# Puzzles

## 10 December

Today's number is the smallest multiple of 24 whose digits add up to 24.

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There are one three-digit even numbers whose digits add to 24: 798, 888 and 996. Of these, only 888 is a multiple of 24.

## 15 December

Today's number is smallest three digit palindrome whose digits are all non-zero, and that is not divisible by any of its digits.

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The number can contain no 1 (as all numbers are multiples of 1). It's first and last digits cannot be 2 (as it would then be even and so a multiple of 2).

The lowest three-digit palindrome containing no 0 or 1 and not ending in 2 is **323**. This is not a multiple of 2 or 3 and so is the solution.

## 20 December

What is the largest number that cannot be written in the form \(10a+27b\), where \(a\) and \(b\) are nonnegative integers (ie \(a\) and \(b\) can be 0, 1, 2, 3, ...)?

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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,

$$219 = 10\times30 - 27\times3$$ $$= 10 + 10\times29 - 27\times3 $$ $$= 1\times10 + 27\times11$$

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":

$$10\times(-1) + 27\times13 = 10\times28 + 27\times(-1) = 233$$

So the answer is 233.

#### Extension

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?

## Elastic numbers

*Throughout this puzzle, expressions like \(AB\) will represent the digits of a number, not \(A\) multiplied by \(B\).*

A two-digit number \(AB\) is called *elastic* if:

- \(A\) and \(B\) are both non-zero.
- The numbers \(A0B\), \(A00B\), \(A000B\), ... are all divisible by \(AB\).

There are three elastic numbers. Can you find them?

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15, 18 and 45 are elastic.

15's factors are 5 and 3. 105, 1005, 10005, etc will all be multiples of 5 (because they end in 5) and multiples of 3 (as their digits add to 6). Hence they are all multiples of 15.

Similarly, 108, 1008, 10008, etc are all multiples of 9 (adding digits) and 2 (they are even), so they are multiples of 18; and 405, 4005, 40005, etc are all multiples of 9 (adding digits) and 5 (last digits are 5), so they are multiples of 45.

#### Extension

How many elastic numbers are there in other bases?

## 14 December

Today's number is the largest number that cannot be written in the form \(27a+17b\), where \(a\) and \(b\) are positive integers (or 0).

## Combining multiples

In each of these questions, positive integers should be taken to include 0.

1. What is the largest number that cannot be written in the form \(3a+5b\), where \(a\) and \(b\) are positive integers?

2. What is the largest number that cannot be written in the form \(3a+7b\), where \(a\) and \(b\) are positive integers?

3. What is the largest number that cannot be written in the form \(10a+11b\), where \(a\) and \(b\) are positive integers?

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

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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 \(m=7\):

Multiple | Remainder |

0 | 0 |

3 | 3 |

6 | 6 |

9 | 2 |

12 | 5 |

15 | 1 |

18 | 4 |

21 | 0 |

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.

#### Extension

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?

## Subsum

1) In a set of three integers, will there always be two integers whose sum is even?

2) How many integers must there be in a set so that there will always be three integers in the set whose sum is a multiple of 3?

3) How many integers must there be in a set so that there will always be four integers in the set whose sum is even?

4) How many integers must there be in a set so that there will always be three integers in the set whose sum is even?

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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:

#### 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?

## Fill in the digits

Can you place the digits 1 to 9 in the boxes so that the three digit numbers formed in the top, middle and bottom rows are multiples of 17, 25 and 9 (respectively); and the three digit numbers in the left, middle and right columns are multiples of 11, 16 and 12 (respectively)?