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Jan
8
comment There exists a power of 2 such that the last five digits are all 3's or 6's. Find the last 5 digits of this number
Once fully understood, I like this answer the most, it breaks the problem down into a plain logic puzzle. For others like myself for whom the possible values for [j+1,k] do not immediately jump to mind: 0:[0|5], 1:[0|5], 2:[1|6], 3:[1|6], 4&5:[2|7], 6&7:[3|8], 8&9: [4|9]
Jan
8
comment There exists a power of 2 such that the last five digits are all 3's or 6's. Find the last 5 digits of this number
Also the relationship between [j+1,k] and [j,k+1] presented in a way that the value you have ([j+1,k]) leads to the odd/even attribute of the value you don't have ([j,k+1]) would be clearer in this case: The choice will determine whether [j,k+1] is odd or even, if the greater of the two options is selected, [j,k+1] will be odd, otherwise [j, k+1] will be even. The [j+1,k] is less than / more than [j,k] doesn't hold when [j,k] is 0 or 9.
Jan
8
comment There exists a power of 2 such that the last five digits are all 3's or 6's. Find the last 5 digits of this number
I see now. It wasn't clicking that this held true generally for all digits [j,k] (when dividing by 2). IMO it wouldn't hurt to add this reasoning into the answer itself somewhere before you start pulling specific numbers out. (maybe a chart of the possible values of [j+1,k] for values of [j,k] at the end)
Jan
8
comment There exists a power of 2 such that the last five digits are all 3's or 6's. Find the last 5 digits of this number
There seems to be some steps/explanation of reasoning missing from this answer. I can follow where the 1's column digits come from, but it isn't obvious from your steps where you are getting the them, and therefore why any given 10's digit must be odd or even. Personally I've failed to make the leap from 'e must be even' to 'the 100's digit of n must be odd'. What information am I missing that makes this obvious?
Sep
30
comment Why is there no “remainder” in multiplication
+1 but there is no reason to avoid using artifact(or any word for that matter) on a six year old, especially one that is asking questions like this one.
Aug
8
comment Why does factoring eliminate a hole in the limit?
+1 for anticipating Walley World
Mar
7
comment Is there a term for the point/line/curve division of concavity of a surface in three dimensions?
@Glougloubarbaki that sounds right, but I've been out of working with any of this stuff in school for years and to be honest I'm having a little difficulty getting my head back around this.
Dec
19
comment How to get more creative in Calculus
+1 'try to do the same exercise in many different ways!' This is a great way to build up a better understanding/instinct as to what works well with which types of problems.
Sep
21
comment What are imaginary numbers?
imaginary numbers have to be the single worst named concept in mathematics. I had some trouble with iuntil I got into upper level calc and engineering classes and realized my problem was all in the name. i is simply another dimension.