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Difficulty: HardNumber Properties and Integer Constraints in Data Sufficiency

If nn is a real number, is nn an integer?

(1) n2+5nn^2 + 5n is an integer.

(2) n2−nn^2 - n is an integer.

  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. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.Answer
  4. D
    EACH statement ALONE is sufficient.
  5. E
    Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

Both statements together are sufficient, but neither statement alone is sufficient.
Evaluating each statement alone shows that non-integer real numbers can produce integer values for n2+5nn^2 + 5n or n2−nn^2 - n. However, combining both statements gives 6n=(n2+5n)−(n2−n)6n = (n^2 + 5n) - (n^2 - n), proving n=k6n = \frac{k}{6} for some integer kk. Substituting n=k6n = \frac{k}{6} back into n2−nn^2 - n requires k2−6k36\frac{k^2 - 6k}{36} to be an integer, which forces k(k−6)k(k-6) to be divisible by 36. Examining remainders modulo 6 shows kk must be a multiple of 6, ensuring nn is an integer.

Step-by-Step Solution

1
Evaluate Statement (1) independently.
Statement (1) is NOT sufficient.
If n2+5n=pn^2 + 5n = p where pp is an integer, nn can be an integer (e.g., n=1  ⟹  n2+5n=6n=1 \implies n^2+5n=6) or a non-integer real number (e.g., n2+5n−1=0  ⟹  n=−5+292n^2+5n-1=0 \implies n = \frac{-5 + \sqrt{29}}{2}, which gives n2+5n=1n^2+5n=1).
2
Evaluate Statement (2) independently.
Statement (2) is NOT sufficient.
If n2−n=qn^2 - n = q where qq is an integer, nn can be an integer (e.g., n=2  ⟹  n2−n=2n=2 \implies n^2-n=2) or a non-integer real number (e.g., n2−n−1=0  ⟹  n=1+52n^2-n-1=0 \implies n = \frac{1 + \sqrt{5}}{2}, which gives n2−n=1n^2-n=1).
3
Combine Statement (1) and Statement (2).
6n6n is equal to an integer k=p−qk = p - q, so n=k6n = \frac{k}{6}.
Subtracting (n2−n=q)(n^2 - n = q) from (n2+5n=p)(n^2 + 5n = p) yields 6n=p−q6n = p - q. Since pp and qq are integers, k=p−qk = p - q must be an integer.
4
Substitute n=k6n = \frac{k}{6} back into Statement (2) to check integer constraints on kk.
kk must be a multiple of 6, which implies n=k6n = \frac{k}{6} is an integer.
Substituting n=k6n = \frac{k}{6} into n2−n=qn^2 - n = q gives (k6)2−k6=k2−6k36=q\left(\frac{k}{6}\right)^2 - \frac{k}{6} = \frac{k^2 - 6k}{36} = q. Thus, 3636 must divide k(k−6)k(k-6). Consequently, 66 divides k(k−6)k(k-6). Since kk and k−6k-6 have the same remainder rr modulo 6, k(k−6)≡r2(mod6)k(k-6) \equiv r^2 \pmod 6. For r2r^2 to be divisible by 6 where r∈{0,1,2,3,4,5}r \in \{0, 1, 2, 3, 4, 5\}, rr must be 00. Hence, kk is a multiple of 6, making nn an integer.

Key Concept

Number properties of non-integer real variables vs. integer constraints when combining polynomial equations.
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Topics

Number Properties and Integer Constraints in Data Sufficiency

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