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42 is composite, 43 is prime.
The number ends in 2, and even digit and so it is divisible by 2. That is, it is divisible be a number other than 1 and itself - ie it is a composite number.
would it will be 15
To be an even number then the final digit = 2. The largest single prime digit is 7 (as both 8 and 9 are composite). The greatest even number fulfilling the conditions is 77777772.
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.