2.4 Parity, Signs & Consecutive Integer Properties

Parity is even versus odd. An even integer has the form 2k; an odd integer has the form 2k + 1. Zero is even because 0 = 2 × 0, and zero is neither positive nor negative. Negative integers have parity too: −4 is even and −3 is odd.

GMAT Problem Solving uses parity to eliminate answer choices without computing a full value, and uses consecutive-integer structure to convert a sum into n times a mean. Official Guide Math Review 3.1 groups even/odd with order and signs; this section is the working toolkit for those items.

Parity of sums and products

Product rule: a product of integers is odd if and only if every factor is odd. One even factor makes the whole product even.

Sum rule: even addends never change parity. The sum is odd exactly when the number of odd addends is odd.

Worked: is 15 × 22 × 37 even or odd? 22 is even, so the product is even — you never need 15 × 37.

Worked: 2^n is even for every integer n ≥ 1. 2^0 = 1 is odd — a real trap if the exponent is allowed to be 0.

Worked: the square of an integer has the same parity as the integer itself, because even^2 is even and odd^2 is odd. That is why a^2 b has the same parity as a b.

Signs

• Positive × positive = positive
• Negative × negative = positive
• Opposite signs = negative

A product of non-zero integers is negative exactly when the number of negative factors is odd.

For a non-zero x:

• (−x) raised to an even power is positive
• (−x) raised to an odd power is negative

Any real square satisfies x^2 ≥ 0, with equality only at x = 0.

On GMAT Quant, “negative even integer” means −2, −4, −6, … — both sign and parity are constraints. “Non-negative” includes 0; “positive” does not. “Least possible integer value” of an even square is 0, not 2.

Consecutive integers

A run of k consecutive integers is n, n+1, n+2, …, n+(k−1). Among any two consecutive integers, one is even. Among any three consecutive integers, one is a multiple of 3. Among any m consecutive integers, one is a multiple of m.

Product of k consecutive integers is divisible by k!

• Two consecutive integers: product divisible by 2! = 2
• Three consecutive integers: product divisible by 3! = 6
• Four consecutive integers: product divisible by 4! = 24

Among four consecutive integers there is a multiple of 4 and a different multiple of 2, giving a factor of 8, and a multiple of 3, giving 24.

Reason: the binomial coefficient C(n, k) = n(n−1)…(n−k+1) / k! is always an integer, so that falling product is always a multiple of k!.

Worked: if P is the product of four consecutive positive integers, P is a multiple of 24, so P + 1 leaves remainder 1 when divided by 24. You do not need to name the four integers.

Sum = n × mean

The sum of n consecutive integers is

S = n × (first + last) / 2 = n × mean

If n is odd, the mean is the middle integer, so S is a multiple of n.

Example: five consecutive integers centered at x give (x−2)+(x−1)+x+(x+1)+(x+2) = 5x.

If n is even, the mean is halfway between the two middle integers (ending in .5 when the terms are integers), so S is not a multiple of n, though it is a multiple of n/2.

Example: four consecutive integers x + (x+1) + (x+2) + (x+3) = 4x + 6 = 2(2x+3). Divisible by 2, never by 4, because 2x+3 is odd.

Worked: the sum of 7 consecutive integers is 161. Then 7 × mean = 161, so the mean (middle term) is 23. The integers are 20, 21, 22, 23, 24, 25, 26. Smallest × largest = 20 × 26 = 520. First + last = 2 × mean = 46, another quick check.

Consecutive even, consecutive odd, consecutive multiples

Consecutive even integers 2k, 2k+2, 2k+4, … are 2 times consecutive integers, so their sum is twice a consecutive-integer sum. An odd count of consecutive even integers has even middle term; the sum is that middle term times the count.

Consecutive odd integers 2k−1, 2k+1, 2k+3, … still average to the middle integer when the count is odd. Three consecutive odds centered at 2k+1 sum to 3(2k+1).

Consecutive multiples of m form an arithmetic sequence with difference m. Counting them in a range uses floor(B/m) − floor((A−1)/m), the same consecutive-multiples count as in the GCF/LCM section. Trap: “how many consecutive multiples of 6 are in 1 through 100” is 16, not 17, and they are not themselves “six consecutive integers.”

What sums are possible?

Three consecutive integers can sum to any multiple of 3. Four consecutive integers can never sum to a multiple of 4. This is a clean “which of the following CANNOT be” template: if the count n is odd, n itself can be a sum of n consecutive integers (the run centered at 1); if n is even, n cannot be a sum of n consecutive integers.

More generally, n consecutive integers summing to S means S = n × mean. For integer terms and odd n, mean is an integer. For even n, mean is a half-integer (k + 1/2), so S = n(k + 1/2) = n k + n/2.

Smart-number checks

When an algebra stem asks only for parity or a remainder, substitute small integers that match the constraints: 0 (even), 1 (odd), 2 (even), −1 (odd). For “must be true,” one counterexample kills an option. For “could be true,” one working example keeps it.

Worked: a^2 b + b^2 c + c^2 a is odd. Squares keep parity, so this has the same parity as ab + bc + ca.

• All even → even + even + even = even
• Two even, one odd → each product includes an even factor → even
• Two odd, one even (say a even, b and c odd) → ab even, bc odd, ca even → odd
• All odd → odd + odd + odd = odd

So the expression is odd precisely when at least two of a, b, and c are odd. “All three are odd” is sufficient but not necessary, so it is not a must-be-true statement. The sum a+b+c is even when two are odd and one is even, and odd when all three are odd, so that sum is not forced to one parity. The only must-be-true claim among typical choices is that at least two are odd.

Consecutive-integer checklist

Remember the three product identities the GMAT likes: even × even = even, even × odd = even, odd × odd = odd. Combined with “the product of k consecutive integers is a multiple of k!,” you can often read a remainder without expanding the product. Combined with S = n × mean, you can often name the whole run from one sum.