Enigmatic Code

Programming Enigma Puzzles

Category Archives: enigma

Enigma 402: A DIY puzzle

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?

by …

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?

[enigma402]

Enigma 1110: Dots and lines

From New Scientist #2266, 25th November 2000

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.

[enigma1110]

Enigma 400: Potential difficulties

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?

[enigma400]

Enigma 1111: Base-age

From New Scientist #2267, 2nd December 2000

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.

Across
1. A palindromic prime.
4. The square of the base being used.
5. A square.

Down
1. Three times my son’s age.
2. A prime.
3. A palindromic square.

How old is my son?

[enigma1111]

Enigma 399: Time, gentlemen, please

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?

[enigma399]

Enigma 1112: Patio zones

From New Scientist #2268, 9th December 2000

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.

[enigma1112]

Enigma 398: Down on the farm

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?

[enigma398]

Enigma 1113: Ten + ten = twenty

From New Scientist #2269, 16th December 2000

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?

[enigma1113]

Enigma 397: All wrong again

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.

[enigma397]

Enigma 1114: Christmas changes

From New Scientist #2270, 23rd December 2000

Christmas is said to change things and so this enigma is an old puzzle with some changes.

Problem: You have to assign a digit to each of the 10 letters in the sum here:

When you have decided on an assignment of digits to the 10 letters, then your assignment is a solution of the problem if it satisfies at least one of the following conditions:

• At least one digit is assigned to more than one letter.
• When the digits are put into the addition sum it is not correct.
• The digit assigned to “U” is smaller than the digit assigned to “H”.

You need to find an assignment of digits to the 10 letters which is NOT a solution of the problem.

What is the value of BLEAT in your assignment?

[enigma1114]

Enigma 396: The hostess’s problem

From New Scientist #1546, 5th February 1987 [link]

At a recent dinner party five men and their wives sat at the 10 places around the table. Men and women alternated around the table and no man sat next to his own wife. No man’s Christian name had the same initial as his surname.

Mrs Collins sat between Brian and David. Colin’s wife sat between Mr Briant and Mr Edwards. Mr Allen sat between Edward’s wife and Mrs Davidson. Brian’s wife sat next to Alan.

In the information which I’ve just given you, if two people were sitting next to each other then I have not told you about it more than once.

Which two men (Christian name and surname of each) sat next to Mrs Edwards?

[enigma396]

Tantalizer 482: Lapses from grace

From New Scientist #1033, 6th January 1977 [link]

An air of rare humility pervades the Common Room at St. Aletheia’s tonight. The seven inmates overdid the post-prandial gin and rashly confessed their sins to one another. Each owned to a different pair of the deadly ones and each sin turned out to have claimed a different pair of victims.

Constance, Emily and Flavia have no sin in common to any two of them. Beatrice, Deborah, Emily and Gertrude confessed to all seven between them. Alice and Gertrude admitted to sloth; Deborah and Emily to lust. Alice is not given to pride nor Beatrice to avarice nor Flavia to either pride or intemperance. Constance, who owned to anger, has a sin in common with Deborah, who did not.

Which pair has fallen prey to intemperance and which pair to envy?

[tantalizer482]

Enigma 1115: New Christmas star

From New Scientist #2270, 23rd December 2000

enigma-1115

Here is another “magical” Christmas star of twelve triangles, in which can be seen 
six lines of five triangles (two horizontal and two in each of the diagonal directions). Your task is to place a digit in each of the twelve triangles so that:

all six digits in the outermost “points” of the star are odd;

the total of the five digits in each line is the same,
 and it is the same as the total of the six digits in the points of the star;

each of the horizontal lines of digits, when read as a 5-digit number, is a perfect square.

What are those two perfect squares?

Thanks to Hugh Casement for providing the source for this puzzle.

[enigma1115]

Enigma 395: By Jove, it figures!

From New Scientist #1545, 29th January 1987 [link]

In the addition sum below, each of the digits from 0 to 9 has been replaced by a letter whenever it occurs. Different letters stand for different digits. You are asked to reproduce the original sum.

[enigma395]

Enigma 1116: United win at last

From New Scientist #2272, 6th January 2001

Albion, Borough, City, Rangers and United played a tournament in which each team played each of the other teams once. Two matches took place in each of five weeks, each team having one week without a match.

One point was awarded for winning in the first week, 2 points for winning in the second week, 3 points for winning in the third week, 4 points for winning in the fourth week and 5 points for winning in the fifth week. For a drawn match each team gained half the points it would have gained for winning it. At any stage, teams that had gained the same number of points were regarded as tying.

After the first week A led, with B tying for second place. After the second week B led, with C tying for second place. After the third week C led, with R tying for second place. After the fourth week R led, with U tying for second place. After the fifth week U had won the tournament with more points than any of the other teams.

(1) Which team or teams finished in second place after the fifth week?

(2) Give the results of Albion’s matches, listing them in the order in which they were played and naming the opponents in each match.

This completes the archive of Enigma puzzles from 2001. There are now 1065 Enigma puzzles on the site, the archive is complete from the beginning of Enigma in February 1979 to January 1987, and from January 2001 to the final Enigma puzzle in December 2013. Altogether there are currently 59.5% of all Enigmas published available on the site, which leaves 726 Enigmas between 1987 and 2000 left to publish.

[enigma1116]

Enigma 394: Unwinding

From New Scientist #1544, 22nd January 1987 [link]

Professor Kugelbaum was unwinding at the Maths Club with a cigar after lunch when a wild-looking man burst in and introduced himself thus:

“My name is TED MARGIN. Juggling with the letters of my name one obtains both GREAT MIND and GRAND TIME. But I digress. Each letter of my name stands for one digit exactly from 1 to 9 inclusive and vice versa. And, do you know,

(A × R × M) / (A + R + M) = 2,

MAD is greater than ART (though RAT is greater than either), and TED = MAR + GIN.

If, in addition, I were to tell you the digit which corresponds to M then you could deduce the one-to-one correspondence between the letters of my name and the digits 1 to 9″.

At this point a helpful butler removed the man, but Kugelbaum was amused to find that the information was quite consistent.

Write the digits 1 to 9 in the alphabetical order of the letters to which they correspond.

[enigma394]

Enigma 1117: Reapply as necessary

From New Scientist #2273, 13th January 2001 [link]

Recently I read this exercise in a school book:

“Start with a whole number, reverse it and then add the two together to get a new number. Repeat the process until you have a palindrome. For example, starting with 263 gives:

leading to the palindrome 2662.”

I tried this by starting with a three-figure number. I reversed it to give a larger number, and then I added the two together, but my answer was still not palindromic. So I repeated the process, which gave me another three-figure number which was still not palindromic. In fact I had to repeat the process twice more before I reached a palindrome.

What number did I start with?

[enigma1117]

Enigma 393: Decode the sum

From New Scientist #1543, 15th January 1987 [link]

In the following addition sum, different letters stand for different digits and the same letter stands for the same digit throughout.

Decode the sum.

[enigma393]

Enigma 1118: 2001 – A specious oddity

From New Scientist #2274, 20th January 2001 [link]

George is planning to celebrate the new millennium — the real one — by visiting Foula, the most remote of the Shetland Islands. It is one of the few places in the world where the inhabitants still live by the old Julian calendar rather than the now almost universal Gregorian calendar.

In order to correct the drift of the Julian calendar against the seasons, Pope Gregory decreed that in 1582, Thursday 4th October (Julian) should be immediately followed by Friday 15th October (Gregorian), and in order to prevent a recurrence of the drift, years divisible by 100 would henceforth only be leap years if divisible by 400. Previously all years divisible by four were leap years. Catholic countries obeyed immediately, others — apart from Foula — fell into line in later centuries.

While planning his visit George programmed his computer to print 12-month calendars for the required year, showing weekdays, under both Julian and Gregorian styles. But when he ran the program he was surprised to find that the two printouts were identical.

He then realised that he had entered the wrong year number — the Julian and Gregorian calendars for the year 2001 are not the same.

What is the first year after 1582 for which they are the same?

[enigma1118]

Enigma 392: Nothing written right

From New Scientist #1542, 8th January 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.

[enigma392]