2009 CEMC Gauss (Grade 8) Problems/Problem 13

Problem

In how many ways can $101$ can be expressed as the sum of two integers, both greater than zero, with the second integer greater than the first?

$\text{ (A) }\ 50 \qquad\text{ (B) }\ 51 \qquad\text{ (C) }\ 101 \qquad\text{ (D) }\ 102 \qquad\text{ (E) }\ 25$

Solution 1

We want to look for a pair of integers $(a, b)$ such that $a + b = 101$ , and $0 < a < b$ .

We can notice that $101 = 50 + 51$ . This shows us that $a$ can be any positive integer from $1$ to $50$ , which gives $\boxed {\textbf {(A) } 50}$ ways.

~anabel.disher

Solution 2 (unrecommended)

We can list all of the possible combinations from $(1, 100), (2, 99), ..., (50, 51)$ , and count them to get the number of ways that $10$ can be expressed as the sum of two integers, which is $\boxed {\textbf {(A) } 50}$ . However, this takes a long time, and is unrecommended.

~anabel.disher

Solution 3 (formula)

We can use the fact that the number of positive integer pairs must be $\lfloor \frac{n - 1}{2} \rfloor$ , where $\lfloor x \rfloor$ is the floor function of the number $x$ , and $n$ is the number given in the problem.

Using this, we can plug in $101$ for $n$ to get:

$\lfloor \frac{101 - 1}{2} \rfloor = \lfloor \frac{100}{2} \rfloor$

$=\lfloor 50 \rfloor =\boxed {\textbf {(A) } 50}$

~anabel.disher

2009 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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CEMC Gauss (Grade 8)

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2009 CEMC Gauss (Grade 8) Problems/Problem 13

Contents

Problem

Solution 1

Solution 2 (unrecommended)

Solution 3 (formula)