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May 22 20

by Brian Gladman

Sunday Times Teaser 3009 – Head Count

by Victor Bryant

Published Sunday May 24 2020 (link)

My grandson and I play a simple coin game. In the first round we toss seven coins and I predict how many “heads” there will be whilst my grandson predicts the number of “tails”. After the tossing I score a point for each head plus a bonus of ten if my prediction was correct — my grandson scores likewise for the tails. We then repeat this with six coins, then five, and so on down to a single coin. After each round we keep cumulative totals of our scores. In one game, for over half of the pairs of cumulative scores, my grandson’s total was like mine but with the digits in reverse order. In fact he was ahead throughout and at one stage his cumulative total was double mine — and he had an even bigger numerical lead than that on just one earlier occasion and one later occasion.

List the number of heads in each of the seven rounds.

May 15 20

by Brian Gladman

Sunday Times Teaser 3008 – Three-Fruit Fractions

by Stephen Hogg

Published Sunday May 17 2020 (link)

The owner of the old curiosity shop repaired an antique mechanical fruit machine having three wheels of identical size and format. Afterwards each wheel was independently fair, just as when new. Each wheel’s rim had several equal-sized zones, each making a two-figure whole number of degrees angle around the rim. Each wheel had just one zone showing a cherry, with other fruits displayed having each a different single-figure number (other than one) of zone repetitions.

Inside the machine were printed all the fair chances (as fractions) of getting three of the same fruit symbol in one go. Each of these fractions had a top number equal to 1 and, of their bottom numbers, more than one was odd.

What was the bottom number of the chance for three cherries?

May 13 20

by Brian Gladman

Radio Times Puzzle for Today No. 272

by Professor Michael Paterson, University of Warwick

Published on Thursday 19 July 2018 (link)

I have a tray of length 5 and width 2 so 10 round coins of width 1 will fit in it snugly without overlaps. No room for another. Similarly, a try of length 50 will accommodate only 100 coins. Things get more interesting with a longer tray! A tray of length 500 and width 2 can accommodate at least 1001 coins. Show how this can be done. To be clear, this is just about fitting non-overlapping circles in rectangles, so no trickery with funny-shaped coins or thermal expansion coefficients! A useful hint is to start packing coins from the middle not from one end.

For a more mathematical challenge, what is the smallest \(n\) so that \(2n+1\) circles of diameter 1 can be packed without overlap in an \(n\) by 2 rectangle? If you think you have a good answer to this, do let me know, but note that the really hard problem, still I think unsolved, is to prove the best optimum result.

May 8 20

by Brian Gladman

Sunday Times Teaser 3007 – Paving Stones

by Howard Williams

Published Sunday May 10 2020 (link)

James has decided to lay square block paving stones on his rectangular patio. He has calculated that starting from the outside and working towards the middle that he can lay a recurring concentric pattern of four bands of red stones, then three bands of grey stones, followed by a single yellow stone band. By repeating this pattern and working towards the centre he is able to finish in the middle with a single line of yellow stones to complete the patio.

He requires 402 stones to complete the first outermost band. He also calculates that he will require exactly 5 times the number of red stones as he does yellow stones.

How many red stones does he require?`

May 1 20

by Brian Gladman

Sunday Times Teaser 3006 – Raffle Tickets

by Andrew Skidmore

Published Sunday May 03 2020 (link)

At our local bridge club dinner we were each given a raffle ticket. The tickets were numbered from 1 to 80. There were six people on our table and all our numbers were either prime or could be expressed as the product of non-repeating primes (eg 18 = 2x3x3 is not allowed). In writing down the six numbers you would use each of the digits 0 to 9 once only. If I told you the sum of the six numbers (a perfect power) you should be able to identify the numbers.

List the numbers (in ascending order).

Apr 25 20

by Brian Gladman

Sunday Times Teaser 3005 – Tubular Bales

by Stephen Hogg

Published Sunday April 26 2020 (link)

Ten equal-length, rigid tubes, each a different prime-valued external radius from 11mm to 43mm, were baled, broadside, by placing the 43mm and 11mm tube together and the third tube, not the largest remaining, touching both of these. Each subsequent tube touched the previous tube placed and the 43mm tube. A sub-millimetre gap between the final tube placed and the 11mm tube, made a near perfect fit.

The radius sum of the first three tubes placed against the 43mm tube was a multiple of one of the summed radii. Curiously, that statement remains true when each of “four”, “five”, “seven” and “eight” replaces “three”. For “two” and “six” tubes placed their radius sum was a multiple of an as yet unplaced tube’s radius.

What radius tube, in mm, was placed last?

Apr 21 20

by Brian Gladman

Sunday Times Teaser 3004 – Going Up

by John Owen

Published Sunday April 19 2020 (link)

In our football league, the teams all play each other once, with three points for a win and one for a draw. At the end of the season, the two teams with most points are promoted, goal difference being used to separate teams with the same number of points.

Last season’s climax was exciting. With just two games left for each team, there were several teams tied at the top of the league with the same number of points. One of them, but only one, could be certain of promotion if they won their two games. If there had been any more teams on the same number of points, then none could have guaranteed promotion with two wins.

How many teams were tied at the top of the league, and how many of the remaining matches involved any of those teams?

Apr 14 20

by Brian Gladman

New Scientist Enigma 544b – Little Puzzlers

by Keith Austin

From Issue #1696, 23rd December 1989

Amy, Beth, Jo and Meg decided to give each other pots of marmee-lade for Christmas. Each girl made a number of pots and then divided her pots into three piles, which were not necessarily equal; then she wrapped up each pile, labelled each parcel, and put them under the Christmas tree. The total number of pots involved was between 50 and 100.

We will let the letters A, B, C, …, I stand for the digits 1, 2, 3 , …, 9 in some order. Amy gave D/F of her pots to Jo, and C/B to Meg. Beth gave H/G of her pots to Amy, H/A to Jo, and D/G to Meg. Jo gave F/G of her pots to Amy. Meg gave A/B of her pots to Beth, and D/H to Jo.

On Christmas day, each girl opened the three parcels she had received. Amy received H/E of her pots from Beth, and F/E from Jo. Jo received D/I of her pots from Amy, D/H from Beth and D/A from Meg

Note that all the fractions were in reduced form before letters were substituted (1/2 and 2/3 are in reduced form, whereas 4/8 and 6/9 are not).

What was the total number of pots that were given?

Apr 11 20

by Brian Gladman

Sunday Times Teaser 3003 – All That Glitters

by Nick MacKinnon

Published Sunday April 12 2020 (link)

My aunt has a collection of sovereigns, and she set me a challenge. “You can have the coins if you can work out the dates, which (in increasing order) are equally spaced and all in the 20th century. The number of coins is an odd prime. The highest common factor of each pair of dates is an odd prime. The sum of the number of factors of each of the dates (including 1 and the date itself) is an odd prime.” I worked out the dates, though the gift was much less valuable than I’d hoped.

What were the dates?

Apr 4 20

by Brian Gladman

Sunday Times Teaser 3002 – Short Cut

by Victor Bryant

Published Sunday April 05 2020 (link)

To demonstrate a bit of geometry and trigonometry to my grandson, I took a rectangular piece of paper whose shorter sides were 24 cm in length. With one straight fold I brought one corner of the rectangle to the midpoint of the opposite longer side. Then I cut the paper along the fold, creating a triangle and another piece. I then demonstrated to my grandson that this other piece had double the area of the triangle.

How long was the cut?