Showing posts with label mindbender. Show all posts
Showing posts with label mindbender. Show all posts

Thursday, 26 July 2012

Mindbender 5


Another puzzle that caught my eye from the Western Morning News "Mindbender" series.

See solution.

- Ray

Wednesday, 15 February 2012

Mindbender 4


I've been following for a few weeks the daily Mindbender puzzles in the Western Morning News, and documenting my solutions for some: not through any claim to brilliance, but because they often present interesting angles on techniques for solving mathematical problems.

This one proved a classic example for a very powerful technique that's not well-known outside the maths/sciences circuit.

See solution.

- Ray

Thursday, 26 January 2012

Mindbender 2


As mentioned previously, I've been getting into the Western Morning News's Mindbender daily mathematical puzzles. Here's another one, which raises interesting points regarding my earlier thoughts on solution methods. Have a go, then go to this backdated post (SPOILER: solution).

- Ray

Sunday, 18 December 2011

Mindbender 1


I regularly do the Mindbender mathematical puzzles in the Western Morning News. Usually they involve fairly simple algebra, but sometimes they need number theory. This one proved even more difficult than usual. The problem comes down to finding a number that's one more than 17 times the sum of its digits. I did solve it eventually, but by a lot of intuitive jumps. Can anyone see a straightforward method? Have a go, and then go here (spoiler with solution) to compare.

- Ray

Thursday, 28 July 2005

Mindbender 5: solution


Solution backdated to avoid spoiler to Mindbender 5.

At first this looked like a case for trying an iterative solution, but it turned out for once to be simple arithmetic. In terms of strings, the puzzle is:

ABCDEF x 6 = DEFABC

This problem reduces since you can call the two blocks P and Q, so numerically:

(1000P + Q) x 6 = 1000Q + P

so expanding and collecting:

6000P + 6Q = 1000Q + P
5999P = 994Q
Q = (5999/994)P

... which simplifies on inspection to Q = (857/142)P

This means P has to be a multiple of 142. And further, given the condition that the number must lie between 100,000 and 200,000, 142 is the only possibility within range.

So the number is 142857.

142857 x 6 = 857142 QED.