Enigmatic Code

Programming Enigma Puzzles

Enigma 210: Football: Letters for digits (4 teams)

From New Scientist #1356, 5th May 1983 [link]

Four football teams A, B, C and D played each other once. After some (or perhaps all) of the games had been played a table giving some details of the matches played, won and lost etc. was drawn up.

But unfortunately (Uncle Bungle again!) the digits have been replaced by letters. Each letter stands for the same digit (from 0 to 9) wherever it appears, and different letters stand for different digits. The table looked like this:

Enigma 210

(Two points are given for a win and one point to each side in a drawn match).

Find the score in each match.

Enigma 1306: Three all

From New Scientist #2465, 11th September 2004

I have in mind three numbers each of three digits (no leading zero) in each of which one digit is 3. Of the following statements about them, three are true and three are false.

(a) The number is a prime.
(b) The number is (appropriately) a cube.
(c) The middle digit is the average of the other two digits.
(d) The third digit differs from the second by 3.
(e) The number has as a factor a two-digit prime the difference of whose digits is 3, or whose sum is a cube.
(f) The number belongs (appropriately) to the set of triangular numbers 1, 3, 6, 10, 15, 21…

What is the sum of the three numbers?

Enigma 209: Double number slab

From New Scientist #1355, 28th April 1983 [link]

Enigma 209

A double number slab is just two rows of squares in a rectangle, with the correct number in each square. The numbers in the top row are just 1, 2, 3, …, n. In each bottom row square is written the number of times the number above it occurs in the completed slab.

So, for instance, if n=5, you fill in the top row as shown, and in the bottom row you replace A by the number of 1’s, B by the number of 2’s, …, E by the number of 5’s in the whole slab.

Given that n is at least 4, can you say:

(a) for what value of n is it impossible to complete the slab properly?
(b) for what value of n is each second-row number less than or equal to every number to the left of it?

Enigma 1307: KO, OK?

From New Scientist #2466, 18th September 2004

Sixteen players numbered 1 to 16 entered a men’s knockout tennis tournament. In each round the numbers of the remaining players were drawn at random to decide who played whom.

At the end of the tournament each player wrote down the number or numbers of the players he had competed against, in the order in which he had played them. The lists of the two finalists had their numbers in increasing order.

Also, each player worked out the total of the numbers in his list. The highest total was four times the lowest.

Which two players were in the final?

Enigma 208: Snooker doubles

From New Scientist #1354, 21st April 1983 [link]

In a recent frame at a snooker match the were no penalty points and after each of the 15 reds was potted a colour was potted (each of the six colours following two or three of the reds). The surprising thing about the result was that the winner’s total score was twice that of the loser, and yet they had both potted the same total number of balls.

What was the loser’s total and how many reds, how many yellows, how many greens, how many browns, how many blues, how many pinks and how many blacks did he pot?

(In snooker the potting of a red is followed by the potting of one of the other colours, the red remaining down but the other colour returning to the table. After 15 such events the remaining six colours are potted in the order stated above, reds are worth 1 point and the rest 2-7 in the order stated above).

Enigma 1308: Passing through

From New Scientist #2467, 25th September 2004

Enigma 1308 - 8x5

Enigma 1308 - 10x4

If you are told to draw a rectangle along the lines of a sheet of graph paper such that its area is 40 squares you could choose rectangles measuring 8×5, 10×4, 20×2 or 40×1. Whether you chose the 8×5 or the 10×4 you would find that a diagonal drawn across your rectangle would pass though 12 of the squares.

(1) What is the smallest number of squares of the graph paper that can be the area of THREE different rectangles whose diagonals each pass through the same number of squares?

(2) How many squares does each of those diagonals pass through?

Enigma 207: Read all about it!

From New Scientist #1353, 14th April 1983 [link]

Hook, Line and Sinker were the judges for the Cooker Book prize this year as usual. They assembled a short list of six and then, as usual, could not agree on the order of merit. In the end each did his own ranking, giving 6 points for first place, 5 for second and so on (no ties). Then they totalled the points, which produced a final order (also without ties). Hook gave 5 points to the book which in fact came out second and 1 point to the book which finished third. He ranked “Stuff” above “Nonsense” and gave “Umph” the number of points which Line gave to “Impenetrables”. Line ranked “Gawp” above “Elements” and placed “Impenetrables” below “Umph”. Sinker ranked “Nonsense” third and “Stuff” fifth. No book totalled 13 or 10 or received the same number of points from any two judges. One of the books totalled 8.

Can you spell out the final order?

Enigma 1309: Fill, cut and fit

From New Scientist #2468, 2nd October 2004

Amber and Ben have a new game which they play on this board.

Enigma 1309

Starting with Amber and going alternately, they each write their initial in an empty square until the board is full. They each then make a paper copy of the board; Amber cuts hers into three vertical strips and Ben cuts his into two horizontal strips. The board is then wiped clean. Starting with Amber and going alternately, they each place one of their own strips onto the board; each strip must retain its orientation and be placed so as to fit neatly over the two or three squares of the board.

They must not overlap their own strips but when they overlap a strip already placed by the other child then the letters in the overlapping squares must agree. If both children place all their strips then the game is a draw; otherwise a child wins when the other child cannot place a strip.

If both children play as well as possible, who wins or is it a draw?

Enigma 206: Division. Some letters for digits, some digits missing

From New Scientist #1352, 7th April 1983 [link]

In the following division sum some of the digits are missing, and some are replaced by letters. The same letter stands for the same digit wherever it appears:

          - - -
- - ) p k m k h
      p m d
        x p k
          - -
          k h h
          m b g

Find the correct sum.

Enigma 205: Doing it by halves

From New Scientist #1351, 31st March 1983 [link]

I was half-expecting something out of the ordinary when Archie presented me with the cross-number puzzle below. So I was not altogether surprised to see that half of the clues were missing, nor to hear his half-baked explanation of how he had rescued the paper they were written on from the grate in his living room.

Enigma 205


1. Seven times 3 down. Also starts or ends with the same two digits as 7 across.
7. Nine times (4 across minus 5 down).


2. Three times 6 across.
4. A multiple (not 1) of 2 down.

All I want to know is: what is the difference between 6 across and 7 across?

Enigma 1310: Addition corrected

From New Scientist #2469, 9th October 2004

In the following sum, digits have been replaced consistently by letters, with different letters being used for different digits. The number DELETION is more than twice as big as ADDITION and their sum is given by:


But before solving the problem the total needs to be corrected because two adjacent letters in it have been mistyped. What is the correct numerical value of the total?

Enigma 1311: Minimal clockspan

From New Scientist #2470, 16th October 2004

I have a clock with a sweep second hand, and I watch how the three hands – hour (H), minute (M) and second (S) – come together and spread apart.  Let us call their span, at any given moment, the smallest fraction of the clock face to contain all three hands.

I ignore the hands’ thickness, so at 12.00.00 the span is zero. At 12.00.01 it is just under 1/60, because H has just past 0 (the vertical); S is in position 1/60 (that is 1/60th of the clock face from 0); and M is between them.  After that the span gets bigger, but by 12.01.00 it is down again to exactly 1/60, with S at 0, M at 1/60 and, this time, H in between.

(a) Between 12.00.01 and 18.00.00, at what time is the span a minimum?   (Give your answer to the nearest minute)

(b) What exact fraction is this minimum span?

Enigma 204: Three mental ages

From New Scientist #1350, 24th March 1983 [link]

Every Christmas I assemble my three nephews and ask each to state his mental age, which must be an exact whole number of years not greater than 100. Then I give to each, for every year of his stated age, as many pounds as he says he is years old.

Both last year and this year I noticed that my total outlay could be more simply calculated by just multiplying the three ages together.

Last year all three claimed the same age. This year the three ages were all different, and none was the same as last year’s.

Can you tell me last year’s and this year’s trios of ages? The answers, I believe, are unique. Trial and error will be shortened once you notice that if (A, B, C) is a possible trio of ages, so is …

Enigma 1312: Hampshire v Middlesex

From New Scientist #2471, 23rd October 2004

Next season Hampshire and Middlesex will be among nine English counties in Division 1 of cricket’s league competition and also among the nine in Division 1 of the game’s county championship.

At the end of each season, at present, three counties are relegated to Division 2 of the league and three are relegated to Division 2 of the Championship, but it has been suggested that the number should be cut from three to two in each competition.

If we assume that in both competitions all counties have an equal chance of being relegated, what, as a fraction in its lowest terms, is the probability that in the season after next Hampshire and Middlesex will be in the same division as each other in both competitions if the number of counties relegated from Division 1 of each competition at the end of next season is (a) three, and (b) two?

Enigma 203: Round walk

From New Scientist #1349, 17th March 1983 [link]

Enigma 203

The plan shows the layout of paths in our local park. At each of the six junctions of paths there is a statue. The shortest route (on the paths) from Disraeli to Victoria is bound to take you past George V only. From Disraeli it is the same distance to Eros as it is to the Unknown Warrior.

I usually choose a walk which starts and finishes at Churchill and passes each of the other statues only once, but I choose it so that the is no other shorter “round walk”.

Last week one stretch of path was closed for repair. This meant that the shortest distance (on paths) from George V to Disraeli was over twice what it was previously. Also, I could not go on any of my usual “round walks” so I chose a new shortest route in the circumstances. As usual it visited each statue just once except for starting and finishing at Churchill. Also I chose the walk and the direction which ensured I passed Eros as soon as possible.

What was the order in which I passed the statues? (C – – – – – C)

Enigma 1313: Triangles

From New Scientist #2472, 30th October 2004

Draw a triangle ABC.  On the side AB mark the point P such that AP=(2/5)AB, on BC mark Q such that BQ=(2/5)BC and on CA mark R such that CR=(2/5)CA.

Draw the lines AQ, BR and CP.  Call the point where AQ and BR cross X, the point where BR and CP cross Y and the point where CP and AQ cross Z.

If you did the appropriate calculations you would find that the area of triangle XYZ is 1/19 of the area of triangle ABC.

I went through the whole procedure above again, but, this time with each occurrence of the number 2/5 replaced by the number k, which is between 0 and 1/2.  This time I found that the area of triangle XYZ was 1/37 of the area of triangle ABC.

What was the number k?

Enigma 202: Football: 4 teams: new method

From New Scientist #1348, 10th March 1983 [link]

“We go to watch football to see goals scored,” as so many people have said. And it is in the hope of attracting more goals that the new method has been devised.

In this method a team gets 10 points for a win, 5 points for a draw, and 1 point for each goal they score.

In a recent competition between four teams, A, B, C and D, in which they all played each other once, the points were as follows:

A – 13
B – 22
C – 39
D – 10

The success of this method can be seen from the fact that each side scored at least one goal in every match.

You are told that B scored the same number of goals against C as they did against D, and that no match was won by a margin of more than one goal.

Find the score in each match.

Enigma 1314: Times table

From New Scientist #2473, 6th November 2004

Enigma 1314

From a full set of dominoes, I have taken just those that have a 1, 2, 3 or 4 at each end. I then arranged them into a rectangle, as shown, with each domino occupying two adjacent squares. The numbers are the product of the entries in that row or column. Just four dominoes are horizontal.

Which ones?

Enigma 201: Secret ballot

From New Scientist #1347, 3rd March 1983 [link]

The Progressive Party has just chosen its prospective parliamentary candidate for Mudbridge North. Six persons were short-listed, and, having duly confided their belief in peace, progress and plenty, were voted upon by the members present. The members have been tight-lipped about the voting figures but the candidates more (apparently) forthcoming. Tabulated below are the votes which the candidates (listed down the left) allege were got by various candidates (listed across the top). For instance D says that F scored 10. You will need to know, however, that there were three candidates of each sex, that each has overstated the votes for anyone of his or her own sex and understated those for anyone of the other sex, and that the same number of votes was cast in total for men as for women. Can you reveal the true voting figures?

Enigma 201

Enigma 1315: Going for gold

From New Scientist #2474, 13th November 2004

Messrs Archer and Bowman each fired three arrows. The arrows recorded six different scores between 2 and 10.

The cumulative score of each of them at any stage was always a prime number. Archer had the greater cumulative score after each had fired one arrow and again after each had fired three arrows, Bowman had the greater cumulative score after each had fired two arrows.

What did Bowman score with each of his arrows? Give the scores in the order in which they were recorded.

When I did archery I was told you couldn’t “fire” an arrow, as there was no fire involved. So, we used to “shoot” arrows.


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