Guess my Number

 

252 

 

816 

642

440

512

3145

9876

8125

Square Number Triangle

Solution

This can be deduced through exhaustion. Shortcuts can be taken, such as A not being a multiple of 10 (otherwise B has a leading 0) and each number only being able to end in certain digits because they are square.

A=144, B=441 and C=121. 12 is a factor of A and 21 is a factor of B.

A+B+C= 706.

Here's another problem which is more open-ended.

Solution

A and C have the same first digit and are both 3 digits long. If A is a factor of C then A=C. By the same logic, B=C as they share the same last digit. So A=B=C.

But what's special about C? 

Well, the first and last digit of B are the same, because A and C end in the same digit. Then, as A=B=C, we know that C is of the form XYX where X and Y represent digits and X is not 0.

Here's a similar problem that's a bit less open-ended. 

Solution

If A+B=C then A must end with the digit 0, because B and C share a common last digit. But B cannot start with 0 as it's a properly written 3 digit number.

For the subtraction proof, B is a 3 digit number, meaning that at least 100 is being subtracted from A. The first digit of C must be less than the first digit of A, but A and C share a common first digit so this cannot be true. Hence, A-B=C cannot be true.

Ones and Zeros

Solution

Triangular Numbers

The triangular numbers form a quadratic sequence, but interestingly the last digit of each triangular number forms it's own sequence.

 

The sequence begins with 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1. This is a palindrome!

Including the "zero-th" term and 19th term, we get 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0. The next 20 terms follow the same sequence and given that the last digits of \(n\) repeat every 10 terms, these 20 terms will repeat indefinitely.