Parity (Even/Odd) Math Example 5

Follow the full solution, then compare it with the other examples linked below.

medium

In a group of

50

people, each person shakes hands with every other person exactly once. Is the total number of handshakes odd or even? (Hint: use the formula

\frac{n(n-1)}{2}

1
Total handshakes $= \dfrac{50 \times 49}{2} = \dfrac{2450}{2} = 1225$ .
2
Parity: $50$ is even, $49$ is odd, product $50 \times 49 = 2450$ is even; divided by $2$ gives $1225$ , which is odd.

Answer

1225

handshakes — an odd number.

The handshake formula

\frac{n(n-1)}{2}

counts combinations of

2

from

n

people. For

n=50

: one of

n

n-1

is always even (consecutive integers), so the product

n(n-1)

is divisible by

2

, giving an integer — whose parity we then check.

About Parity (Even/Odd)

The classification of integers as even (evenly divisible by 2, with no remainder) or odd (not divisible by 2).

Without fully computing, determine the parity (odd or even) of [formula] and of [formula].

Prove that the sum of any two consecutive integers is always odd.

Prove that the sum of two odd numbers is always even.

Classify each as odd or even without fully computing: (a) [formula], (b) [formula], (c) [formula].