Enigmatic Code

Programming Enigma Puzzles

Tag Archives: by: Richard England

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.

[enigma1030]

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.

[enigma1034]

Enigma 1044: Seven-digit squares

From New Scientist #2200, 21st August 1999 [link]

I have found a 2-digit number that can form (without at any stage reversing the order of the digits) both the first 2 digits and the last 2 digits of both a 5-digit and a 6-digit and a 7-digit perfect square.

Harry and Tom have each found a 2-digit number that can form (without reversing the order of the digits) the first 2 digits of a 5-digit and a 6-digit and a 7-digit perfect square, the last 2 digits of these squares in each case being the same as the first 2 but in reverse order.

Harry’s 3 squares are all palindromes, but Tom’s squares are not all palindromes.

What are the 7-digit squares formed by (1) me, (2) Harry, (3) Tom?

[enigma1044]

Enigma 1046: Albion yet again

From New Scientist #2202, 4th September 1999 [link]

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

The points scoring system was: 1 point to each team winning 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. But a draw was an allowable result: if a match was drawn each team was awarded half the points it would have been awarded for winning it.

The final table was: Albion won with 7 points, Borough, City and Rangers all tied with 6 points, and United finished last with 5 points. But if they had used the traditional points scoring system that awards a team 2 points for each win and 1 point for each draw the order of the teams would have been reversed: United would have finished with more points than any other team and Albion would have finished with fewest points, the other three teams still tying for second place.

If Borough beat City on the fourth Saturday give the results of Rangers’ matches, listing them in the order in which they were played and naming the opponents in each match.

[enigma1046]

Enigma 1050: Find the link

From New Scientist #2206, 2nd October 1999 [link]

I was trying to construct a chain of 2-digit and 3-digit perfect squares such that each square in the chain had at least two digits in common with each of its neighbours (or with its sole neighbour, if it was at either end of the chain). If a square had a repeated digit that digit only counted more than once in calculating the number of digits that the square had in common with another square if it also appeared more than once in the other square; so 121 had only one digit in common with 100, but 100 had two [digits] in common with 400.

I found that I could construct three totally different chains each consisting of at least five squares; and I made each of these chains as long as possible, consistent with the stipulation that no square should be used more than once. But I could not link these chains into a single long chain until I used one particular 4-digit square as the means of linking one end of my first chain to one end of my second chain and also the other end of my second chain to one end of my third chain. In each place where it was used, this 4-digit square had at least two digits in common with each of its chain neighbours.

(1) Identify this 4-digit square.
(2) Which were the squares at opposite ends of the single long chain?

[enigma1050]

Enigma 1052: Half-dozen doubled

From New Scientist #2208, 16th October 1999 [link]

SIX × TWO = TWELVE

In 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. SIX, appropriately enough, must be a multiple of 6.

What is the 6-digit number represented by TWELVE?

[enigma1052]

Enigma 1054: Transferred number

From New Scientist #2210, 30th October 1999 [link]

The players in a football team used to wear the numbers 1 to 11, but nowadays each member of a club has a personal number. So Albion’s players are numbers from 1 to 37, except for one number which is unused because the player who wore it has just been transferred.

Last Saturday two Albion teams were playing. In one team the 11 players all wore different prime numbers, in the other team the 11 players all wore different semi-prime numbers. A semi-prime number is the product of two prime numbers; the square of a prime counts as a semi-prime (but 1 is neither prime nor semi-prime).

The sum of the numbers worn by one team last Saturday was the same as the sum of the numbers worn by the other team. If you knew the number of the player who has been transferred, you could deduce with certainty the numbers worn by all the players in each team.

What was the number of the player who has been transferred?

[enigma1054]

Enigma 1057: Recycled change

From New Scientist #2213, 20th November 1999 [link]

The denominations of coins currently in circulation are 200, 100, 50, 20, 10, 5, 2 and 1p. When we pay for an item we quite often exchange fewer coins when change is given than when the exact amount is offered. For instance, an item costing 91p would require at least four coins (50+20+20+1) for the exact amount, but the purchase can be made with the exchange of only three coins (100+1–10) if change is given.

Harry, Tom and I each bought an identical item that cost less than 100p. None of us offered the exact amount, but we each exchanged fewer coins than if we had done so. In fact, we each exchanged the minimum number of coins possible for an item of that price even though we each offered a different amount of money in payment.

I paid first, and Harry and Tom each included a different one of the coins I had received in my change among the coins that they offered.

How much did the item cost?

[enigma1057]

Enigma 1059: Century break

From New Scientist #2215, 4th December 1999 [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 and 7 for black.

Davies potted his first red ball, followed by his first coloured ball, then his second red ball, and so on until he had potted all 15 red balls, each followed by a coloured ball.

After potting 15 red balls and 15 coloured balls, Davies had scored exactly 100 points; but it was interesting because in calling his score after each pot the referee had called every perfect square between 1 and 100.

Question 1: If in achieving this Davies had potted as few different colours as possible, which of the coloured balls would he have potted?

In fact Davies had brought a greater variety to the choice of coloured balls potted: for instance the 2nd, 5th, 8th, 11th and 14th coloured balls potted were all different and if I told you what they were you could deduce with certainty which ball was potted on each of his other pots.

Question 2: What (in order) were the 2nd, 5th, 8th, 11th and 14th coloured balls potted?

(In answering both questions give the colours).

[enigma1059]

Enigma 1061: Par is never prime

From New Scientist #2217, 18th December 1999 [link]

The local golf course has 18 holes; each of them has a par of 3, 4 or 5; no two consecutive holes have the same par. If you play the holes in order from 1 to 18 and score par for each hole, the number of strokes that you have played after you have completed any hole is never a prime number.

If I told you the par for one particular hole you could deduce with certainty the par of each of the first 15 holes.

Question 1: What is the number of the hole whose par I would tell you? And what is par for that hole?

If I told you the number of the hole whose par I was going to tell you so that you could deduce with certainty the par for each of the last three holes you could in fact make that deduction even before I had told you the par for that hole.

Question 2: What is the number of the hole whose par I would be going to tell you? And what is par for that hole?

[enigma1061]

Enigma 1068: Triangular Fibonacci squares

From New Scientist #2224, 5th February 2000 [link]

Harry, Tom and I were trying to find a 3-digit perfect square, a 3-digit triangular number and a 3-digit Fibonacci number that between them used nine different digits. (Triangular numbers are those that fit the formula ½n(n+1); 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 and we each created a second valid solution by retaining two of the numbers of our first solution but changing the other one. Our six solutions were all different.

List in ascending order the numbers in the solution that none of us found.

[enigma1068]

Enigma 1073: Cross-country match

From New Scientist #2229, 11th March 2000 [link]

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]

Enigma 1075: No factors

From New Scientist #2231, 25th March 2000 [link]

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]

Enigma 1077: Identical square sums

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]

Enigma 1081: Prime cuts

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]

Enigma 1083: Trios of semi-primes

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]

Enigma 1086: Stacking trays

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]

Enigma 1096: Prime break

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]

Enigma 1099: Unconnected cubes

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]

Enigma 1104: Odd and even squares

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]