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How many $3$-digit positive integers are there whose middle digit is equal to the sum of the first and last digits?

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closed as off-topic by choco_addicted, colormegone, Harish Chandra Rajpoot, user228113, user91500 Mar 16 '16 at 4:36

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  • $\begingroup$ Does $011$ count? $\endgroup$ – Henry Apr 2 '15 at 21:10
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    $\begingroup$ 9 that begin with 1, 8 that begin with 2, 7 that begin with 3... $\endgroup$ – Mike Apr 2 '15 at 21:14
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For middle digit n=9 (odd number) you have (n-1)/2 = 4 pairs and 1 single.
198, 891
297, 792
396, 693
495, 594
990

For middle digit n=8 (even number) you have (n-2)/2 = 3 pairs and 2 singles.
187, 781
286, 682
385, 583
484
880

Try to generalize these 2 cases, and you'll get the answer.
Also, be careful with the smallest ones: middle digits: 3,2,1.
I mean, for them the above observations/formulas may not hold.
So check them one by one by hand.

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the answer is 45. we should start with the middle digit of 9, then 8, then 7,.....the last will be 1. If the middle digit is 9, we will have 9 numbers: 198,891,297,792,396,693,495,594, and 990. The number of middle digit is 8, we will have 8 numbers, and so on. so the answer will be 9+8+7+6+5+3+2+1=45

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  • $\begingroup$ This question has a well-accepted answer. You are not contributing anything new. Please refrain from answering such old questions. There are many new unanswered questions $\endgroup$ – Shailesh Mar 15 '16 at 7:24

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