Yes, Kandice the Kangaroo will reach Joey without getting caught!
How will Kandice the Kangaroo reach Joey without getting caught?
- First, note that two consecutive cubes are always separated by one more than a multiple of 6. To see this, note that (n+1)3 = n3 + 3n2 + 3n + 1, so (n+1)3 - n3 = 3n(n+1) + 1. n must be either even or odd; if it is even, then n(n+1) is even, and if it is odd, then n+1 is even, so again, n(n+1) is even. Therefore, 3n(n+1) is divisible by both 2 and 3, so it must be a multiple of 6, so the difference (n+1)3 - n3 is one more than a multiple of 6.
- Now, it's easy to check that Kandice will safely jump over the first two guards (at 1 and 8). Suppose Kandice later gets caught and Gary the Guard is the first guard who catches her. Since passing the guard previous to Gary, Kandice would have finished one jump of size 7 and several jumps of size 6.
Because the distance between the guards is one more than a multiple of 6 (as shown in this illustration), counting backwards we see this pattern would have forced Kandice to land on the previous guard also. But this is impossible, since Gary was the first guard to catch Kandice. So Gary does not exist and Kandice will never be caught!
Key: cube numbers are the positions of the guards.
Evens and odds alternate, so if some number n isn't even, then the next number (n+1) must be even.
Illustration of a number line starting at 0 and going to infinity with 32 tick marks labeled at every 3rd tick mark. The mark at 0 is labeled K and the marks at 1 8 and 27 are labeled G. There are arched arrows above the number line indicating the bounce pattern of Kandice. It illustrates that Kandice jumps from 0 to 6 (6 units) 6 to 13 (7 units) 13 to 20 (7 units) 20 to 26 (6 units) and 26 to 32 (6 units).