A *lattice point* is a point with integer coordinates such as (2,3).

An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at (0,0) and takes a 10-step path, how many different points could be the final point?

Guest Mar 29, 2015

#2**+10 **

Wow....that 's pretty neat, Alan.....I've never looked at that in this way before.....!!!

Notice that Alan has pointed out the fact that the final "landing points" have coordinates whose sum * MUST be even!!!! It is impossible to land on a point whose coordinate sum is ODD!!!*...Try it for yourself....I dare you!!!

Notice that, along the x-axis "row," there are 11 possible landing points.

And, on each successive "row" above the x axis, we have (11-n) possible landing points where n is the y value associated with that "row"......so 10 + 9 + 8 +........+2 + 1 = 55 possible landing points * above* the x axis. And, by symmetry, we have the same number

So, there are 2(55) + 11 = 121 possible "landing points."

Also notice that we could have gotten the same answer by just "squaring" the 11 points on each side of the.....well.....**square !!!! **** **[Hey....I like to do things the hard way !!!]

Thanks, Alan.....that's a very clear "visual" answer !!!!

[I'm recommending this one for the Daily Wrap. ]

CPhill
Mar 29, 2015

#2**+10 **

Best Answer

Wow....that 's pretty neat, Alan.....I've never looked at that in this way before.....!!!

Notice that Alan has pointed out the fact that the final "landing points" have coordinates whose sum * MUST be even!!!! It is impossible to land on a point whose coordinate sum is ODD!!!*...Try it for yourself....I dare you!!!

Notice that, along the x-axis "row," there are 11 possible landing points.

And, on each successive "row" above the x axis, we have (11-n) possible landing points where n is the y value associated with that "row"......so 10 + 9 + 8 +........+2 + 1 = 55 possible landing points * above* the x axis. And, by symmetry, we have the same number

So, there are 2(55) + 11 = 121 possible "landing points."

Also notice that we could have gotten the same answer by just "squaring" the 11 points on each side of the.....well.....**square !!!! **** **[Hey....I like to do things the hard way !!!]

Thanks, Alan.....that's a very clear "visual" answer !!!!

[I'm recommending this one for the Daily Wrap. ]

CPhill
Mar 29, 2015

#3**+5 **

I already answered this question before

I don't know why people post the same answered question more than once.

**Anyway, Chis, your explanation is better than mine. Yours is a great answer especially as it is accompanied by Alan pic. :)**

Maybe the anon asker will be happy now :)

**Anon**, if you are not happy with an answer please question it on the original thread. We like people to question our answers it means you are actually trying to learn from us.

If you want to repost, that is fine but please do not pretend that it is a brand new question.

Here are some intructions on how it should be done.

Melody
Mar 30, 2015