Puzzles

The sum of 13 consecutive integers is 13 times the middle number, so today's number is a multiple of 13. The sum of 12 consecutive integers is 6 times the sum of the two middle numbers, so today's number is also a multiple of 6. Therefore today's number is a multiple of 78.

78 and 156 do not work (the numbers would not all be strictly positive), so today's number is 234.

The digits of the number must be 1, 2 and 3, so the largest number is 321.

The only way to make 208 by adding consecutive numbers is 10+11+12+13+14+15+16+17+18+19+20+21+22. Therefore next year Noel will give his grandchildren a total of £221.

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?

Yes: 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16.

It is clearly possible for the numbers 1 to 15 (remove the 16 from the end of the sequence above) and 1 to 17 (add 17 to the start of the sequence above).

The OEIS sequence A090461 gives other numbers for which this is possible. It starts 15, 16, 17, 23, then includes every number from 25 onwards. It is conjectured, but not proven, that it is possible for every number above 25.

415

1. 7

2. 11

3. 90

First, if \(n\) and \(m\) share a common factor other than 1, there will be no largest number: any number that is not a multiple of the common factor will be impossible to make.

If \(n\) and \(m\) are coprime, then considered the remainders when the multiples of \(n\) are divided by \(m\). For example, if \(n=3\) and \(m=7\):

Once a number with a given remainder is reached, then all other numbers with that remainder can be reached by repeatedly adding \(m\). Once \(mn\) is reached, the remainders column will repeat itself. Before \(mn\), all remainders will appear (this can be shown by showing that there are \(m\) rows which much all have different remainders). Hence above \(mn\) all numbers can be made.

In the 3,7 example, the last remainder to be hit is 4. The highest number that cannot be made will be the highest number with remainder 4 that is less than 18 (when remainder 4 is hit).

In general, the last remainder will be hit at \(mn-n\). The number before this with the same remainder will be \(mn-n-m\). This will be the highest number that cannot be made.

Given \(n\), \(m\) and \(k\), what is the largest number that cannot be written in the form \(na+mb+kc\), where \(a\), \(b\) and \(c\) are positive integers?

1) Yes, there must be either at least two even integers or at least two odd integers. The sum of two even integers is even. The sum of two odd integers is even.

2) Five.

3) Five.

4) An infinite number: in a list of odd integers, there will never be three integers which add up to an even number.

This puzzle leads naturally into the following extension:

How many integers must there be in a set so that there will always be \(a\) integers in the set whose sum is a multiple of \(b\)?

For which values of \(a\) and \(b\) will the answer to this be finite?