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Programming Enigma Puzzles

26 February 2018

Posted by on **From New Scientist #2229, 11th March 2000**

In cross-country matches, teams consist of six runners. The team scores are decided by adding together the finishing positions of the first four runners to finish in each team. The team with the lowest score is the winner. Individuals never tie for any position and neither do teams because if two teams have the same score the winner is the team with the better last scoring runner.

The fifth and sixth runners to finish in each team do not score. However if they finish ahead of scoring runners in another team they make they make the scoring positions of those scoring runners, and the corresponding team score, that much worse.

In a recent match between two teams, I was a non-scorer in the winning team. Each team’s score was a prime number, and if I told you what each team’s score was you could deduce with certainty the individual positions of the runners in each team. I won’t tell you those scores, but if you knew my position you could, with the information given above, again deduce with certainty the individual positions of the runners in each team.

(1) What was my position?

(2) What were the positions of the scoring runners in my team?

[enigma1073]

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12 February 2018

Posted by on **From New Scientist #2231, 25th March 2000**

I have found a five-digit number such that it is impossible to factorise the numbers formed by its first digit or last digit or first two digits or last two digits or first three digits or last three digits or first four digits or last four digits or all five digits. In other words all those numbers are prime except that either or both of the single digit numbers may be unity.

Identify the five-digit number.

[enigma1075]

29 January 2018

Posted by on **From New Scientist #2233, 8th April 2000** [link]

I have found three examples of a three-digit number that can be the sum of two three-digit perfect squares in two different ways. One particular perfect square contributes to all three of my examples.

Everything stated above about me is also true of both Harry and Tom, but each of us has a different perfect square contributing to all three examples. One of my examples is the same as one of Harry’s, and another of my examples is the same as one of Tom’s.

(1) Which three-digit number is the sum of each pair of squares in the example that I found but neither Harry or Tom found?

[There is a further example of a three-digit number than can be the sum of two three-digit perfect squares in two different ways that none of us found].

(2) Which three-digit number is the sum of each pair of squares in the example that none of us found?

[enigma1077]

1 January 2018

Posted by on **From New Scientist #2237, 6th May 2000** [link]

My supermarket was offering a discount of £2 off the cost of shopping if the bill exceeded a certain number of pounds. I qualified for the discount though my bill exceeded the minimum required for it by less than £1. My bill, both before and after the discount was applied, was for a prime number of pence.

The same discount offer applied the following week and everything stated above was again true of my new bill. Over the two weeks the total cost of my shopping was an exact number of pounds, prime whether or not you take the discounts into account and less than £30.

How much did each of my two bills amount to before the discounts were applied? Remember, there are 100 pence in a pound.

[enigma1081]

18 December 2017

Posted by on **From New Scientist #2239, 20th May 2000** [link]

A semi-prime is the product of two prime numbers. Harry, Tom and I each chose a set of three two-digit semi-primes which formed an arithmetic progression, such that the six factors of the three semi-primes in any one set were all different. The first (lowest) number in each of our sets was identical.

We then each chose another such set. This time the middle number in each of our sets was identical.

We then each chose another such set. This time the last (highest) number in each of our sets was identical, and of the other two semi-primes in my set one also appeared in Harry’s set and the other also appeared in Tom’s set.

The nine sets that we chose between us were all different, but one semi-prime appeared in all three sets that I chose.

List the three sets that I chose in the order in which I chose them, with the numbers in each set in ascending order.

[enigma1083]

27 November 2017

Posted by on **From New Scientist #2242, 10th June 2000** [link]

Harry owns fewer than 500 snooker balls; he also owns four differently sized rectangular trays in which he can stack the balls, each tray being large enough to accommodate at least four balls along each side.

Harry starts the stacking by filling the base of the tray with as many balls arranged in rows and columns as its size allows; he then stacks further balls in layers, the balls in each layer covering the cavities between the balls in the layer below. If he starts with a square tray (though he does not necessarily have one that is square) the number of balls in each successive layer remains square until the top layer consists of just one ball. If he starts with a tray that is not square the top layers consists of a single row of two or more balls. So on top of a layer of 15 balls arranged 5 × 3 would rest a layer of eight (= 4 × 2) and then a top layer of three (= 3 × 1) balls.

If Harry stacks any one of his four trays as described above, he uses all his snooker balls to complete the task.

Everything stated above about Harry is also true for Tom. But Tom owns more snooker balls than Harry.

How many more?

[enigma1086]

25 September 2017

Posted by on **From New Scientist #2252, 19th August 2000** [link]

At snooker a player scores 1 point for potting one of the 15 red balls, but scores better for potting any of the 6 coloured balls: 2 points for yellow, 3 for green, 4 for brown, 5 for blue, 6 for pink, 7 for black.

Davies potted his first red ball, followed by his first coloured ball, then his second red ball followed by his second coloured ball, and so on until he had potted all 15 red balls, each followed by a coloured ball. Since the coloured balls are at this stage always put back on the table after being potted, it is possible to pot the same coloured ball repeatedly.

Davies’ break was interesting as after he had potted each of the 15 coloured balls his cumulative score called by the referee was always a prime number.

After potting the 15 red balls and 15 coloured balls, a player’s final task is to attempt to pot (in this order) yellow, green, brown, blue, pink and black. I won’t tell you how many of those Davies managed to pot, nor could you be sure how many of them he potted even if I told you his total score for the break.

What was that total score?

[enigma1096]

4 September 2017

Posted by on **From New Scientist #2255, 9th September 2000** [link]

I have constructed a cyclical chain of four-digit perfect cubes such that each cube in the chain has no digits in common with either of its neighbours in the chain. The chain consists of as many different four-digit cubes as is possible, consistent with the stipulation that no cube appears in it more than once.

If I were to tell you how many cubes lie between 1000 and 1331 either by the shorter route or by the longer route round the chain you could deduce with certainty the complete order of the cubes in the chain.

Taking the longer route round the chain from 1000 to 1331, list in order the cubes that you meet (excluding 1000 and 1331).

[enigma1099]

31 July 2017

Posted by on **From New Scientist #2260, 14th October 2000** [link]

In the following statement digits have been consistently replaced by capital letters, different letters being used for different digits:

ONE and NINE are odd perfect squares, FOUR is an even perfect square.

Find the numerical value of the square root of (NINE × FOUR × ONE).

[enigma1104]

17 July 2017

Posted by on **From New Scientist #2262, 28th October 2000** [link]

Harry, Tom and I each found a set consisting of a 4-digit perfect square, a 3-digit perfect square and a 2-digit perfect square that between them used nine different digits; but none of us could add a 1-digit square with the unused digit because 9, 4, 1 and 0 all appeared in each of our three sets. No two of us found exactly the same set; none of the squares in my set appeared in either Harry’s set or Tom’s set. There is one further set that none of us found whose unused digit is again not itself a perfect square.

List in ascending order the three squares in this set that none of us found.

[enigma1106]

26 June 2017

Posted by on **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).

[enigma1109]

8 May 2017

Posted by on **From New Scientist #2272, 6th January 2001** [link]

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]

17 April 2017

Posted by on **From New Scientist #2275, 27th January 2001** [link]

I invite you to select a three-digit prime number such that if you reverse the order of its digits you form a larger three-digit prime number. Furthermore the first two digits and the last two digits of these two three-digit prime numbers must themselves be four two-digit prime numbers, each one different from all the others.

Which three-digit prime number should you select?

[enigma1119]

20 March 2017

Posted by on **From New Scientist #2279, 24th February 2001** [link]

In the following statement digits have been consistently replaced by capital letters, different letters being used for different digits:

VIER and NEUN are both perfect squares.

If I told you the number represented by VIER you could deduce with certainty the number represented by NEUN. Alternatively, if I told you the number represented by NEUN you could deduce with certainty the number represented by VIER.

What is the numerical value of the square root of (VIER × NEUN)?

[enigma1123]

6 February 2017

Posted by on **From New Scientist #2285, 7th April 2001** [link]

The coins from 1p to 100p in circulation are for 1p, 2p, 5p, 10p, 20p, 50p and 100p. If one receives 40p in change for a purchase it can be paid in a minimum of 2 coins (20p + 20p) but could be paid in one coin more than that minimum number (20p + 10p + 10p). But a few amounts of change cannot be paid for in one coin more than the minimum possible number of coins, for instance 1p and 50p.

Harry, Tom and I each paid 100p for an item priced such that the change due could not be paid in one coin more than the minimum possible number of coins. Each item cost a different amount.

Next day Harry and Tom bought the same items at the same price as on the previous day, whereas I bought a different item at a different price from any of the others, which again was such that the change due from 100p could not be paid in one coin more than the minimum possible number of coins. Each day the total cost of the three purchases was a prime number of pence.

How much did each of my purchases cost?

[enigma1129]

19 December 2016

Posted by on **From New Scientist #2292, 26th May 2001** [link]

Triangular numbers are those that fit the formula ½n(n+1), like 1, 3, 6 and 10.

In the following statement digits have been consistently replaced by capital letters, different letters being used for different digits:

“ONE, THREE, SIX and TEN are all triangular numbers, none of which starts with a zero”.

Which numbers are represented (in this order) by ONE, THREE, SIX and TEN?

[enigma1136]

14 November 2016

Posted by on **From New Scientist #2297, 30th June 2001** [link]

In the following statements digits have been consistently replaced by capital letters, different letters being used for different digits:

TWO is a prime number,

FOUR is a perfect square,

EIGHT is a [perfect] cube.You should assume that neither TWO nor FOUR nor EIGHT starts with a zero.

Find the number represented by EIGHT.

[enigma1141]

26 September 2016

Posted by on **From New Scientist #2304, 18th August 2001** [link]

After they had each played four rounds in the golf tournament Bernhard, Colin, Darren and Ernie all ended up with the same total score even though the scores for the 16 individual rounds that they played were all different. Each player’s score for each round was in the 60s or 70s (that is to say between 60 and 79 inclusive). Bernhard’s score in each of his four rounds was a prime number, Colin’s score in each of his four rounds was a semi-prime (the product of two prime numbers), Darren’s and Ernie’s scores in each of their four rounds were numbers that are neither prime nor semi-prime. Darren’s best (lowest) round was better than Ernie’s best round, and Darren’s worst round was better than Ernie’s worst round.

List in ascending order Ernie’s scores for the four rounds.

[enigma1148]

22 August 2016

Posted by on **From New Scientist #2309, 22nd September 2001** [link]

In the Fibonacci sequence the first two terms are 1 and 1, and each subsequent term is the sum of the previous two terms; so the sequence starts 1, 1, 2, 3, 5. Less well known is the Lucas sequence, whose first two terms are 1 and 3, and each subsequent term is the sum of the previous two terms; so the sequence starts 1, 3, 4, 7, 11. In the Tribonacci sequence (so named by Mark Feinberg) the first three terms are 1, 1 and 2, as in the Fibonacci sequence, and each subsequent term is the sum of the previous three terms; so it starts 1, 1, 2, 4, 7.

Harry, Tom and I were looking to find a 2-digit Fibonacci number, a 2-digit Lucas number and a 2-digit Tribonacci number that used six different digits. We each found a different solution; our three Fibonacci numbers were all different from each other; our three Lucas numbers were all different from each other; and our three Tribonacci numbers were all different from each other. None of the numbers in my solution appeared in either Harry’s or Tom’s solution.

List in ascending order the numbers in my solution.

[enigma1153]

25 July 2016

Posted by on **From New Scientist #2313, 20th October 2001** [link]

Albion, Borough, City, Rangers and United have played another tournament. This time each team played each of the other teams twice. The two matches that any two teams played against each other had different results, even in the case of City, which did not draw any of its matches.

Two points were awarded for a win and one point for a draw. After each team had played each of the other teams once, United was in the lead, one point ahead of Rangers, which was one point ahead of City, which was one point ahead of Borough, which was one point ahead of Albion.

But by the end of the tournament the positions were completely the opposite: Albion finished one point ahead of Borough, which was one point ahead of City, which was one point ahead of Rangers, which was one point ahead of United.

If you knew the results of the first match between Borough and United you could deduce with certainty the results of all the other matches played.

Give the result of the first match Borough played against each of the other four teams, naming the opponents in each match.

[enigma1157]

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