Pamela, we have two responsrs for you.

Hi Pamela,

This is an interesting question because you don't need the second sentence at all. Suppose that the digits of the number are a and b with b the units digit. Thus the number is ab. Since the number is 5 times the units digit

ab = 5 b

But the units digit of any multiple of 5 is either 5 or 0.

Can b be 0?
If b is 5, what is a?

Thus there is only one two digit number that satisfies the first sentence in your message to us. Is the second sentence true for this number?

Penny

Hi there,

Let's set up the words of the problem algebraically:

"A two digit number is five times its units digit":

If we call the units digit "y" and the other digit "x", we can write the two-digit number as "xy". Then, if we assume we are working in a base 10 number system:

10x + y = 5(y)

Which means that the sum of the ys (units) and the xs (tens) equals five times the units digit, as specified in the problem.

This doesn't give us enough information to solve the problem, however. Let's look at the second part of the problem:

"If the digits are reversed, the resulting number is 27 more than the original number":

So, reversing the digits of number xy would give a number yx. This number is supposed to be 27 more than the original, so let's write:

10x + y + 27 = 10y + x

Now we can use these two equations to solve for x and y, the digits of our number!

Hope this helps.
Gabriel

Footnote:

Pamela,

Penny's technique uses the property that the units digit of any multiple of 5 is either 5 or 0. For the similar problem

A two-digit number is six times its units digit. If the digits are reversed, the resulting number is 18 more than the original number. Find the original number.

Penny's method won't work but Gabriel's will.

Harley