3 ms·
The eleven numbers are n,n+1,..., n+10. Their sum is 11n + 55. Let the number removed be n + k, then the sum of the remaining is just 10n + 55 -k = 2012. Or 10
by abdullahkhalids 7y ago
The eleven numbers are n,n+1,..., n+10. Their sum is 11n + 55. Let the number removed be n + k, then the sum of the remaining is just 10n + 55 -k = 2012. Or 10n-k = 1957. Then it is obvious that k=3 so that n is an integer. Hence, n=196
The eleven numbers are 196,197,..., 206, out of which 199 was removed.
- chongli 7y agoThat’s a very good solution. Thanks. I began with n being the largest of the sum so that 11n - 55 = 2012 + k. I solved for k in terms of n and then used an inequality to find n which gave me k. Now I see that the method I used relies on math beyond elementary school, which makes the problem a bit more complicated than it ought to be. On the other hand, the elementary school only method relies on the fact that you have 10 numbers remaining and so you can easily see the remainder. It wouldn’t generalize as well, especially in cases where there may be multiple solutions and you’re asked to find all of them.