Question

Difficulty: MediumNumber Properties and Integer Constraints in Data Sufficiency

If rr and ss are positive integers, is r2s2r^2 - s^2 divisible by 8?

(1) rr and ss are both odd integers.
(2) r+sr + s is divisible by 4.

  1. A
    Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
  2. B
    Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
  3. C
    BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
  4. EACH statement ALONE is sufficient.Answer
  5. E
    Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

Each statement alone is sufficient to answer the question with a definitive 'Yes'.
The correct response identifies that each statement alone gives a definitive 'Yes' answer. Statement (1) shows that the square of any odd integer is 1(mod8)1 \pmod 8, so subtracting two odd squares gives 0(mod8)0 \pmod 8. Statement (2) shows r+sr+s is a multiple of 4, which forces rr and ss to share parity, making rsr-s even and their product (r+s)(rs)(r+s)(r-s) a multiple of 4×2=84 \times 2 = 8.

Step-by-Step Solution

1
Rephrase the question target using algebraic factorization.
Note that r2s2=(r+s)(rs)r^2 - s^2 = (r + s)(r - s). The question asks whether (r+s)(rs)(r + s)(r - s) is a multiple of 8.
Factoring difference of squares isolates the sum and difference components to easily test parity and divisibility.
2
Evaluate Statement (1): rr and ss are both odd integers.
Any odd integer can be written as 2k+12k + 1. Squaring gives (2k+1)2=4k2+4k+1=4k(k+1)+1(2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k + 1) + 1. Since k(k+1)k(k + 1) is the product of two consecutive integers, it is always even, making 4k(k+1)4k(k + 1) a multiple of 8. Thus, any odd integer squared leaves a remainder of 1 when divided by 8. Therefore, r21(mod8)r^2 \equiv 1 \pmod 8 and s21(mod8)s^2 \equiv 1 \pmod 8, so r2s211=0(mod8)r^2 - s^2 \equiv 1 - 1 = 0 \pmod 8.
Statement (1) alone yields a definitive 'Yes', so Statement (1) is sufficient.
3
Evaluate Statement (2): r+sr + s is divisible by 4.
If r+sr + s is divisible by 4, then r+s=4mr + s = 4m for some integer mm. Since r+sr + s is even, rr and ss must have the same parity (either both even or both odd). Consequently, their difference rs=(r+s)2s=4m2s=2(2ms)r - s = (r + s) - 2s = 4m - 2s = 2(2m - s) is an even integer (a multiple of 2). Multiplying (r+s)(r + s) by (rs)(r - s) yields (multiple of 4)×(multiple of 2)=multiple of 8(\text{multiple of } 4) \times (\text{multiple of } 2) = \text{multiple of } 8.
Statement (2) alone yields a definitive 'Yes', so Statement (2) is sufficient.

Key Concept

Divisibility Rules of Difference of Squares and Parity of Integers
Estimated Time:2m 0s
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