for X in the above two equations is called “solving a system of linear congruences.” In order to do this via the Chinese Remainder Theorem, the modulus of the two congruences (i.e. the numbers that come after “mod”) must be relatively prime (i.e. has no common divisors besides 1). There’s another theorem that states that if two integers a and b are relatively prime, there must exist a multiple of a and b such that their sum is equal to 1 (i.e. ap + bq = 1, where p,q are integers). Through trial and error we find that 7*(3) + 10*(-2) = 1.
Skipping the proof, the Chinese Remainder Theorem states that since 7 and 10 are relatively prime,
X = 3*(7*3) + 0*(10*-2) + (7*10)k = 63 + 70k
for any integer k will solve the above two equations. Since we know from the problem that X must be a two-digit number, k must equal zero, so X = 63.
Note: The sum of X’s units being one-seventh of X is a necessary consequence of the other conditions given in the problem (i.e. that X is a two-digit number, has 3 in the single’s unit, and is a multiple of 7).