Summer Problem Solving Marathon Solution #33

For any choice of n, we can write 2ⁿ in the form nk + m, for some k and some m < n. (That is, we write 2ⁿ as some multiple of n, plus a remainder on division by n.)

We then rewrite 2⁽²ⁿ⁾ as 2^{nk + m} = 2^nk2^m.

We are now interested in the remainder of this when divided by 2ⁿ - 1. Since m < n, the remainder of 2^m when divided by 2ⁿ -1 is 2^m. 2^nk can be rewritten as (2ⁿ)^k. But the remainder of 2ⁿ when divided by 2ⁿ -1 is clearly 1, so the remainder of (2ⁿ)^k when divided by 2ⁿ -1 is 1^k = 1.

Thus the remainder of 2⁽²ⁿ⁾ when divided by 2ⁿ -1 is 2^m. We want the remainder not to be a power of 4, which means we want m to be odd.

We are thus looking for n such that 2ⁿ, when divided by n, produces an odd remainder. Examination of cases then shows that 25 is the first value of n that works.

Summer Problem Solving Marathon Question #36

[Value = 3 points]

How many integers n, 100 < n < 200, have the same remainder when divided by 6 and when divided by 8?

Summer Problem Solving Marathon Question #35

[Value = 3 points]

The interior of a right circular cone is 8 inches tall with a 2 inch radius at the opening. The interior of the cone is filled with ice cream, and the cone has a hemisphere of ice cream exactly covering the opening of the cone. What is the volume of the ice cream?

Summer Problem Solving Marathon Solution #32

Given any S meeting the conditions, let s be the number of elements in S, and let N be the sum of all the elements in S.

Given any n∈S, the mean of the elements in S other than n is an integer. There are s-1 other elements in S, and their sum is N - n. So s-1 is a factor of N-n. Thus N and n are congruent mod s-1 (that is, they have the same remainder when divided by s-1).

Since 1∈S, N and 1 are congruent mod s-1, and in fact all elements of S are congruent to 1 mod s-1. Since 2002∈S, we know that 2002 is congruent to 1 mod s-1, so 2001 is a multiple of s-1. Since 2001 = 3 x 23 x 29, the possible values for s are 4, 24, 30, 70, 88, 668, and 2002.

If we list the numbers that are congruent to 1 mod s-1, we get 1, s, 2s-1, 3s-2, ... . The s^th such number is 1 + (s-1)(s-1) = s² - 2s + 2. So the largest element of S must be of size at least s² - 2s + 2. Since the largest element is 2002, this rules out 70, 88, 668, and 2002 as values of s.

Thus the largest possible value of s is 30. It remains to see that this value of s can be made to work. We will need a set of 30 elements, each of which is congruent to 1 mod 29. There are many sets meeting this constraint, including {1, 30, 59, 88, ..., 813, 2002}.

Summer Problem Solving Marathon Question #34

[Value = 2 points]

A right triangle similar to a 3:4:5 right triangle is inscribed in a circle of radius 7. What is the area of the triangle?

Summer Problem Solving Marathon Question #33

[Value = 10 points]

Let N be

(a) the smallest integer n > 1 such that when 2⁽²ⁿ⁾ is divided by 2ⁿ-1, the remainder is not a power of 4

or

(b) 0, if there is no such n.

What is N?

Summer Problem Solving Marathon Question #32

[Value = 9 points]

Let S be a set of positive integers, such that 1 ∈ S, 2002 ∈ S, and for all n ∈ S, n < 2003.

Suppose S has the following feature: for any n ∈ S, the mean of the elements of S - {n} is an integer.

What is the greatest number of elements S could have?