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Difficulty: HardNumber Properties and Integer Constraints in Data Sufficiency

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

(1) n2−3nn^2 - 3n is an integer.
(2) n2+5nn^2 + 5n 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 to answer definitively that nn is an integer, but neither statement alone is sufficient.
Neither statement alone is sufficient because quadratic equations of the form n2−3n=kn^2 - 3n = k and n2+5n=mn^2 + 5n = m have non-integer real roots for certain integer values of kk and mm. When both statements are combined, subtracting the two equations reveals that 8n8n is equal to an integer pp, so n=p8n = \frac{p}{8}. Substituting n=p8n = \frac{p}{8} into n2−3nn^2 - 3n requires p(p−24)64\frac{p(p-24)}{64} to be an integer. Examining the power of 2 in p(p−24)p(p-24) proves that pp must be divisible by 8, which guarantees that nn is an integer. Thus, the two statements together provide a definitive 'Yes' answer.

Step-by-Step Solution

1
Evaluate Statement (1) independently.
Statement (1) is NOT sufficient.
If n2−3n=−1n^2 - 3n = -1 (an integer), then n2−3n+1=0n^2 - 3n + 1 = 0, giving n=3+52n = \frac{3 + \sqrt{5}}{2}, which is not an integer. However, if n=3n = 3, n2−3n=0n^2 - 3n = 0, which is an integer. Thus, nn can be an integer or a non-integer.
2
Evaluate Statement (2) independently.
Statement (2) is NOT sufficient.
If n2+5n=1n^2 + 5n = 1 (an integer), then n2+5n−1=0n^2 + 5n - 1 = 0, giving n=−5+292n = \frac{-5 + \sqrt{29}}{2}, which is not an integer. However, if n=1n = 1, n2+5n=6n^2 + 5n = 6, which is an integer. Thus, nn can be an integer or a non-integer.
3
Combine Statement (1) and Statement (2).
Deduce that 8n8n must be an integer.
Let n2−3n=kn^2 - 3n = k and n2+5n=mn^2 + 5n = m, where kk and mm are integers. Subtracting the first equation from the second yields (n2+5n)−(n2−3n)=m−k  ⟹  8n=m−k(n^2 + 5n) - (n^2 - 3n) = m - k \implies 8n = m - k. Since m−km - k is an integer, let 8n=p8n = p for some integer pp. Hence, n=p8n = \frac{p}{8}.
4
Analyze whether n=p8n = \frac{p}{8} can be a non-integer fraction.
nn MUST be an integer, making BOTH statements TOGETHER sufficient.
Substitute n=p8n = \frac{p}{8} back into Statement (1): n2−3n=p264−3p8=p2−24p64n^2 - 3n = \frac{p^2}{64} - \frac{3p}{8} = \frac{p^2 - 24p}{64}. For this value to be an integer, p2−24p=p(p−24)p^2 - 24p = p(p - 24) must be divisible by 64=2664 = 2^6. If pp is odd, p−24p - 24 is odd, so p(p−24)p(p-24) is odd and not divisible by 64. If pp is a multiple of 2 but not 4 (p=2bp = 2b with bb odd), p(p−24)=4b(b−12)p(p-24) = 4b(b-12), containing only 222^2 factors of 2. If pp is a multiple of 4 but not 8 (p=4bp = 4b with bb odd), p(p−24)=16b(b−6)p(p-24) = 16b(b-6), containing only 242^4 factors of 2. Thus, for p(p−24)p(p-24) to contain 262^6, pp must be a multiple of 8. If pp is a multiple of 8, then n=p8n = \frac{p}{8} must be an integer.

Key Concept

Evaluating integer constraints vs. real numbers in Data Sufficiency through algebraic elimination and prime factorization of rational expressions.
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Topics

Number Properties and Integer Constraints in Data Sufficiency

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