What number is it when the the ones digit is one more than its tens digit and it's a composite number has two digits and it's even and has two less than a square number?

Answer 1

Let's list the requirements for the mystery number: One's digit more than the ten's digit. Composite number. Two digits. Even number. Is two less than a square. Now that we've listed the requirements, let's look for a place to "attack" the problem. The easiest is probably to look at squares since our mystery number is two less than a square. Here are the first few squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 and 121. Since our mystery number is 2 less than a square, let's go through that. -1, 2, 7, 23, 34, 47, 62, 79, 98, 119. Now we'll look at two things at once; we'll look only at the 2-digit numbers that are even. (We don't have to look at the "composite number issue" because every even number except 2 is composite since it can be divided by 2.) 34, 62, and 98. Now the number with the one's digit greater than the ten's digit - and there's only one of them: 34. The number 34 has a one's digit greater than the ten's digit, is composite, has two digits, is even, and is 2 less than a square.