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

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

## closed as off-topic by choco_addicted, colormegone, Harish Chandra Rajpoot, user228113, user91500Mar 16 '16 at 4:36

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• Does $011$ count? – Henry Apr 2 '15 at 21:10
• 9 that begin with 1, 8 that begin with 2, 7 that begin with 3... – Mike Apr 2 '15 at 21:14

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.

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

• 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 – Shailesh Mar 15 '16 at 7:24