How many five-digit numbers are there that do not contain the numbers 3 and 5, and are multiples of 4? (It cannot start with 0, ie. 01234 is not a 5 digit number)
Answer: As the solutions below explain he answer is 9408. Very well done to, in order of correct reply: Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; John Bowen, rubber industry consultant, Bromsgrove, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany. And, as this was quite a tricky teaser, we should also give an honourable mention to the following readers who came very close and/or were on the right track: Laurent Simonet, Michelin, France; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Michael Easton, sales and marketing director, Globus Group, Manchester, UK.
To be divisible by 4 there is a choice of 21 possible combinations of the last 2 digits if 5 and 7 are excluded
Possible first digits are 1to 9 excluding 5 and 7 , ie 7
Possible second and third digits are 8 [ie 0 to 9 less 5 and 7]
And there are 21 possible combinations for [fourth and fifth] digit.
So possible numbers is given by 1st x 2nd x 3rd x [4th + 5th] = 7 x 8 x 8 xd 21 = 9408
In total, there are 90.000 numbers (five digits), thereof 70.000 without 3 or 5 as the first digit. 25% of them are divisible by 4 (very 4th one).
Of these 17.500, 80% have no 3 or 5 in the 2nd digit, the same for the 3rd digit. Of the remaining 11.200 numbers, 4 out of 25 need to be removed (for numbers ending with 32, 36, 52, 56). This leaves 84% of 11200 -> 9408.
The first digit is a number from 1 to 9 excluded 3 and 5, there are 7 possibilities
The second and third digits are numbers from 0 to 9 excluded 3 and 5, there are 8 possibilities- the 4th and 5th digit must form a multiple of 4, there are 25 possibilities, but we must exclude 32,36, 52 and 56, so the possibilities are 21.
The numbers are therefore 7x8x8x21 = 9408
There are 9408 5-digits numbers with the mentioned rules, no number 3, no number 5 and multiples of four.
Well, using an Excel worksheet,
1. Starting with 10000 and concluding with 99999
2. Eliminating all number containing 3 and 5
3. Making the rest of the number divide by four
4. Eliminating all fractional results and keeping only integer numbers.
The result is 9408 different numbers.
New teaser on Monday