A professor writes N consecutive natural numbers, beginning with 1, on the blackboard. One of the students in the class deletes one of the numbers (exactly one number), from that list.
Now, given that the average of the remaining N-1 numbers is 271/16.
Can you find out the number that has been deleted from the list ?

19 was removed from the first 33 counting numbers.

The average of a set of numbers is their sum divided by the number of terms.

The average of the first n counting numbers is simply (n+1)/2; their sum is n(n+1)/2.
With a little manipulation, this becomes n²/2 + &n/2.
For the purposes of this problem, the average can also be rewritten as (n²/2 + n/2)/n.

If a some term b is removed from the set, the sum is reduced by the value of b, and the total number of terms is reduced by one. The average now will be (n²/2 + n/2 - b)/(n-1).

We are given that this is equal to 271/16:
(n²/2 + n/2 - b)/(n-1) = 271/16
16(n²/2 + n/2 - b) = 271(n-1)
8n² + 8n - 16b = 271n - 271
8n² - 263n +271 - 16b = 0

There are an infinite number of solutions to this, now, except that we know that n and b are both integers. Furthermore, given the form of the average, n is one more than a multiple of 16.

Solving for b and substituting (16i+1) for n, where i is some positive integer, we get:
16b = 8(16i+1)² - 263(16i+1) +271
16b = 8(256i² + 32i + 1) - 263(16i) - 263 + 271
16b = 8(256)i² + 8(32)i - 263(16)i +16
b = 128i² - 247i + 1

Now, we can just plug in values for i until we get a result for b that is a positive integer 1<b
Trying i=1, we get n=17 and b=-118. In other words, for this to work, we need to 118 to the sum to get the desired average. Obviously, not what we are looking for.

If i=2, we get n=33 and b=19. This fits our criteria, and indeed 1+2+3+...+17+18+20+21+...+31+32+33=542, which when divided by the number of remaining terms, 32, reduces to 271/16.

Let's try i=3, just to see if there are any more possibilities. This yields n=49, while b=412. That means you would have to subtract 412 from the sum of the first 49 terms to obtain the desired average. Making i and larger will make b even larger than n, so we have found the only solution.

**
Another method occured to me as I was checking the answer to this.

The formula for the sum of the counting numbers from a to b is just
(b-a)(b+a)/2.

If we have the first n counting numbers, and some term b is removed, then we have two such series, one from 1 to b-1, and one from b+1 to n.

The total sum, then, will be: [(b-1-1)(b-1+1) + (n-(b+1))(n+b+1)]/2.

Multiplying this out and canceling terms yields the sum as [n²-4b-1]/2.

The average of the remaining n-1 terms, then, will be [n²-4b-1]/2.

This is equal to 271/16: [n²-4b-1]/2 = 271/16.
I see that this will yield the same second-order equation as found by the previous method. Oh well, it looked like a promising way to a simpler method..
Edited on August 29, 2003, 12:04 am

Posted by DJ on 2003-08-27 16:39:21