From New Scientist #1554, 2nd April 1987 [link]
My daughter has a regular hexagonal clock without numerals, as illustrated. I tried to fool her recently by rotating it and standing it on a different edge, but she recognised that the hands did not look quite right.
On the other hand, my son, has a clock on a regular polygon, again without numerals, which I can stand on any different edge and make the clock show the wrong time with its hands in apparently legitimate positions.
How many edges does this regular polygon have?
From New Scientist #1028, 25th November 1976 [link]
I overheard Professor Foresight discussing the results of a small precognition test the other day. It emerged that he had tossed a penny five times, inviting the thirteen members of his class to write down what was coming before each throw. Six students had done better than the rest, all scoring the same number, although no two had produced identical lists of guesses. Nor had any two of the remaining students produced identical lists.
It also emerged that the penny had not come up Heads all five times. Nor was the actual series Head, Tail, Tail, Tail, Head. Nor was it Tail, Tail, Head, Tail, Tail. At this point the discussion broke up and I was left wondering just what the actual series was. Given that each of these series just mentioned was the guess of one of the unsuccessful seven, can you oblige?
From New Scientist #2264, 11th November 2000 [link]
George was nominated for president of the Golf Club. There was only one other candidate, and the president was elected by a simple ballot of the 350 members, not all of whom in fact voted.
The ballot papers were taken from the ballot box one at a time and placed in two piles — one for each candidate — with tellers keeping a count on each pile.
George won (what did you expect?), and furthermore his vote was ahead of his opponent’s throughout the counting procedure.
“That must be a one-in-a-million chance,” said the demoralised loser.
“No,” said George. “Now that we know the number of votes we each received, we can deduce that the chance of my leading throughout the count was exactly one in a hundred.”
How many members did not vote?
29 June 2017
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From New Scientist #1553, 26th March 1987 [link]
“What animals have you in that barn there?” said the man from the ministry.
The farmer beamed. “Pigs, cows and ducks, sir.”
“How many are there, though?”
“Oh, quite a few, really, sir.”
“I need figures, man!” persevered the would-be census taker.
“If it’s figures you’ll be wanting, sir,” replied the farmer, “I can tell you that multiplying the number of horns by the number of legs by the number of wings gives 720.”
“Yes, but how many of each animal are there?” snapped the other exasperatedly.
“Telling you the number of cows alone wouldn’t enable you to deduce the number of ducks and pigs. Telling you the number of ducks alone wouldn’t enable you to deduce the number of pigs and cows. But telling you the number of pigs would enable you to deduce the number of cows and ducks right enough. So I reckon you can work out how many cows and ducks there be in yonder barn even if I don’t tell you the number of pigs in it.”
With that he ducked into a hen-house.
How many cows, ducks and pigs were there in the barn?
28 June 2017
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From New Scientist #1029, 2nd December 1976 [link]
King Ethelweed needed a new champion. So he commanded his three doughtiest knights to appear before him on the first Monday of the new year and bade them fight one another. They fought all day long until the eventide, when the king called a respite and awarded x ducats to the winner, y ducats to the second knight and z to the third. x, y and z are positive descending whole numbers.
To the valiant knights’ dismay, the same happened on the next and each following day, until King Ethelweed at length declared himself satisfied. One each day the same prizes of x, y and z were awarded, the being no ties on any day.
Thus it befell that Sir Kay gained the most ducats and became the king’s champion, even though he fared worse on the second day than on the first. Sir Lionel took home twenty ducats in all and Sir Morgan, despite winning top prize on the third day, amassed a mere nine.
Which was the final day and who won how many ducats on it?
26 June 2017
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From New Scientist #2265, 18th November 2000 [link]
At tennis a set is won by the first player to win 6 games, except that if the score goes to 5 games all, the set is won either by 7 games to 5 or by 7 games to 6. The first person to win three sets wins the match.
Sometimes at the end of a match each player has won exactly the same number of games. This happened when André beat Boris in a match in which no two sets contained the same number of games.
You will have to work out whether André won by 3 sets to 1 or by 3 sets to 2, but if I told you how many games in total each player won you would be able to deduce with certainty the score in each of the sets that André won and in the set or each of the sets that Boris won.
What was the score in each of the sets that André won? (Give each set’s score in the form x-y, André’s score always comes first).
23 June 2017
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From New Scientist #1552, 19th March 1987 [link]
In the puzzle below, select some of the guest lists and discard the remainder. The lists you keep should make a puzzle which has exactly one solution and involves no more lists than is necessary.
Who did it?
There has been a series of robberies at house parties recently. Each was clearly a one-man job — the same man each time — and each was an inside job. The possible suspects are Alan, Bryan, Chris, David, Eric, Fred, George, Harry, Ian, Jack, Ken and Len. The male guest list at the parties were as follows:
1. All but David, George and Len.
2. Bryan, Chris, Eric, Harry, Ian and Ken.
3. All but Bryan and Ken.
4. Chris and Ian.
5. All but Alan, Fred, Jack and Len.
6. Bryan, Chris, Ian and Ken.
7. All but Eric and Harry.
8. All but David and George.
9. All but Chris and Ian.
10. All but Alan and Fred.
Who carried out the robberies?
Which lists should you use in your puzzle?
What is the answer to your puzzle?
21 June 2017
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From New Scientist #1122, 28th September 1978 [link]
A couple of one’s, a couple of two’s and a six;
All wrong, all wrong!
If only I thought that the puzzle was one I could fix,
I’d sing a song.
But as I feel sure that it’s rather too much for me,
My voice is muted.
Uncle Bungle’s my name and I fear that you must agree,
I’m rather stupid.
So please, I implore,
Continue the fight,
With tooth and with claw,
With main and with might,
To make wrong sums right.
The figures given are all incorrect. Write out the whole division sum.
19 June 2017
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From New Scientist #2266, 25th November 2000 [link]
Matthew and Ben are playing a game. The board is a 1-kilometre square divided into 1-centimetre squares. The centre of each small square is marked by a red dot.
Matthew begins the game by choosing a number. Ben then selects that number of red dots. Finally Matthew chooses two of Ben’s selected dots and draws a straight line from one to the other. Matthew wins if his line passes through a red dot other than those at its ends; otherwise Ben wins.
What is the smallest number that Matthew can choose to be certain of winning?
In the magazine this puzzle was incorrectly labelled Enigma 1104.
16 June 2017
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From New Scientist #1550, 5th March 1987 [link]
I asked Electrophorus what he was working on.
“You know that joining unlike terminals of a pair of batteries produces a voltage across the two free terminals equal to the sum of the voltages of the separate batteries. And connecting unlike terminals produces a voltage equal to the difference of the voltages of the separate batteries?”
“Yes”, I replied. “With a battery of 2 volts and one of 5 volts one obtains 3 volts (sources opposing) or 7 volts (sources reinforcing).”
“Well, before lunch I had three batteries, none of which had zero voltage, and a voltmeter with a holder that would accommodate only two batteries. So I measured the voltages across the free terminals of all possible pairwise combination of these three batteries both in the case where the voltages reinforced and where they opposed. I wrote on my blackboard the resulting six positive numbers in order of increasing magnitude.”
“When I returned from lunch eager to calculate the ratings of the three batteries, I found the three batteries gone and my blackboard wiped clean. I remember that the second smallest reading occurred twice. It was either 13 or 17 volts, I forget which. I had noticed, rather inconsequentially perhaps, that reversing the digits of this double reading produced another reading which occurred in my measurements.”
What were the ratings of the three batteries?
14 June 2017
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From New Scientist #1030, 9th December 1976 [link]
Someone let the cat out. Who was it? That is rather hard to decide. Delia says it was one of the twins, meaning Bert or Claud. Alice says it was Bert; and Bert (shame on him!) says it was Claud. Meanwhile Claud says it was Delia; and Emma says it was not Claud.
So it is all a bit of a puzzle and you will be expecting to be told how many of them are right in what they say. But that would make it all much too easy, as you could then deduce who the culprit was. So you will just have to manage with what information you have.
Who let the cat out?
12 June 2017
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From New Scientist #2267, 2nd December 2000 [link]
Fill in the following cross-figure. No answer begins with a zero. The same base is used for all the entries, but it is not necessarily 10.
1. A palindromic prime.
4. The square of the base being used.
5. A square.
1. Three times my son’s age.
2. A prime.
3. A palindromic square.
How old is my son?
From New Scientist #1549, 26th February 1987 [link]
The beer-mats at our local pub have puzzles on them. Here is one in which the digits are consistently replaced by letters.
BEER – MAT = TEST
NINE is a perfect square
IT is a number
THIS is odd!
What is TIME gentlemen (and ladies) please?
From New Scientist #1031, 16th December 1976 [link]
When the Olympic games were last held in Patagonia, the Famous torch entered the country at a point exactly 35.27 km from its pedestal in the Olympic stadium. The honour of transporting it from the frontier fell to two Patagonian athletes, Pita and Pata, who were to carry it in turns for the 35.27 km. By presidential decree each was to carry it at each turn any distance he pleased not less than 1 km and not more than 2 km.
Each secretly resolved that he would be the one to carry it the final awesome metre. Since there was nothing in the decree to forbid a different choice of distance at each turn much calculation went on before Pita and Pata tossed for the privilege of having the first turn. In fact Pita won the toss and chose second turn.
Did he chose right?
From New Scientist #2268, 9th December 2000 [link]
George is building a patio, which will be covered using one-foot-square concrete slabs of seven different colours. He has divided the rectangular patio into seven rectangular zones, without any gaps.
Each zone will be covered by slabs of one colour, with five different colours appearing around the perimeter of the patio, and four different colours at the corners. The seven rectangular zones are all different shapes, but all have the same perimeter, which is less than 60 feet.
What are the dimensions of the patio that George is building?
This puzzle is referenced by Enigma 1221.
From New Scientist #1548, 19th February 1987 [link]
Farmer O. R. Midear has crossed a turnip with a mangel to get a tungel, and he now has many fields of tungels. As a tungel expert, he looks at a tungel to see if it is red or not, if it is smooth or not, and if it is firm or not. He knows that if a tungel is red then it is smooth.
Regulations have just been introduced which put fields of tungels into classes A, B, C, D, E; a field may be in more than one class.
Class A: All fields containing no red tungel.
Class B: All fields containing no smooth tungel.
Class C: All fields in which every red tungel is firm.
Class D: All fields in which every smooth tungel is firm.
Class E: All fields containing a red tungel which is not firm.
To test Farmer Midear’s understanding of the regulations he was asked to say which of the following statements are true.
1. If a field is in A then it is in B.
2. If a field is in B then it is in A.
3. If a field is in C then it is in B.
4. If a field is in B then it is in C.
5. If a field is in C then it is in D.
6. If a field is in D then it is in C.
7. If a field is in D then it is not in E.
8. If a field is not in E then it is in C.
9. In every field there is a tungel such that if it is not red then the field is in A.
What should Farmer Midear’s answer be?
From New Scientist #1123, 5th October 1978 [link]
In the addition sum below, letters have been substituted for digits. The same latter stands for the same digit whenever it appears and different letters stand for different digits.
Write the sum out with numbers substituted for letters.
From New Scientist #2269, 16th December 2000 [link]
In the multiplications shown, where the combined products of multiplications (I) and (II) (both identical) equal the product of multiplication (III), each letter consistently represents a specific digit, different letters being used for different digits while asterisks can be any digit.
The multiplications in fact are not difficult to solve, and easier still if I told you that TEN is even.
How much is TWENTY?
From New Scientist #1547, 12th February 1987 [link]
In the following addition sum all the digits are wrong. But the same wrong digit stands for the same correct digit wherever it appears, and the same correct digit is always represented by the same wrong digit.
Find the correct addition sum.
From New Scientist #1032, 23rd December 1976 [link]
Oops! What the message is meant to say is of course:
HAPPY CHRISTMAS TO YOU FROM THE NEW SCIENTIST.
Perhaps you would like to put it right by sliding on word at a time along a line into a vacant oval. If you are not too saturated with Christmas pud, you should manage it in 26 moves.