Advent calendar 2017

Alex lives in Maidstone, Kent (1) and wants French hens (3).

Ben lives in Leighton Buzzard, Beds (8) and wants gold rings (5).

Carol lives in Burcot, Worcestershire (2) and wants calling birds (4).

If \(p\) and \(q\) are prime numbers, then the number \(p^a\times q^b\) will have \((a+1)(b+1)\) factors. This is because all its factors are of the form \(p^\alpha\times q^\beta\), with \(\alpha=0,1,...,a\) and \(\beta=0,1,...,b\). The same idea can be used on numbers with three or more prime factors; in general the number \(p_1^{a_1}\times...\times p_n^{a_n}\) has \((a_1+1)\times...\times(a_n+1)\) factors.

28 can be written as: 28, 14×2, 7×4, or 7×2×2. Therefore the following numbers have 28 factors:

and any other number with 28 factors will have larger prime factors, so will be larger.

These numbers are 134217728, 24576, 1728 and 960. Therefore the smallest number with 28 factors is 960.

120

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.

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.

512

529

343

106

196

495

216

Adding up all three sums of pairs (284) will give twice the sum of the three numbers. So the sum of the three numbers is 142.

123