# New Scientist Enigma 1714 – Building Sand Castles

*by Susan Denham*

#### From Issue #1714, 28th April 1990

You can add a letter at a time to build up words from ‘A’ to ‘CASTLE’ in the following way:

**A → AT → CAT → CAST → CASTE → CASTLE**

Now I can substituted digits for letters in many ways, one being:

**5 → 56 → 256 → 2576 → 25761 → 257631**

in which there are numbers divisible by 2, by 3, by 5 … and in fact only the first number is prime.

Your task today is to find a different substitution of non-zero digits in that word-chain (as always different digits consistently replacing different letters) so that none of the resulting numbers is divisible by 2 or 3 or 5 — and in fact so that they are all prime and so that **SANDS** is prime too.

You don’t have to dig too hard — there are lots of short-cuts and, of course, there is an answer.

What is the value of **CASTLE**?

# Sunday Times Teaser 3020 – Baz’s Bizarre Arrers

*by Stephen Hogg*

#### Published Sunday August 09 2020 (link)

“Bizarrers” dartboards have double and treble rings and twenty sectors ordered as on this normal dartboard. However, a sector’s central angle is [100 divided by its basic score]°. The 20 sector incorporates the residual angle to complete 360º.

Each player starts on 501 and reduces this, eventually to 0 to win. After six three-dart throws, Baz’s seventh could win. His six totals were consecutive numbers. Each three-dart effort lodged a dart in each of three clockwise-adjacent sectors (hitting, in some order, a single zone, a double zone and a treble zone). The three-sector angle sum (in degrees) exceeded that total.

The sectors scored in are calculable with certainty, but not how many times hit with certainty, except for one sector.

Which sector?

# Sunday Times Teaser 3019 – William’s Prime

*by Bill Scott*

#### Published Sunday August 02 2020 (link)

William was searching for a number he could call his own. By consistently replacing digits with letters, he found a number represented by his name: WILLIAM.

He noticed that he could break WILLIAM down into three smaller numbers represented by **WILL**, **I** and **AM**, where **WILL** and **AM** are prime numbers.

He then calculated the product of the three numbers **WILL, I** and **AM**.

If I told you how many digits there are in the product, you would be able to determine the number represented by **WILLIAM**.

# Sunday Times Teaser 3018 – Debased

*by Howard Williams*

#### Published Sunday July 26 2020 (link)

Sarah writes down a four-digit number then multiplies it by four and writes down the resultant five-digit number. She challenges her sister Jenny to identify anything that is special about these numbers. Jenny is up to the challenge as she identifies two things that are special. She sees that as well as both numbers being perfect squares she also recognises that if the five-digit number was treated as being to base 7 it would, if converted to a base 10 number, be the same as the original four- digit number.

What is the four-digit number?

# Sunday Times Teaser 3017 – Mr Green’s Scalene Mean Machine

*by Stephen Hogg*

#### Published Sunday July 19 2020 (link)

My maths teacher, Mr. Green, stated that the average of the squares of any two different odd numbers gives the hypotenuse of a right-angled triangle that can have whole-number unequal sides, and he told us how to work out those sides.

I used my two sisters’ ages (different odd prime numbers) to work out such a triangle, then did the same with my two brothers’ ages (also different odd prime numbers). Curiously, both triangles had the same three-figure palindromic hypotenuse. However, just one of the triangles was very nearly a 45° right-angled triangle (having a relative difference between the adjacent side lengths of less than 2%).

In ascending order, what were my siblings’ ages?

# Sunday Times Teaser 3016 – Eureka

*by Andrew Skidmore*

#### Published Sunday July 12 2020 (link)

Archimedes’ Principle states that the upward force on a floating body equals the weight of water displaced by the body. In testing this, Liam floated a toy boat in a cylindrical drum of water. The boat had vertical sides and took up one tenth of the surface area of the water. He also had some identical ball bearings (enough to fit snugly across the diameter of the drum), which were made of a metal whose density is a single-figure number times that of water.

Liam loaded the boat with the ball bearings and noted the water level on the side of the drum. When he took the ball bearings out of the boat and dropped them in the water, the water level changed by an amount equal to the radius of a ball bearing.

How many ball bearings did Liam have?

# Sunday Times Teaser 3015 – Quid Quo Pro

*by Victor Bryant*

#### Published Sunday July 05 2020 (link)

In Readiland the unit of currency is the quid. Notes are available in two denominations and with these notes it is possible to make any three-figure number of quid. However, you need a mixture of the denominations to make exactly 100 quid. Furthermore, there is only one combination of denominations that will give a total of 230 quid.

What are the two denominations?

# Sunday Times Teaser 3014 – Family Business

*by Danny Roth*

#### Published Sunday June 28 2020 (link)

George and Martha run a company with their five daughters. The telephone extensions all have four positive unequal digits and strangely only four digits appear in the seven extensions:

\[\begin{array}{|l|r||}

\hline Andrea & abcd \\

\hline Bertha & acdb \\

\hline Caroline & bacd \\

\hline Dorothy & dabc \\

\hline Elizabeth & dbca \\

\hline George & cabd \\

\hline Martha & cdab \\

\hline \end{array}\]

They noticed the following:

Andrea’s and Bertha’s add up to Dorothy’s.

Bertha’s and Elizabeth’s add up to George’s.

Caroline’s and Dorothy’s add up to Martha’s.

What is Andrea’s extension?

# Sunday Times Teaser 3013 – Arian Pen-Blwydd

*by Howard Williams*

#### Published Sunday June 21 2020 (link)

When I thought that my daughter was old enough to be responsible with money I gave her on her next, and all subsequent birthdays, cash amounts (in pounds) which were equal to her birthday age squared.

On her last birthday her age was twice the number of years for which she received no such presents. I calculated at this birthday that if I had made these gifts on all of her birthdays then she would have received 15% more than she had actually received. I then decided that I would stop making the payments after her birthday when she would have received only 7.5% more if the payments had been made on all of her birthdays.

What was the amount of the final birthday payment?

# New Scientist Enigma 553 – Playing For Time

*by Chris Maslanka*

#### From Issue #1706, 3rd March 1990

One disadvantage of doing recurrence relations in your head, Kugelbaum told himself, was that all to often you were still calculating away when the clock showed it was time to give a class. To buy himself a little more time he jotted down on the whiteboard:

\[\frac{A\widehat{X}B}{B\widehat{X}A}=\frac{AB}{BA}\]

‘In this relation’, explained Kugelbaum to the class, ‘\(A\), \(B\) and \(X\) are all non-zero digits and \(AB/BA\) is a proper fraction in its lowest terms. The hat above the X in the numerator and that above the X in the denominator means that the digit \(X\) is used but it must be repeatable an arbitrary number of times without affecting the validity of the equation. So the \(X\) can be there once, twice, … 37 times … but, no matter how many times you use it, the equation must remain true. It is, or course, to be taken for granted that the number of \(X\)s interposed between the A and B must be the same for both the numerator and the denominator”.

So saying, the professor returned to his mental calculations while the class set about solving the problem.

How many distinct solutions does Kugelbaum’s equation have?

NB. A proper fraction has its numerator smaller than its denominator; if it’s in its lowest terms, numerator and denominator have no common factor other than 1.