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Programming Enigma Puzzles

3 April 2020

Posted by on **From New Scientist #2121, 14th February 1998**

At tennis a set is won by the first player to win 6 games, except that if it goes to 5 games all it is won either 7 games to 5 or 7 games to 6. (As far as this puzzle is concerned this applies even to the final set).

The match that we are considering went to 5 sets and no two sets contained the same number of games. At the end of each set the total number of games played up to that point was always a prime number. From this information the score in one or more of the five sets can be deduced with certainty.

Which sets had a score that can be deduced with certainty, and what was the score in each of the sets concerned?

[enigma966]

20 March 2020

Posted by on **From New Scientist #2123, 28th February 1998**

Albion, Borough, City, Rangers and United have played another tournament in which each team played each of the other teams once. Three points were awarded for a win and one point to each team for a draw.

On points gained, the teams finished in the order: Albion, Borough, City, Rangers, United. Each team gaining a different number of points. Each team won at least one match and lost at least one match.

If I told you the result of one particular match in the tournament you would be able to deduce with certainty the results of all the other matches.

Which particular match, and what was the result?

[enigma968]

28 February 2020

Posted by on **From New Scientist #2126, 21st March 1998**

Harry, Tom and I were trying to find a two-digit perfect square, a three-digit perfect square and a four-digit perfect square, such that each of the nine digits from 1 to 9 was used once.

We each found a different valid solution. Both Tom and Harry found the only valid solution for their two-digit squares. Our solutions had no squares in common, but our two-digit squares were consecutive.

List in ascending order the perfect squares used in my solution.

[enigma971]

7 February 2020

Posted by on **From New Scientist #2129, 11th April 1998** [link]

Each pony club competing at the gymkhana entered one boy and one girl for the showjumping. In order to determine the order of competition, Mary put all the riders’ names in a hat and drew them out one at a time.

When Mary had drawn precisely half of the names she was surprised to find she had drawn one rider from each club. She calculated that if she did such a draw a thousand times this would only be likely to happen with this number of competing clubs on three occasions (to the nearest whole number). She then realised in amazement that the riders she had drawn in that first half were all the girls.

1. How many clubs were competing?

2. If the number of clubs competing is still as in question 1, on how many occasions (to the nearest whole number) would Mary be likely to draw all the girls first if she carried out the draw five million times?

[enigma974]

20 December 2019

Posted by on **From New Scientist #2134, 16th May 1998** [link]

Triangular numbers are those that fit the formula

n(n + 1)/2,so that the sequence starts: 1, 3, 6, 10, 15, 21, 28, …From the first 30 triangular numbers select a set that uses each of the ten digits 0 to 9 once.

1. What are the largest and smallest numbers in your set?

In the Fibonacci sequences the first two terms are 1 and 1 and each succeeding term is the sum of the previous two terms, so that sequence starts: 1, 1, 2, 3, 5, 8, 13, …

From the first 30 Fibonacci numbers select a set that uses each of the ten digits 0 to 9 once.

2. What are the largest and smallest numbers in your set?

[enigma979]

22 November 2019

Posted by on **From New Scientist #2138, 13th June 1998** [link]

Albion, Borough, City, Rangers and United have played another tournament in which each team played each of the other teams once. Two matches took place on each of the five consecutive Saturdays, and each of the five teams having one Saturday without a match.

A new points scoring system was introduced to try to maintain interest to the end:

1 point was awarded to each team that won on the first Saturday;

2 points to each team that won on the second Saturday;

3 points to each team that won on the third Saturday;

4 points to each team that won on the fourth Saturday; and

5 points to each team that won on the fifth Saturday.The use of penalty shoot-outs eliminated the possibility that matches might end in a draw.

Albion did not play on the fifth Saturday, and also lost two matches, but still ended up with more points than any other team. United ended up with the fewest points.

If Rangers beat City on the second Saturday give the results of Borough’s matches, listing them in the order in which they were played.

[enigma983]

25 October 2019

Posted by on **From New Scientist #2142, 11th July 1998** [link]

A semi-prime is the product of two prime numbers; the square of a prime counts as a semi-prime.

Harry, Tom and I were trying to find pairs of 2-digit semi-primes such that if we added the two semi-primes together we formed a 2-digit prime. We each found three such pairs; the 18 semi-primes we used and the 9 primes that were formed were all different.

Harry’s three odd semi-primes were all greater than 50; Tom’s three even semi-primes were all greater than 50.

What were my three pairs of semi-primes?

[enigma987]

11 October 2019

Posted by on **From New Scientist #2144, 25th July 1998** [link]

Harry and Tom have been investigating sets of positive integers that form arithmetic progressions where all the members of the set are prime numbers.

When they looked for a set of five such prime numbers, Harry found the set whose final (largest) prime is the smallest possible for the final prime in such a set. Tom found the set whose final (largest) prime is the next smallest possible.

Exactly the same thing happened when they looked for a set of six such prime numbers, and again when they looked for a set of seven such prime numbers.

What were the smallest and the largest primes in each of the three sets that Tom was able to find?

[enigma989]

20 September 2019

Posted by on **From New Scientist #2147, 15th August 1998** [link]

I have constructed a cyclical chain of 4-digit perfect cubes such that each cube has at least two digits (but not necessarily the same digits) in common with each of its neighbours in the chain. The chain consists of as many different 4-digit cubes as is possible consistent with the stipulation that no cube appears in it more than once.

If a cube has a repeated digit, that digit only counts more than once in calculating the number of digits it has in common with another cube if it also appears more than once in the other cube; so 1000 has three digits in common with 8000, but only one digit in common with 4096.

Which two cubes are the neighbours of 9261 in the chain if:

1. Somewhere in the chain two neighbouring cubes have more than two digits in common?

2. Nowhere in the chain do two neighbouring cubes have more than two digits in common?

[enigma992]

23 August 2019

Posted by on **From New Scientist #2151, 12th September 1998** [link]

You may think you have seen this puzzle before but the solution this time is different. Just as before, each letter stands for a different digit, the same letter represents the same digit wherever it appears and no number starts with zero. But this time THREE is an even number.

What is your WEIGHT?

[enigma996]

9 August 2019

Posted by on **From New Scientist #2153, 26th September 1998** [link]

The denominations of coins in circulation which are less than a pound are 50, 20, 10, 5, 2 and 1p.

Harry, Tom and I went into a shop recently and each made a purchase costing less than £1 (100p). The cost of each of these purchases was different. We each paid with a £1 coin and each received four coins in change — in each case the change due could not be given in fewer than four coins; but equally if we had paid the exact price for our purchases it would have been possible for each of us to have done so with four coins, but not with fewer.

The total cost of our three purchases was not only more than, but also an exact multiple of, the total amount of change than between us we received.

What did each of our purchases cost?

[enigma998]

26 July 2019

Posted by on **From New Scientist #2155, 10th October 1998** [link]

Since

Mis the Roman numeral for 1000, we can say that with this puzzleNew Scientisthas published its EnigmaMtimes — which is significant because:

ENIGMA ÷ M = TIMESIn 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.

I reckon that, with the extra puzzles that are sometime published under the same number at Christmas time, by the time **Enigma 1000** was published there had actually been 1011 *Enigma *puzzles in *New Scientist*.

However, a number of the puzzles in that range were flawed (I have found 17 so far, and there are 494 puzzles remaining to add to the site).

**Enigma 401** is unusual, as not only was the flaw acknowledged by *New Scientist*, but a corrected version of the puzzle was published as **Enigma 405**. Also **Enigma 9** is identical to **Enigma 83**. Together these reduce the count by 2 to give 1009 puzzles published.

[enigma1000]

12 July 2019

Posted by on **From New Scientist #2157, 24th October 1998** [link]

Albion have been playing in Europe, where the result of any fixture is decided by the aggregate scores achieved by the teams on two matches, each team being at home for one match and away for the other match. If the aggregate scores over the two matches are equal the rule is that the team that has scored more goals away from home wins.

Albion played in and won five fixtures in Europe. In four of them the aggregate scores were equal and Albion won each time on the “away goals” rule, even though in their five away matches Albion scored fewer goals than in their five home matches.

No two matches out of the ten that Albion played had the same score and no team scored more than three goals in any of the matches.

What were the scores in Albion’s five home matches? Give each score in the form

x-y, with Albion’s score first in each match.

I was going to mention that there are now “only” 500 *Enigma* puzzles left to post, but when I counted it up it looked like there were “only” 498 left, so I should have mentioned this with **Enigma 1003**. Although it is hard to be exact (there are duplicate puzzles, corrections, and sometimes extra puzzles at Christmas) I think it is safe to say that there are about 1294 *Enigma* puzzles posted so far, and about 498 left, so we are about 72% of our way through the *Enigma* puzzles.

[enigma1002]

21 June 2019

Posted by on **From New Scientist #2160, 14th November 1998** [link]

At tennis a set is won by the first player to win 6 games, except that if it goes to 5 games all it is won either 7 games to 5 or 7 games to 6. (As far as this puzzle is concerned, this applies even to the final set).

The match between André and Boris went to a deciding fifth set. At the end of each set that André won, the total number of games played up to that point was always a prime number. At the end of each set that Boris won, the total number of games played up to that point was always a perfect square. No two sets contained the same number of games. If I told you the total number of games played at the end of one particular set you could deduce with certainty who won each individual set and what the score was in each set.

In giving the score of a tennis match, it is the convention to give the score in each set in order, always giving the match-winner’s score first, so that if the score is given as (say) 7-6, 5-7, 6-4, 3-6, 6-2, it shows that the two sets the match-winner lost were the second and fourth, won by the opponent by 7 games to 5 and 6 games to 3, respectively.

Using this convention, tell me who won my match and give me its score.

[enigma1005]

7 June 2019

Posted by on **From New Scientist #2162, 28th November 1998** [link]

Harry, Tom and I were trying to find a 3-digit triangular number, a 3-digit Fibonacci number and a 3-digit perfect cube that between them used 9 different digits. (Triangular numbers are those that fit the formula n(n+1)/2; in the Fibonacci sequence the first two terms are 1 and 1, and every succeeding term is the sum of the previous two terms).

We each found a valid solution. Mine had one number in common with Harry’s solution and a different number in common with Tom’s solution; otherwise, the numbers used in our solutions were all different.

List in ascending order the numbers used in my solution.

[enigma1007]

3 May 2019

Posted by on **From New Scientist #2168, 9th January 1999** [link]

In this sum each letter represents a different digit. The same letter represents the same digit wherever it appears and no number starts with a zero.

What is the 5-digit number represented by EIGHT?

This is the first puzzle that was published in 1999, so there is now a complete 15 year archive of *Enigma* puzzles from the start of 1999 to the final *Enigma* puzzle published in December 2013. There is also a complete 11 year archive of earlier puzzles from October 1977 to January 1989. As well as *Tantalizer* puzzles from 1976 and 1977. This brings the total number of archived puzzles to over 1400. I will continue to expand the archive by posting puzzles on a regular schedule.

[enigma1012]

5 April 2019

Posted by on **From New Scientist #2172, 6th February 1999** [link]

A semi-prime is the product of two prime numbers. Sometimes when the digits of a semi-prime are reversed, the resulting number is also a semi-prime: 326 and 623 (2 × 163 and 7 × 89 respectively) are both semi-primes. More rarely the sum of these two semi-primes is itself a semi-prime, as with 326 + 623 = 949 = 13 × 73.

In fact, 949 is the largest of very few three-digit semi-primes that have these characteristics; but two of the three-digit semi-primes that can be the sum of two semi-primes, of which one is the reverse of the other, are consecutive numbers.

Identify these two consecutive three-digit numbers and the sums that lead to them. (Give your answer in the form

a = b + candd = e + f, whered = a + 1).

[enigma1016]

15 March 2019

Posted by on **From New Scientist #2175, 27th February 1999** [link]

If Britain doesn’t join the common European currency, the British will have to get used to converting pounds into euros or vice versa.

Maybe pounds will be double the value of euros, maybe even four times the value:

(a) EUROS × 2 = POUNDS

(b) EUROS × 4 = POUNDSThese two multiplications are entirely distinct — any letter may or may not have the same value in one sum as the other. But within each sum each letter represents a different digit, the same letter represents the same digit wherever in the sum it appears, and no number starts with a zero.

Just as the multiplier in (b) is double that in (a), so is the number of solutions — there is just one solution to (a) but there are two to (b).

Please find the six-digit numbers represented by POUNDS in the solution to (a) and in each of the solutions to (b).

A correction was published with **Enigma 1023**, that: *the phrase “any letter may not have the same value” should have read “any letter may or may not have the same value”.* I have made the correction in the text above. It think all it’s trying to say is that the two alphametic sums should be treated separately, and don’t necessarily use the same mapping of letters to digits.

[enigma1019]

15 February 2019

Posted by on **From New Scientist #2179, 27th March 1999** [link]

Harry and Tom have been investigating sequences of positive integers that form arithmetic progressions where each member of the sequence is the product of two different prime numbers and no members of the sequence have any factors in common. Harry has found the sequence of four such integers whose final (largest) member is the smallest possible for the final member of such a sequence; Tom has found the sequence of five such integers whose final (largest) member is the smallest possible for the final member of such a sequence.

What are the smallest and largest integers in:

(1) Harry’s sequence,

(2) Tom’s sequence.

[enigma1023]

18 January 2019

Posted by on **From New Scientist #2183, 24th April 1999** [link]

Harry, Tom and I were asked by Mary each to select a five-digit perfect cube that consisted of five different digits and to tell her (in secret) which one had been selected. After we had each done so she said:

“If I now told any one of you individually how many digits his cube has in common with each of the other two cubes he could deduce with certainty one, but not both of them”.

That in itself was enough information to enable me to deduce with certainty both the cubes that Harry and Tom had between them selected.

What were those two cubes?

In the magazine this puzzle was published as **Enigma 1026** (despite that number having been used the previous week).

[enigma1027]

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