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Jul 10 20

by Brian Gladman

Sunday Times Teaser 3016 – Eureka

by Andrew Skidmore

Published Sunday July 12 2020 (link)

Archimedes’ Principle states that the upward force on a floating body equals the weight of water displaced by the body. In testing this, Liam floated a toy boat in a cylindrical drum of water. The boat had vertical sides and took up one tenth of the surface area of the water. He also had some identical ball bearings (enough to fit snugly across the diameter of the drum), which were made of a metal whose density is a single-figure number times that of water.

Liam loaded the boat with the ball bearings and noted the water level on the side of the drum. When he took the ball bearings out of the boat and dropped them in the water, the water level changed by an amount equal to the radius of a ball bearing.

How many ball bearings did Liam have?

Jul 3 20

by Brian Gladman

Sunday Times Teaser 3015 – Quid Quo Pro

by Victor Bryant

Published Sunday July 05 2020 (link)

In Readiland the unit of currency is the quid. Notes are available in two denominations and with these notes it is possible to make any three-figure number of quid. However, you need a mixture of the denominations to make exactly 100 quid. Furthermore, there is only one combination of denominations that will give a total of 230 quid.

What are the two denominations?

Jun 29 20

by Brian Gladman

Sunday Times Teaser 3014 – Family Business

by Danny Roth

Published Sunday June 28 2020 (link)

George and Martha run a company with their five daughters. The telephone extensions all have four positive unequal digits and strangely only four digits appear in the seven extensions:
\[\begin{array}{|l|r||}
\hline Andrea & abcd \\
\hline Bertha & acdb \\
\hline Caroline & bacd \\
\hline Dorothy & dabc \\
\hline Elizabeth & dbca \\
\hline George & cabd \\
\hline Martha & cdab \\
\hline \end{array}\]

They noticed the following:

Andrea’s and Bertha’s add up to Dorothy’s.

Bertha’s and Elizabeth’s add up to George’s.

Caroline’s and Dorothy’s add up to Martha’s.

What is Andrea’s extension?

Jun 19 20

by Brian Gladman

Sunday Times Teaser 3013 – Arian Pen-Blwydd

by Howard Williams

Published Sunday June 21 2020 (link)

When I thought that my daughter was old enough to be responsible with money I gave her on her next, and all subsequent birthdays, cash amounts (in pounds) which were equal to her birthday age squared.

On her last birthday her age was twice the number of years for which she received no such presents. I calculated at this birthday that if I had made these gifts on all of her birthdays then she would have received 15% more than she had actually received. I then decided that I would stop making the payments after her birthday when she would have received only 7.5% more if the payments had been made on all of her birthdays.

What was the amount of the final birthday payment?

Jun 15 20

by Brian Gladman

New Scientist Enigma 553 – Playing For Time

by Chris Maslanka

From Issue #1706, 3rd March 1990

One disadvantage of doing recurrence relations in your head, Kugelbaum told himself, was that all to often you were still calculating away when the clock showed it was time to give a class. To buy himself a little more time he jotted down on the whiteboard:
\[\frac{A\widehat{X}B}{B\widehat{X}A}=\frac{AB}{BA}\]
‘In this relation’, explained Kugelbaum to the class, ‘\(A\), \(B\) and \(X\) are all non-zero digits and \(AB/BA\) is a proper fraction in its lowest terms. The hat above the X in the numerator and that above the X in the denominator means that the digit \(X\) is used but it must be repeatable an arbitrary number of times without affecting the validity of the equation. So the \(X\) can be there once, twice, … 37 times … but, no matter how many times you use it, the equation must remain true. It is, or course, to be taken for granted that the number of \(X\)s interposed between the A and B must be the same for both the numerator and the denominator”.

So saying, the professor returned to his mental calculations while the class set about solving the problem.

How many distinct solutions does Kugelbaum’s equation have?

NB. A proper fraction has its numerator smaller than its denominator; if it’s in its lowest terms, numerator and denominator have no common factor other than 1.

Jun 12 20

by Brian Gladman

Sunday Times Teaser 3012 – Number Blind Rage

by Stephen Hogg

Published Sunday June 14 2020 (link)

After school, angry at getting “50 lines”, I kicked my satchel around. Impacts made my 11-digit calculator switch on. An 11-digit number was also entered and the display was damaged. Strangely, I found “dYSCALCULIA” displayed and saved this to memory (as shown).

After various tests I confirmed that all arithmetic operations were correct and the decimal point would appear correctly if needed. No segments were permanently “on”, two digits were undamaged, and for the other digits, overall, several segments were permanently “off”.

Retrieving “dYSCALCULIA”, I divided it by 9, then the result by 8, then that result by 7, then that result by 6. No decimal point appeared and the last result (at the right-hand side of the display) had three digits appearing as numerals.

What number was “dYSCALCULIA”?

Jun 11 20

by Brian Gladman

Project Euler Problem 917 – Number Splitting

We define an S-number to be a natural number, \(n\), that is a perfect square and its square root can be obtained by splitting the decimal representation of n into 2 or more numbers then adding the numbers.

For example, 81 is an S-number because \(\sqrt{81} = 8 + 1\).
6724 is an S-number: \(\sqrt{6724} = 6 + 72 + 4\).
8281 is an S-number: \(\sqrt{8281} =8 + 2 + 81 = 82 + 8 + 1\).
9801 is an S-number: \(\sqrt{9801} = 98 + 0 + 1\).

Further we define \(T(N)\) to be the sum of all S numbers \(n ≤ N\).  You are given \(T(10^4) = 41333\).

Find \(T(10^{12})\).

Jun 7 20

by Brian Gladman

New Scientist Enigma 957 – Open The Box!

by Susan Denham

From Issue #2112, 13th December 1997

Ten executives each rented two of the 20 safe-deposit boxes shown:

Each of the 10 had the same number of gold coins to share out between her two boxes. The first executive put 1 coin in one of her boxes and the rest in her other box. The second executive put 2 coins in one of her boxes and the rest in her other box. The third executive put 3 coins in one of her boxes and the rest in her other box. And so on, up to the tenth executive, who put 10 coins in one of her boxes and the rest in her other box. The overall effect of this was that each of the four rows of boxes contained in total an equal number of the gold coins. Also each of the five columns of boxes contained in total an equal number of coins.

One night a burglar broke into and emptied four of the boxes, three of them being in the same row. His total haul of coins was a three-figure number with no zeros.

How many coins did each executive have in the first place?

Jun 5 20

by Brian Gladman

Sunday Times Teaser 3011 – Optical Illusion

by Danny Roth

Published Sunday June 07 2020 (link)

George and Martha are studying optics. If you place an object a specific distance from a lens, an image will appear at a distance from that lens according the following formula:

The reciprocal of the object distance plus the reciprocal of the image distance is equal to the reciprocal of the focal length of the lens.

The object distance was a two-digit whole number of cm (ab). The image distance and the focal length of the lens were also two-digit whole numbers (cd and ef respectively). The six digits were all different and non-zero. Also, the object distance and the focal length were of the same parity and b was an exact multiple of d. Martha pointed out that the sum of those three two-digit numbers was a prime number.

What was that prime number?

May 30 20

by Brian Gladman

Sunday Times Teaser 3010 – Putting Game

by John Owen

Published Sunday May 31 2020 (link)

A putting game has a board with eight rectangular holes, like the example (not to scale), but with the holes in a different order.

If you hit your ball (diameter 4cm) through a hole without touching the sides, you score the number of points above that hole. The sum of score and width (in cm) for each hole is 15; there are 2cm gaps between holes.

I know that if I aim at a point on the board, then the ball’s centre will arrive at the board within 12cm of my point of aim, and is equally likely to arrive at any point in that range. Therefore, I aim at the one point that maximises my long-term average score. This point is the centre of a hole and my average score is a whole number.

(a) Which hole do I aim at?

(b) Which two holes are either side of it?