Question

For how many two- digit positive numbers will tripling the tens digit give us a two-digit number that is triple the original number?

Answer 1

well, recall that a two digit number can be expressed \(N=10A+B\) for digits \(A,B\) (\(A\) corresponds to our ten's place, \(B\) our one's place)... anyways, observe that tripling \(N\) we get:$$3N=30A+3B$$yet tripling the ten's digit only gives:$$N'=10(3A)+B=30A+B$$

Answer 2

now it's just algebra that tells us that \(N'=3N\) means \(30A+3B=30A+B\) and so \(3B=B\) i.e. \(B=0\) i.e. our one's digit must be \(0\)

Answer 3

lastly for us to be able to 'triple' our first digit and still get a valid digit as a result (i.e. a number from \(1\) to \(9\)) it must be that our digits cannot be bigger than \(3\) (since \(3\times3=9\) but \(4\times 3=12\) which is no longer a valid digit) hence we have that \(A\) (our ten's place digit) can be either \(1,2,\text{ or }3\) while our one's digit must be \(0\) hence we have \(3\times1=3\) possibilities: \(10,20,30\)