Puzzles

Consider the top, front, and right sides of the cube.

If the top number is \(a\) more than a multiple of six, then the front and right numbers must both be \(a\) less than a multiple of 6 (so that when added to the top number they make a multiple of 6). But when the front and right numbers are added, they make \(2a\) less than a multiple of 6; but this must also be a multiple of 6.

This is only possible if \(a=0\) or \(a=3\). So the numbers must be either all multiples of 6, or all 3 more than multiples of 6.

The smallest set of numbers that are all 3 more than multiples of 6 is 3,9,15,21,27,33. The sum of these is 108. The smallest set of numbers that are all multiples of 6 is the set with each number three more than these, so 108 is the smallest possible total.

Six numbers are written on the faces of a cube. The sum of the numbers on any two adjacent faces is a multiple of \(n\).

What is the smallest possible sum of the six numbers?

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