A comment
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@audrey Maybe it is because...that all of the rows have pattern, like they repeat (Ex. 12321, 1221, 1234321).
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@victorioussheep haven't gotten here yet, but is this pascal's triangle? I love when a math patter lines up like that.
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@triumphantox Yes! It is a pascal's triangle, which is really cool when you get to Module 3
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@triumphantox yep, it is!
@victorioussheep that's definitely true! Also, if you take a look at those terms:
11, 121, 1331, 14641...
do you see anything interesting how you get from one term to another? (Hint: what is, say, 121 * 11? What is 1331*11?)
And remember, we're trying to show that all of these are divisible by 11... -
@audrey Ooo! So it is like 11*(Ex.1331, 14641, ...)11 times any number=the next number?
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@victorioussheep Yes, exactly--nice job! So notice that because the next term is the previous term times 11, then as long as the first term (11) is divisible by 11, then all of them are, like you said!
Of course, we could also prove it more rigorously. A hint on how to do that: say you have some number ABA. What do you get when you "literally" multiply ABA by 11, assuming that no carrying happens? (in other words, treat ABA like a normal number, like 171)
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@audrey oh...so it like ABA · 11 = A(B+1)(B+1)A?
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@victorioussheep That's definitely true, if A = 1! More generally, though, look what happens if we just try to do multiplication the "normal" way (again, as long as no carrying happens):
But if A B A is one row of Pascal's triangle, the next row is exactly A (A+B) (A+B) A, based on the definition of Pascal's triangle!! And we can show the exact same thing for a number AA (11 in this case), or a number ABBA (1331). So basically, we've proved that
- the second row of Pascal's triangle, seen as a "number," is divisible by 11
- going from one row to the next (as long as no carrying happens, as in 14641 to 1 5 10 10 5 1) is just multiplication by 11
So your observation can be proved to be true! (By the way, the official term for this sort of thing, where you prove that a condition is true for one term, and then prove that if the condition is true for one term, it is true for the next, is called induction) Again, very interesting observation!
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@audrey Wow...I didn't even think of A (A+B) (A+B) A. Thank you for the comments and observations!
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@victorioussheep No problem, it was a very intriguing question!
