Puzzles

768

414

Incredibly, the result will always be \(n^2\).

To see why, imagine writing every number, \(n+1\leq k\leq 2n\), in the form $$k=2^ab$$ where \(b\) is an odd number and also the \(k\)'s largest odd factor. The next largest number whose largest odd factor is \(b\) will be \(2^{a+1}b=2k\). But this will be larger than \(2n\), so outside the range. Therefore each number in the range has a different largest odd factor.

Each of the largest odd factors must be one of \(1, 3, 5, ..., 2n-1\), as they cannot be larger than \(2n\). But there are \(n\) odd numbers here and \(n\) numbers in the range, so each number \(1, 3, 5, ..., 2n-1\) is the highest odd factor of one of the numbers (as the highest odd factors are all different).

Therefore, the sum of the odd factors is the sum of the first \(n\) odd numbers, which is \(n^2\).

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?

720720 has 240 factors.

Many more puzzles like this one can be found in my post on the Chalkdust blog.

(b) implies that the digits of \(N\) are all 1 or 7, so \(N\) can only be 111, 117, 171, 177, 711, 717, 771 or 777. These are all divisible by 3, so no such integers \(N\) exist.

Consider 21-digit integers \(N\) such that:

(a) \(N\) is not exactly divisible by 2, 3 or 5.

(b) No digit of \(N\) is exactly divisible by 2, 3 or 5.

How many such integers \(N\) are there?