Five numbers are selected and deleted from a set of consecutive positive integers beginning with 1. The arithmetic mean of the remaining numbers in the set is 45.1.

Find the largest possible single number that could have been deleted from the original set.

Let the set of consecutive positive integers be 1, 2, ...., (p+5).

Then, by conditions of the problem, the sum of all the numbers of the set after the deletion would be 45.1*p = M(say)

Thus, 45.1*p must correspond to a whole number.

Now,

Max M = (1+2+ ....+(p+5)) - (1+2+3+4+5)

= p(p+11)/2

and, Min M = (1+2+ ....+(p+5)) - (p+1+p+2+p+3+p+4+p+5)

= (1+2+....+ p)

= p(p+1)/2

Now, Min M <= 45.1*p = M <= Max M

Or, p(p+1)/2 <= 45.1*p <= p(p+11)/2

Or, 79.2<= p<= 89.2.......(#)

We recall that 45.1*p is an integer. The only value of p for which this is possible occurs at p= 80.......(##)

Let the maximum number which was actually deleted from the set be x.

This is only possible if the remaining four numbers which were deleted corresponds to 1,2,3,4.

Thus, we have:

(p+5)(p+6)/2 -(10+x) = 45.1*p

Or, 85*86/2 - (10+x) = 45.1*80 [Since by (##), p=80]

Or, 3645 - x = 3608

Or, x = 37.

Consequently, the maximum amongst the numbers that could have been deleted is 37.

*Edited on ***May 21, 2007, 11:16 am**