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, user91500 Mar 16 '16 at 4:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – choco_addicted, colormegone, Harish Chandra Rajpoot, Community, user91500
For middle digit n=9 (odd number) you have (n-1)/2 = 4 pairs and 1 single.
For middle digit n=8 (even number) you have (n-2)/2 = 3 pairs and 2 singles.
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