Enigmatic Code

Programming Enigma Puzzles

Category Archives: enigma

Enigma 483: Undigital … Unfathomable?

From New Scientist #1634, 15th October 1988 [link]

Below is an addition sum with letters substituted for digits. The same letter stands for the same digit wherever it appears and different letters stand for different digits.

Enigma 483

Write the sum out with numbers substituted for letters.


Enigma 1028: A perfect pass

From New Scientist #2184, 1st May 1999

This is part of a football pitch; C is a corner, CE is a goal-line, CD is a side-line and AB is a side of the penalty area. Rovers have been awarded an indirect free-kick at the point F on AB and the ball is placed at F. Two players, Fay and Patricia, got to G on CD to discuss their plan. Then together they set off running, Fay towards F and Patricia towards P, each at a steady speed. After 10 seconds Fay reaches F and Patricia reaches P. Fay immediately takes the free-kick and kicks the ball along FA, so that it travels at a steady speed. Patricia carries on running at the same speed and in the same straight line. At the moment Patricia reaches AF, the ball reaches Patricia. Our problem is to find the speed of the ball, as follows:

Draw a line which passes through two of the labelled points, A, B, C, … Select a point where your line crosses an existing line and mark it X. Select a labelled point and mark it Y. You are to do this so that the distance between X and Y is the distance the ball travels in 10 seconds.

Which to labelled points should you choose to draw the line through? Which point is Y?




Enigma 482: Hopscotch

From New Scientist #1633, 8th October 1988 [link]

Enigma 482

I remember playing a version of hopscotch when I was a child. We used a chalked outline like the one shown, and there were various games we could play on it. The simplest one was to start where shown and throw a pebble towards number 1 and then hop to the pebble: then throw it towards number 2 and hop to it, and so on, finally throwing the pebble towards number 9 and hopping to the pebble. You scored 1 if the pebble landed on the correct number, ½ if it missed by one, ⅓ if it missed it by two, and so on, making 9 the maximum possible total score. You were disqualified if the pebble didn’t land on a numbered square.

When I first tried the game I was pretty hopeless. When throwing at number 1 the pebble went past it. I hopped to the pebble, threw the pebble towards number 2, and continued in this way to complete the game. On no occasion was I standing on the square which I should be aiming at and, apart from when the pebble went past number 1 on my first throw, on only one other occasion did the pebble go too far and go past the square I was aiming at. I ended up with the pebble landing on all the squares from 1 to 9 (albeit in the wrong order) and my score was a whole number.

In what order did I visit the squares?

The issue date of New Scientist that this puzzle was published in falls on a Saturday, the issue date of previous magazines fell on a Thursday, so the date of this issue is 9 days after the date of the previous issue.


Enigma 1029: Chancelot

From New Scientist #2185, 8th May 1999

The company Chancelot has been asked to set up a lottery for a foreign country. It will work a bit like Britain’s own lottery with participants choosing some numbers: then the winning numbers will be decided by the company choosing some numbered balls at random.

The government has laid down some strict guidelines:

1. It wants participants to have to choose six numbers from 1, 2, …, N, where the top number N has not yet been decided. Then six of the numbered balls will be chosen and the winner’s choices must match all six.

2. It believes that the public is always suspicious when the winning selection includes two consecutive numbers. Therefore of all the combinations of six numbers from the N, it wants more than half of them not to include two consecutive numbers.

3. To give the public a fair chance of winning, it wants N to be the lowest possible satisfying the above conditions.

How many balls will there be in Chancelot’s lottery?


Enigma 481: Seconds out?

From New Scientist #1632, 29th September 1988 [link]

Professor E. Mit was somewhat puzzled, as the experiment he had been conducting had been a complete failure. After some investigation the cause was pinpointed to a faulty timing device, which measure time in minutes and hours only, in a 12 hour format with no distinction between am and pm.

Having replaced the offending clock, Mit decided to ascertain the nature of the error. This he did as follows; placing the faulty clock alongside an identical one which was known to keep the correct time and starting both clocks at the same time, he observed that the erroneous clock gained one minute in the first minute, two minutes in the second minute, three minutes in the third minute, and so on. For example, if both clocks start at 12:00:

Enigma 481

If both clocks start as above at 12:00 precisely, what time will it be when the faulty clock next shows the correct time?

This puzzle was originally published in the last issue of New Scientist to carry an issue date that falls on a Thursday. Subsequent issues have an issue date that falls on a Saturday.


Enigma 1030: Uncommonly different progressions

From New Scientist #2186, 15th May 1999

Certain 5-digit perfect squares can be formed by coupling a pair of smaller squares: a 1-digit square followed by a 4-digit square, or a 2-digit square followed by a 3-digit square, or a 4-digit square followed by a 1-digit square. A leading zero in front of the second square makes it ineligible; that means that 64009 can only be regarded as 80² followed by 3² (not at 8² followed by 3²) and squares such as 10000 are eliminated — so don’t waste time looking for a 3-digit square followed by a 2-digit square; it should be obvious that no such 5-digit square can exist.

Harry, Tom and I each chose three eligible 5-digit squares, and on each of our squares we multiplied the square roots of the pair of coupled smaller squares. We each found that our three products could be arranged to form an arithmetic progression; in addition the common difference of our three progressions could themselves be arranged to form another arithmetic progression whose common difference was different from that of any of the three previous progressions.

1. List in ascending order those 5-digit squares that were chosen by just one of us.
2. List in ascending order those eligible 5-digit squares that none of us chose.


Enigma 480: An irrational question

From New Scientist #1631, 22nd September 1988 [link]

Kugelbaum was running through a geometrical proof with some of his students when he suddenly went off at a tangent.

“What an extraordinary rectangle I have just drawn!” he remarked out loud. “Why, the number of inches in the perimeter is an integer equal to the number of square inches in its area. And yet, no one of its sides is a rational number of inches long”. (A rational number is one which can be expressed as the ration of two definite integers: for example 1.5, but not √2).

What is the smallest possible perimeter of such a rectangle, measured in inches?

Happy Christmas from Enigmatic Code.


Enigma 1031: Security

From New Scientist #2187, 22nd May 1999

George’s new office has a security lock in which you have to key in a series of digits — all different — before you can open the door. Unfortunately, George is having difficulty committing this security number to memory.

So far, he has memorised a number comprising a selection of different digits from the security number. He has made a note of the memorised number on the memo sheet on which he was given the security number.

After studying the sequence yet again, he absent-mindedly left the memo lying around his house. When his wife found the piece of paper bearing just two numbers, she thought George has been trying to devise an “Alphametic” Enigma for New Scientist. Just for fun, she multiplied the numbers together and found that the product was a seven-digit number in which all the digits are the same.

What is George’s office security number?


Enigma 479: Road island

From New Scientist #1630, 15th September 1988 [link]

On the faraway island of Roadio, the quaint villages with the curious one letter names A, B, C, …, S are joined by a network of roads as shown:

Enigma 479

The numbers indicate distances between villages in miles. I have written on all the distances I remember, however I also recall that, by road, no village is more than 20 miles from D and no village is more than 29 miles from P. What is the maximum distance between any two villages on the island?


Enigma 1032: Colonial powers

From New Scientist #2188, 29th May 1999

The old colonial powers had simply drawn lines on the map when they established the colonies of Abongo, Ebongo, Ibongo, Obongo and Ubongo. As a result, the map of the five colnies was a rectangle with each colony being a right-angled triangle. Abongo, Ebongo and Ibongo each had the same area. Obongo was bigger and Ubongo was bigger still.

After independence, Abongo, Ebongo and Ibongo united to form a single triangular country, larger than Ubongo.

On an old map of the five colonies, the shorter side was 60 centimetres long.

What was the length of the longer side?

When it was originally published the information that the combined country was the largest was omitted. A correction was published along with Enigma 1043 (along with the reassurance: “All Enigmas are checked to ensure they have a unique answer”). Without this fact there are three possible solutions.


Enigma 478: Football, addition, letters and a twist

From New Scientist #1629, 8th September 1988 [link]

In both the football league table and addition sum below, letters have been substituted for digits. Each letter stands or should stand for a different digit (from 0 to 9), and different letters should stand for different digits. And so they do except for the fact that one of the letters is incorrect on one of the occasions on which it appears (if indeed it appears more than once).

The three teams are eventually going to play each other once, or perhaps they have already done so.

Enigma 478

Which letter was wrong? What should it be?

Find the scores in the football matches and write the addition sum out with numbers substituted for letters.


Enigma 1033: Squirrels up and down

From New Scientist #2189, 5th June 1999

Samantha and Douglas are counting the squirrels visiting their garden. They record the monthly total for each of the 15 months January 1998 to March 1999. Samantha calculates that the number of squirrels is increasing, for she divides the 15 months into three five-month periods and finds the five-month totals are increasing with time. For example, there were more squirrels June to October 1998 than January to May 1998. But Douglas says the number of squirrels is decreasing, for he divides the 15 months into five three-month periods and finds the three-month totals decreasing with time, for example there were fewer squirrels April to June 1998 than January to March 1998.

1. Could there have been a total of 4 squirrels for the two months April and May 1998 and fewer than 75 squirrels for the whole of the 15 months? If so, how many squirrels were there for the 15 months?

2. Could there have been a total of 5 squirrels for the two months April and May 1998 and a total of fewer than 65 squirrels for the 15 months?

3. Could there have been 107 squirrels in June 1998 and 100 squirrels in October 1998? If so, how many squirrels were there in the 15 months?

4. Could there have been 108 squirrels in June 1998 and 100 squirrels in October 1998? If so, how many squirrels were there in the 15 months?


Enigma 477: Gap ‘n enigma

From New Scientist #1628, 1st September 1988 [link]

In the following long multiplication I’ve replaced digits with letters in some places and left gaps in the rest. Where letters are used, different letters are used for different digits.

Enigma 477

That’s all you actually need, but to avoid hours of work I can also tell you that GAP is divisible by 9.

What is the value of IMPINGE?


Enigma 1034: Double-digit squares

From New Scientist #2190, 12th June 1999

Four of us were trying to find a one-digit perfect square, a two-digit perfect square, a three-digit perfect square and a four-digit perfect square, such that five different digits were each used twice to form them. To make matters harder for ourselves we agreed that one of those five digits must be 7. Since we remembered that both 0 and 1 are perfect square we were each able to chose a different one-digit square.

We each found a valid solution, and our solutions had no squares in common.

List in ascending order the other squares in the solution that had 9 as its one-digit square.


Enigma 476: A curious question

From New Scientist #1627, 25th August 1988 [link]

Kugelbaum wandered into a history lecture by mistake and, almost as quickly, but not by mistake, wandered out again. “1210 may well have been a dull year,” he said to himself, “but it’s an interesting number. The first digit gives the number of 0s in it, the next the number of 1s in it, the next the number 2s in it and so on, quite consistently, right up to the very last digit. And there are other such numbers too, such as 2020 and 3211000! Curiously, I can’t find one with six digits, though.”

As he sought vainly the room where he was to give his lecture on number theory he amused himself by calculating all the numbers having this property. “I wonder,” he remarked as he looked into broom cupboard, “if one were to take all the numbers having this property and add them together, what the result would be?”

What is the sum of all the numbers having the property that their first digit gives the number of 0s in the number, the next the number of 1s in the number, the next the number of 2s in the number and so on, consistently right through to and including the last digit of the number?

(Since 10, 11, 12 and so on are not digits, any such number containing more than 10 digits must have zeros in its 11th place and any other places after this).

Note: The original puzzle statement gave 211000 as an example, not 3211000.


Enigma 1035: Connected numbers

From New Scientist #2191, 19th June 1999 [link]

Take a large sheet of paper and write on it the numbers, 5, 6, 7, …, 999998, 999999, 1000000. You are now going to draw lines that connect pairs of the numbers as follows. Start with 5. Split 5 into two numbers, both larger than 1, in as many ways as you can. So 5 = 2 + 3. Multiply the two numbers together. Now 2 × 3 = 6, so draw a line connecting 5 and 6. The next number is 6. Now 6 = 2 + 4 = 3 + 3 and 2 × 4 = 8 and 3 × 3 = 9, so draw a line connecting 6 and 8 and another connecting 6 and 9. The next number is 7 = 2 + 5 = 3 + 4, and so we draw a line that connects 7 and 10 and another connecting 7 and 12.

Repeat the procedure for 8, 9, 10, …, in turn. Note that when a product is larger than 1000000 then no line is drawn, for example: 500002 = 2 + 500000 = 3 + 499999 = …, so a line is drawn connecting 500002 and 1000000 but no line is drawn for 3 × 499999 = 1499997.

1) When your diagram is complete, are there two numbers, both less than 250000, such that there is no path along the lines connecting one to the other?

Now for your second task, take another piece of paper. You want to write the numbers 5, 6, 7, …, 98, 99, 100 on it, and then copy onto it, from your first piece of paper, all the lines connecting numbers which are both less than or equal to 100.

2) Can you complete your second task in such a way that no two of the lines in fact cross?


Enigma 475: Dance hall

From New Scientist #1626, 18th August 1988 [link]

At the dance there are 10 girls, Ann, Babs, Cath, Dot, Emma, Fay, Gwen, Hazel, Irene and Jane, and 10 boys. Jane knows one boy and Tom knows one girl, but I cannot tell you who they know. However, I can tell you all the other acquaintances:

Ken knows D, F;
Len knows E, H, I;
Mac knows B, F, G, I;
Ned knows A, B, H;
Owen knows E, G, I;
Pat knows A, B, C, D;
Quentin knows E, G;
Ray knows A, C;
Sam knows C, E.

The first dance pairs them off as follows:

Ann takes the hand of the first boy she knows, Ned (first always means first in alphabetical order), Babs does the same to Mac, then Cath, to Pat, then Dot, to Ken, then Emma, to Len. When Fay approaches Ken she finds he is holding Dot’s hand and the procedure becomes more complicated.

They form a line of the dance floor, F, K-D and ask the first boy who knows any girl on the floor to come out, bringing any girl that is holding his hand. The line becomes F, K-D, M-B. The procedure is repeated to give F, K-D, M-B, N-A. Repeating adds P-C to the line. Repeating again adds Ray to the line and the procedure stops as he has his hands free. The we have the line F, K-D, M-B, N-A, P-C, R.

Now Ray takes the hand of the girl he knows who is nearest to Fay in the line, Ann, and she releases Ned’s hand. Ned repeats the procedure Ray used and takes the hand of Babs who releases Mac. Mac repeated the procedure and takes the hand of Fay. That completes the pairing for Fay’s round.

The procedure is repeated for Gwen, Hazel, Irene, and Jane in turn. The pairing obtained after Jane’s round includes C-S, D-K, F-M, H-N and I-L.

Who do Jane and Tom know?

Note: I corrected a typo in the original puzzle while transcribing this (and I hope I didn’t introduce any more myself).


Enigma 1036: Multiple shifts

From New Scientist #2192, 26th June 1999 [link]

George’s young son has been playing with numbered cards and has discovered an interesting curiosity. Having placed six cards on the table to show the number 179487, he then moved the right-hand card to the left to form the number 717948. A quick check on his calculator confirmed that this number is exactly four times the first number.

George congratulated his son on this discovery, and then challenged him to find the smallest number which can be multiplied by four simply by moving the last digit to the front.

Can you find this number — and the smallest number that can be multiplied by five in similar fashion?


Enigma 474: More goals

From New Scientist #1625, 11th August 1988 [link]

In this football league table, the sides are eventually going to play each other once, but, of the figures given, one is incorrect.

Enigma 474

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

Which figure is wrong? What should it be? Find the score in each match.


Enigma 1037: The perfect shuffle

From New Scientist #2193, 3rd July 1999 [link]

I recently took an ordinary pack of playing cards and placed them on the table, face down. Somewhere in the pack the four aces were together, and in places further down the kings were together, the queens were together, and the jacks were together.

I then did the “perfect shuffle”. In other words I took the top 26 cards in my left hand, and the other 26 in my right, I flicked the bottom left-hand card on to the table, followed by the bottom right-hand card on top of it, followed by the next left on top of them, next right, next left, and so on. The cards remained face down at all times.

My fellow players were so impressed with this performance that I did three more perfect shuffles with the same pack. When I had finished, the arrangement of suits within the pack was exactly the same as when I started (for example, the top card was a heart before and after the four shuffles, the next was a spade before and after the four shuffles, and so on).

Counting from the top, what was the position of the ace of hearts after the four shuffles?