Puzzles

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.

819

755

315

497

The number of 0s at the end of 1000! will be equal to the number of times 10 is a factor of it. For each 10 that is a factor, there will be a 5 and a 2 that are factors. Therfore the number of 0s will be equal to the number of times 5 is a factors of 100! (because 5 is larger than 2).

Therefore the number of 0s at the end of 1000! will be equal to \(\left\lfloor\frac{1000}{5}\right\rfloor+\left\lfloor\frac{1000}{25}\right\rfloor+\left\lfloor\frac{1000}{125}\right\rfloor+\left\lfloor\frac{1000}{625}\right\rfloor\). This is 249.