Brain Teasers
Mad Ade's Kebab Essay
When Mad Ade was at University he completed an essay on "The molecular construction of a standard Kebab". After printing the work he was surprised to find that the number of digits used to number all the pages was a multiple of the number of actual pages the essay contained.
Using this information can you work out how many pages the essay contained in total?
Just to help, the essay was more than 1000 pages but less than 10,000 pages.
Using this information can you work out how many pages the essay contained in total?
Just to help, the essay was more than 1000 pages but less than 10,000 pages.
Answer
1107 pages.The number of digits used to number all the pages is 3321.
1-9: 9 digits
10-99: 180 digits
100-999: 2700 digits
1000-1107: 432 digits
Total: 3321 digits
3321/1107= 3
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Comments
I never knew that valence and taste were related! Great essay, er, teaser Mad.
I agree..great teaser!!!
My personal preference for the answers to Math teasers is to be shown how to arrive at the solution, after all it did say "can you work out". I liked the teaser but the answer was too glib. How do you work it out?
1000 pages = 2893 chars
2893/1000 = 2.893
10000 pages = 38894 chars
38894/10000 = 3.8894
we now know that 3 is the only possible integer that will work in the problem... Now the fun part is reversing the equation to find where 3 fits...
1 way (it may just be coincidence) is: 3 - 2.893 = .107, since we are working on the 1000's range of pages, multiply this by 1000 which = 107. add the 107 to the base (that being the basis of your 2.893 number) of 1000 and you get 1107.
OR - do it the longer (and probably more reliable way).
2889 (the number of chars in 999 pages + 4x (4 chars per page for an unknown number of pages) divided by "999 + x" (the total number of pages) = 3
numerically:
(2889 + 4x) / (999 + x) = 3
2889 + 4x = 2997 + 3x
2889 + x = 2997
x = 108
108 + 999 = 1107
if you see any errors here, please let me know.....
thx
2893/1000 = 2.893
10000 pages = 38894 chars
38894/10000 = 3.8894
we now know that 3 is the only possible integer that will work in the problem... Now the fun part is reversing the equation to find where 3 fits...
1 way (it may just be coincidence) is: 3 - 2.893 = .107, since we are working on the 1000's range of pages, multiply this by 1000 which = 107. add the 107 to the base (that being the basis of your 2.893 number) of 1000 and you get 1107.
OR - do it the longer (and probably more reliable way).
2889 (the number of chars in 999 pages + 4x (4 chars per page for an unknown number of pages) divided by "999 + x" (the total number of pages) = 3
numerically:
(2889 + 4x) / (999 + x) = 3
2889 + 4x = 2997 + 3x
2889 + x = 2997
x = 108
108 + 999 = 1107
if you see any errors here, please let me know.....
thx
since my message did not include carriage returns, it may be difficult to follow the last part... In the end I was simplifying the equation to solve for X.... Since the returns are not present in my post, it just looks like a long jumbled equation w/ a bunch of = signes... oh well, hopefully you can still figure out what I was doing.....
The way I did it is similar.
To number the first 1000 pages, the digits are 3000+1-18-90=2893.
Let p be the number of pages beyond 1000, and m the multiple.
2893+4p = m(1000+p) or
2893 = 1000m + (m-4)p
m must be at least 3 and must be odd. So I try 3 first:
2893 = 3000 - p
p=107
number of digits: 2893+4x107 = 3321 = 3x1107. It works!
I proved that it is a solution. I did not prove that it is the ONLY solution.
To number the first 1000 pages, the digits are 3000+1-18-90=2893.
Let p be the number of pages beyond 1000, and m the multiple.
2893+4p = m(1000+p) or
2893 = 1000m + (m-4)p
m must be at least 3 and must be odd. So I try 3 first:
2893 = 3000 - p
p=107
number of digits: 2893+4x107 = 3321 = 3x1107. It works!
I proved that it is a solution. I did not prove that it is the ONLY solution.
Dec 18, 2008
my freakin math teacher uses logic problems from this web site as homework
Very nice teaser. Gave me more trouble to figure out than such a simple teaser should.
Between gmredneck and canu they have the whole solution.
The reason 3 - 2.893 = .107 worked is because the difference between 3 and 4 is one, so each additional page added reduces the gap by (4-3)/1000 = .001.
It's interesting that 1107 is the only solution greater than 9 and less than 10000. The next solution is 11106 pages which use 44,424 digits (multiple 4), and that there is exactly one solution between 10^d and 10^(d+1) pages where d > 2, which always has a multiple of d.
1107 x 3 = 3321
11106 x 4 = 44424
111111 x 5 = 555555
1111117 x 6 = 6666702
11111124 x 7 = 77777868
111111132 x 8 = 888889056
1111111141 x 9 = 10000000269
. . .
Between gmredneck and canu they have the whole solution.
The reason 3 - 2.893 = .107 worked is because the difference between 3 and 4 is one, so each additional page added reduces the gap by (4-3)/1000 = .001.
It's interesting that 1107 is the only solution greater than 9 and less than 10000. The next solution is 11106 pages which use 44,424 digits (multiple 4), and that there is exactly one solution between 10^d and 10^(d+1) pages where d > 2, which always has a multiple of d.
1107 x 3 = 3321
11106 x 4 = 44424
111111 x 5 = 555555
1111117 x 6 = 6666702
11111124 x 7 = 77777868
111111132 x 8 = 888889056
1111111141 x 9 = 10000000269
. . .
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