by Stephen Hogg
Published August 18 2019 (link)
Using her ordinary 15cm ruler and Zak’s left-handed version (numbers 1 to 15 reading right to left) Kaz could display various fractions. For instance, putting 5 on one ruler above 1 on the other ruler, the following set of fractions would be displayed: 5/1, 4/2, 3/3, 2/4 and 1/5. Zak listed the fifteen sets starting from “1 above 1” up to “15 above 1”.
Kaz chose some fractions with values less than one from Zak’s sets (using just the numerals 1 to 9, each once only in her selection). Of these, two were in simplest form, one of which had consecutive numerator and denominator. Zak correctly totalled Kaz’s selection, giving the answer as a fraction in simplest form. Curiously, the answer’s numerator and denominator were both palindromic.
Give Zak’s answer.
by Eric Emmet
From Issue #1664, 13th May 1989
Four football teams are to play each other once. After some of the matches have been played a document giving some details of the matches played, won, lost and so on looked like this:
|Team||Played||Won||Lost||Drawn||Goals For||Goals Against||Points|
Two points are given for a win and one point to each side in a drawn match.
Find the score in each match.
by Danny Roth
Published August 11 2019 (link)
George and Martha’s garden is a perfect square of length (whole number of metres) a two-digit number ab. The area is a three-digit number cde. In the middle, they have planted a square flowerbed of a length which is a single-digit number f and area a two-digit number gh.
They have called in a gardener, who works for a single-digit i hours. He works for a whole number of minutes on the flowerbed and the remainder on the surrounding lawn. Each square metre of the flowerbed requires n (a single digit) times the time spent on each square metre of the surrounding lawn. I have mentioned nine letters, a-i inclusive, and each stands for a different positive digit.
How many minutes does the gardener work on the lawn?
by Susan Denham
From Issue #2154, 3rd October 1998
To celebrate next week’s 1000th edition of Enigma, we each made up an Enigma. Each one consisted of four clues leading to its own unique positive whole number answer. In each case none of the four clues was redundant. To avoid duplication, Keith made up his Enigma first and showed it to Susan before she made up hers.
The two Enigmas were meant to be printed side-by-side but the publishers have made a (rare) error and printed the clues in a string:
(A) It is a three-figure number;
(B) It is less than a thousand;
(C) It is a perfect square;
(D) It is a perfect cube;
(E) It has no repeated digits;
(F) The sum of its digits is a perfect square;
(G) The sum of its digits is a perfect cube;
(H) The sum of all the digits which are odd in Keith’s
answer is the same as the sum of all the digits
which are odd in Susan’s.
Which four clues should have formed Keith’s Enigma, and what was the answer to Susan’s?
by Angela Newing
Published August 4 2019 (link)
I have done a “long multiplication”, which is reproduced above. [If the multiplication was ABC x DE, then the third line shows the result of ABC x E and the fourth line shows the result of ABC x D]. However instead of writing the actual digits involved, I have written “o” where there is an odd digit and “e” where there is an even digit.
What is the result of the multiplication?
by Oliver Anderson
From Issue #1662, 29th April 1989
Professor Puzzleothers has privately decided to allow his ex-wife a resettlement of precisely one third his current annual salary, but only if she can work out exactly how much she is to get.
He instructs his solicitor to tell her lawyers that he will agree to alimony calculated according to the following formula.
She has to find two numbers A and B which between them contain each of the digits from 1 to 9 exactly once and contain no 0 digit, such that B = 2A, A is divisible by 3, and the quotient when A is divided by 3 is a number which contains all the digits from 1 to 4. Then £A will be the annual settlement.
What does Puzzleothers currently earn?
by Eric Emmet
From Issue #1068, 8th September 1977
A lot of people have been of the opinion for some time that in football competitions some importance should be attached to the number of goals scored as well as to the actual result of the game. It is hoped that this will lead to more goals and therefore to more attractive games.
Three local teams of my acquaintance have been experimenting on these lines. They have had a competition in which they all played each other once, and they have awarded ten points for a win, five points for a draw, no points of course for a loss, and one point for each goal scored.
As a result of this competition one team scored 16 points, the second scored 18 points, and the third scored 10 points. It was interesting to notice that at least one goal was scored by both sides in every match.
What was the score in each match?
by Richard England
From Issue #2155, 10th October 1998
Since M is the Roman numeral for 1000, we can say that with this puzzle New Scientist has published its Enigma M times – which is significant because:
ENIGMA ÷ M = TIMES
In this problem each letter stands for a different digit, and the same letter represents the same digit wherever it appears. No number starts with a zero.
by Howard Williams
Published July 28 2019 (link)
Sarah and Jenny are runners who train together on a circular track, and Sarah can run up to 20 per cent faster than Jenny. They both run at a constant speed, with Sarah running an exact percentage faster than Jenny. To reduce competition they start at the same point, but run round the track in different directions, with Sarah running clockwise.
On one day they passed each other for the seventh time at a point which is an exact number of degrees clockwise from the start. Sarah immediately changed her pace, again an exact percentage faster then Jenny. After a few passes both runners reached the exit, at a point on the track an exact number of degrees clockwise from the start, at the same time.
How fast, relative to Jenny, did Sarah run the final section?
by Victor Bryant
Published July 21 2019 (link)
Last year a two-figure number of teams entered our football competition. With that number, it was impossible to have a straightforward knockout competition (where half the teams are knocked out in each round), so we divided the teams into equal-sized groups, each pair of teams within a group played each other once, and the overall winner from each group went forward to the second stage consisting of a knockout competition. Unfortunately our local team was knocked out in the quarter-finals of that stage.
This year a higher number of teams entered (but not twice as many). Again a straightforward knockout competition was impossible so we copied last year’s model but with smaller groups and more of them. By coincidence the total number of games played this year equalled the number played last year.
How many teams entered last year and how many this year?