Puzzles

294

496

You can find my write up of a solution to this here. Pedro Freitas (@pj_freitas) sent me the following nice alternative solution:

Any number can be written as \(10a+27b\) with integer \(a\) and \(b\), since \(1=3\times27-8\times10\). So the problem may be thought of as asking when one of \(a\) and \(b\) must be negative.

Given one way of writing a number, you can get the others by shifting by 14*29. For example,

So the question now becomes: When does this adjustment fail to eliminate negative numbers?

This is when you are at what Pedro calls "limit coefficients":

So the answer is 233.

Let \(n\) and \(m\) be integers. What is the largest number that cannot be written in the form \(na+mb\), where \(a\) and \(b\) are nonnegative integers?

268

154

194

570

The number of 10 character strings containing at least 3 As is equal to the number of 10 character strings containing exactly 3 As plus the number of 10 character strings containing exactly 4 As plus ... plus the number of 10 character strings containing exactly 10 As. The number of 10 character strings containing exactly \(a\) As is \(\left(\begin{array}{c}10\\a\end{array}\right)\) ("10 choose \(a\)"). Therefore the number we are looking for is

You can work this out by either using the formula \(\left(\begin{array}{c}n\\r\end{array}\right)=\frac{n!}{r!(n-r)!}\), or by remembering that these numbers all appear in the 10th row of Pascal's triangle.

The answer is 968.