The inhabitants of BZ-land have been cutoff from the rest of the world for generations and their culture has drifted. For example, although they write out numerals in the Arabic manner, when counting verbally to nine, they use “biz” for 5, “buz” for 7 and “baz” for 9. Even more confusing, for numbers above 9, their system requires “biz” to be said once for every time the digit 5 appears in the number, and also an additional “biz” for each time 5 is a divisor of that number. Similar rules exist for numbers that contain and/or are divisible by 7 (“buz”) & 9 (“baz”). Numbers that do not result in any biz-es, buz-es or baz-es merely retain their original name. Hence 10 is spoken “biz”, 27, “buz baz”, and 35 is “biz biz buz”. However, 31 is “thirty-one”. Also, due to their simplicity and isolation, the BZ-landers have no need for numbers greater than 5000 ( “biz biz biz biz biz” ). To them, after that there is just “many”.

One consequence of this system is that verbal counting numbers are not always unique; hence math operations that can be done on paper are hard to interpret out loud. Can you answer the following? Analytical solutions preferred, if they exist. (Full disclosure: the author created the questions and found all answers by creating a list, but believes that some or all of the questions can be answered with analysis, logic and insight).

1) Once you reach the sixth “biz biz biz”, what is the largest number of consecutive numbers that are named by their conventional (English) names before you reach the seventh “biz biz biz”?

2) What is the smallest value that is correctly represented by “biz buz baz”?

3) Is there such a number as “biz biz buz buz baz baz”?

4) Is there such a number as “biz biz biz buz buz baz”?

5) Can you find a sequence of three numbers, n, n+1, n+2 where n=biz, n+1=buz & n+2=baz?

Considering biz buz baz less than 500, the number can't end in 5 as that would be at least biz biz. The smallest number which ends with 57 and divisible by 9 is 637 and ends with 59 and divisible by 7 is 259. 259 would seem low, but 252 doesn't end in 7 or 9, and is divisible by 63. All other biz buz baz must be greater than 500 if they don't have a 5 in the tens place.

As for part 3, it is more arduous. Similar to the usage in part 1, multiples of 25 (not divisible by 100) have biz biz biz already, and it easy to see there are none that are divisible by 100. So the number is not divisible by 25 and thus must have a 5 somewhere. It's also clear that the number must have two 7, two 9, or one of each in addition to a 5, as the smallest mutliple of 5*7*7*9 or 5*7*9*9 that contains the other two numbers is 5670, which is greater than 5000.

Since there is no room for two 5 in the number, it must end in a 5. If the number is divisible by 9, its digits must add up to 9 and it must contain one 7, but that would mean the other digits must add up to 6, thus not allowing a 7 or 9 without raising the number above 5000. If it is not divisible by 9, it must contain two of them and be divisible by 49, but no number ending with 2 9s and a 5 is divisible by 49 for quite a while past 5000.

Since we have ran out of possibilities below 5000, we may conclude there are no biz biz buz buz baz baz below 5000.

Posted by Gamer on 2007-01-18 23:20:17