Math Chat: Accurate Census Taking And Orderly Ants
Old marching ants challenge (Luke Somers)
A column of ants marches in single file, until one ant falls away and they then march two by two, until another ant falls away and they then march three by three, and so on until finally they are marching 10 by 10. How many ants were there to start with?
Charles Medler and Andrew Sornborger used computers to find the first possible answer: 2,519 ants. It works because 2,518 is divisible by 2; 2,517 is divisible by 3; 2,516 is divisible by 4; 2,515 is divisible by 5; 2,514 is divisible by 6; 2,513 is divisible by 7; 2,512 is divisible by 8; 2,511 is divisible by 9; and 2,500 is divisible by 10.
There are actually infinitely many other possible solutions: 5,039 ants; 7,559 ants; 10,079 ants; and so on: one less than any multiple of 2,520 ants.
To reason out all solutions, suppose you start with n ants. Then n-1 must be a multiple of 2; consequently n+1 must also be a multiple of 2. Similarly n-2 and n+1 must be multiples of 3. And so on until n-9 and n+1 are multiples of 10. Therefore n+1 must be a common multiple of all the integers from 2 to 10. The least common multiple turns out to be n+1= 2,520, so n=2,519. Other solutions come from multiples of 2,520: n+1= 2,520k, n=2,520k-1. Taking k= 1, 2, 3, and then 4 yields the first four solutions given above.
The easiest way to find the least common multiple of the integers from 2 to 10 is to factor them: 2=2; 3=3; 4=22; 5=5; 6=2x3; 7=7; 8=23; 9=32; 10= 2x5. To be a multiple of them all, n+1 must have 2 as a factor three times because 8 does; 3 as a factor twice because 9 does; and 5 and 7 as factors once. Therefore the least common multiple is 23 x 32 x 5 x 7 = 2,520.
Eric Brahinsky and Glenn Case suggest the fanciful answer of -1 ant, since -2 is a multiple of 2, -3 is a multiple of 3, and so on. Brahinsky also suggests solving the problem by using ANTilogarithms!
Medler computes that for the march to continue through marching 18 by 18 would require 12,252,239 starting ants.
(Other winning solutions by M. Abel, J. Berkovitz, S. Brill, M. Kramer, J. Morrison, and G. Sahagen.)
New census challenge
Gordon Squire notes that planners for the 2000 US Census wonder what to do about missing certain groups of people such as many homeless in New York. Suggestions include statistical random sampling. What approach would you suggest?
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