*by Susan Denham*

#### From Issue #2091, 19th July 1997 (link)

To enter the lottery I chose six lucky numbers: I called this “selection A”. I have added 1 to each of those six numbers to form a new “selection B”, I’ve added 1 again to each to form “selection C”, and repeated this twice more to form “selection D” and “selection E”. So, for example, the lowest number in selection E is four more than the lowest number in selection A. Of course, as always, each selection is in the range 1-49.

In one of the selections (but not selection A) the six numbers are all two-digit numbers with the same first digit.

In one of the selections (but not selection A) there are no prime numbers.

Each week I decide to have two goes at the lottery and I choose at random two of my five selections. I know that whichever pair I choose, it would be possible for one of the selections to contain four of the six winning numbers and the other selection to contain none. On the other hand, whichever pair I choose, it would also be possible for both selections to contain four of the six winning numbers.

Which six numbers form “selection A”?

## Sunday Times Teaser 3103 – Empowered

*by Howard Williams*

#### Published Sunday March 13 2022 (link)

I have a rectangular garden whose sides are whole numbers of feet. Away from the edge, an exposed strip of ground, again a whole number of feet in width, runs straight across (not diagonally) from a shorter side of the garden to the other shorter side. I need to run an electrical cable along the ground, between two opposite corners of the garden. Where the cable crosses the exposed area, it has to be threaded through expensive linear ducting to avoid damage. Because of costs, whilst seeking to minimise the length of cable, my overriding concern is to minimise the length of ducting used.

The straight-line distance between the two corners to be connected is 123 feet, but in minimising costs, the length of cable needed is a whole number of feet longer than this.

What is the length of cable needed?

## Sunday Times Teaser 3102 – Jokers

*by Victor Bryant*

#### Published Sunday March 06 2022 (link)

A group of us wanted to play a card game. We had a standard 52-card pack but we needed to deal out a whole pack with each person getting the same number of cards, so I added just enough “jokers” to make this possible. Then I randomly shuffled the enlarged pack (in this shuffled pack there was more than a 50:50 chance that there would be at least two jokers together). Then I dealt the cards out, each of us getting the same number (more than three). There was more than a 50:50 chance that my cards would not include a joker.

How many of us were there?

*by Mark Valentine*

Published Sunday February 27 2022 (link)

In a charity fundraising drive, two friends completed laps of their local track. Alan, a cyclist, raised £1 for his first lap, £2 for the second, £3 for the third and so on, with each extra lap earning £1 more than the prior. Bob, who walks, raised £1 for his first lap, £2 for the second, £4 for the third and so on, with each extra lap earning double the prior. Starting together, they travelled at constant speeds, and finished their last lap simultaneously.

After they had added up their totals, they were surprised to find they each raised an identical four-figure sum.

Including the beginning and end of their drive, how many times did Alan and Bob cross the lap start-finish line together?

*by Nick MacKinnon*

#### Published Sunday February 20 2022 (link)

Pip and Estella know each other to be honest and clever. At the reading of Miss Havisham’s will they see a large crate on which are placed two envelopes addressed V+S+E and V, and a lipstick. The solicitor gives Estella the V+S+E envelope and Pip the V envelope and says, “Inside this crate Miss Havisham has left a cuboidal gold bar whose dimensions in mm are different whole numbers greater than 1, and whose volume, surface area and total edge length are V mm³, S mm² and E mm respectively. Inside your envelope, which you should now open, is the secret value of your address. Every fifteen minutes a bell will ring, and if at that point you know the other’s value, write both values on the crate with lipstick and the gold is yours.” At the first bell Pip and Estella sat still, but when the bell rang again Estella lipsticked 3177 and Pip’s value on the crate.

What was Pip’s value?

## Polynomial Library Update

I have just updated my Polynomial library to include a very fast cyclotomic polynomial generator.

*by Andrew Skidmore*

#### Published Sunday February 13 2022 (link)

Liam is learning about fractions and decimals. He has been shown that some fractions give rise to decimals with a recurring pattern eg 4/11 = 0.3636… . Each pupil in his class has been given four numbers. The largest (two-figure) number is to be used as the denominator and the others as numerators, to create three fractions that are to be expressed as decimals. Liam found that each decimal was of the form 0.abcabc…, with the same digits but in different orders.

Liam’s largest number contained the digit 7 and if I told you how many of his four numbers were divisible by ten you should be able to work out the numbers.

What, in increasing order, are the four numbers?

*by Michael Fletcher*

#### Published Sunday February 06 2022 (link)

I have two octahedral (eight-faced) dice. There are positive whole numbers (not necessarily all different) on each of their faces. I throw the two dice and add the two numbers that are showing on the top faces. The probability of a total of 2 is 1/64, the probability of a total of 3 is 2/64 and so on. The probabilities of the totals are the same as for a pair of standard octahedral dice (each numbered 1 to 8). Interestingly, however, the highest number on the first die is 11.

What are the eight numbers (in ascending order) on the second die?

## Sunday Times Teaser 3097 – Crazy Golf

*by Angela Newing*

#### Published Sunday January 30 2022 (link)

Ian was trying to entertain John and Ken, his 2 young nephews, in the local park, which had a 9-hole crazy golf course. He decided to play a round with each of them separately, and gave them money for each hole he lost on a rising scale. He would pay £1 if he lost the first hole (or the first hole after winning one), then £2 if he lost the next consecutive hole, and £3 for the third, and so on.

In the event, Ian won only 5 holes in total between the two rounds, including the first hole against John and the last hole against Ken. There were no ties. At the reckoning after both rounds, both boys received equal amounts of money.

How much did it cost Uncle Ian?

*by Victor Bryant*

#### From The Sunday Times, 18th November 1979 (link)

Problems concerning ages have always proved fruitful and entertaining exercises to both mathematicians and non-mathematicians. Trial and error methods, calculators and normal or esoteric mathematical techniques can all be deployed to find the correct solution. The most elegant or the most economical method is naturally the most commendable, but the correct solution, however obtained, is the desideratum.

Our problem concerns six men whose ages are within the range 21 to 89 and any two of them differ by at least 9. If we take the two digits comprising each of the ages of three of the men, and reverse them, we obtain the ages of the other three men.

What is more, if we take the sum of the ages of the first group, we find that it equals the sum of the ages of the second group of three.

Also the sum of the squares of the three ages of the first group equals the sum of the squares of the ages of the second group of three.

Finally, one of the ages in each group is exactly twice an age in the other group.

What are the ages of the six men (in increasing order)?