# Number of 3 digit numbers in AP or GP

How many three digit numbers have the property that their digits taken from left to right form an arithmetic or geometric progression?

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1. If this is homework - please add the homework tag. 2. try to write what you have tried so far. –  nbubis May 8 '12 at 9:11
@ nubis I believe in completing my homework myself. I am stuck at this problem and counted the answer as 54 and given answer is 42. –  Arpit Bajpai May 8 '12 at 10:52

Let us count. The list of geometric progressions is short, though not as short as I first believed! There are the obvious $1, 2, 4$, and $2, 4, 8$, and $1, 3, 9$, and their reverses. Then there is the less obvious $4,6,9$ and its reverse, which I had missed. And of course $a, a, a$ where $a$ is any of $1$ to $9$. These last also happen to be arithmetic progressions. What about "common ratio" $0$? We will go along with Wikipedia's definition and not allow that.

Now let's count the arithmetic progressions. There are the $9$ with common difference $0$.

There are $7$ increasing ones with common difference $1$, and $8$ decreasing ones, since $0$ can be the final digit in that case.

There are $5$ increasing ones with common difference $2$, and $6$ decreasing ones.

There are $3$ increasing ones with common difference $3$, and $4$ decreasing ones.

There is $1$ increasing one with common difference $4$, and there are $2$ decreasing ones.

If we decide to forget about the sequences $a, a, a$ we get a count of $44$, for we listed $6$ geometric progressions and $36$ arithmetic progressions. Adding in the sequences $a, a, a$, which we definitely should, since they are indeed in both categories, gives us $53$.

For whatever it is worth, Wikipedia does not allow common ratio $0$. If we accept that, the correct count is $53$.

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You've missed the geometric progressions 469 and 964. But then I don't understand how, starting with 42, and adding in the aaa-numbers, of which there are 9, you get 53. Did you really get the same answer I got, earlier, by making two cancelling mistakes? –  Gerry Myerson May 8 '12 at 13:45
Yes, I did add $42$ and $9$, getting $53$. And did entirely miss common ratios $3/2$, $2/3$. Coffee had not yet kicked in. –  André Nicolas May 8 '12 at 14:21

111 123 135 147 159 210 222 234 246

258 321 333 345 357 369 420 432 444

456 468 531 543 555 567 579 630 642

654 666 678 741 753 765 777 789 840

852 864 876 888 951 963 975 987 999

should be all the arithmetic progressions, that's 45 right there, so 42 can't be right.

124 139 248 421 469 842 931 964 so I get 53. What did I miss?

Are we counting 100 200 ... 900 as geometric progressions with constant ratio zero?

EDIT: For what it's worth, the number of 3-term geometric progressions with entries from $\{{1,2,\dots,n\}}$ is $(6/\pi^2)n\log n+O(n)$. There's a proof in my paper, Trifectas in geometric progression, Austral. Math. Soc. Gaz. 35 (2008) 189–194, available online at http://www.austms.org.au/Publ/Gazette/2008/Jul08/TechPaperMyerson.pdf. In the paper, I leave counting the 3-term arithmetic progressions as an exercise.

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