A claim says n^2 - n = 0 for every n in {1, 2, 3, 4, 5}. Check inputs in increasing order. What is the first input where the claim fails?

For integers n and m, an odd-number sum becomes 2(n + m + 1). Which fact permits the conclusion that the sum is even?

• The sum n + m + 1 is an integer.
• n + m + 1 is always positive.
• n and m must be equal integers.
• n + m + 1 is always an odd integer.

Which expression always gives an odd number when n is an integer?

• 2n + 1
• 2n + 2
• 2n + 4

Mo confirms that a pattern holds at n = 1, 2 and 3. Mo says these checks alone prove it for all integers n. Is that claim justified?

The inequality n^2 < 10 holds at n = 1, 2 and 3. Someone claims it holds for every positive integer. What is the smallest positive integer counterexample?

To prove a claim about the integers 1 to 5 by exhaustion, you must do what?

• Check only inputs 1 and 5.
• Check a few integers and predict the others.
• Check and verify all five integers.

Ari obtains 2n + 2m + 2 when adding two odd forms, then factors it as 2(n + m + 2). Which repair makes this step correct?

• Change the outside factor from 2 to 1.
• Change the inside constant from 2 to 1.
• Change both variable coefficients from 1 to 2.
• Remove the constant from inside the brackets.

Lee checks n^2 + n for n = 1 to 5 and lists 2, 6, 12, 18, 30. At n = 4, Lee uses 4^2 = 14, then adds 4. Since every listed result is even, this is a valid exhaustion proof.

The calculation (2n + 1) + (2n + 1) = 2(2n + 1), for integer n, covers only equal odd integers. It is insufficient by itself to prove that any two odd integers have an even sum.

For integer n, n^2 + n = n(n + 1). If n is even, n is an even factor. If n is odd, n + 1 is an even factor. Why does this prove evenness for all integers?

• Two examples settle all integers.
• Checking inputs 1 to 5 establishes all other inputs.
• Exhaustion calculates every integer input individually.
• Two general cases cover every integer.