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Zorluk: Çok zorNumber Properties and Integer Constraints in Data Sufficiency

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

(1) x2+xx^2 + x is an integer.
(2) x3+x2x^3 + x^2 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.Cevap
  4. D
    EACH statement ALONE is sufficient.
  5. E
    Statements (1) and (2) TOGETHER are NOT sufficient.

Cevap

Both statements together are sufficient to determine that xx is an integer, but neither statement alone is sufficient.
Evaluating each statement independently reveals that non-integer real numbers can produce integer values for x2+xx^2 + x or x3+x2x^3 + x^2. However, combining both statements allows us to factor x3+x2x^3 + x^2 as x(x2+x)x(x^2 + x). Setting x2+x=mx^2 + x = m and x3+x2=kx^3 + x^2 = k for integers mm and kk, we find x⋅m=kx \cdot m = k. If m=0m = 0, xx is 0 or -1 (integers). If m≠0m \neq 0, x=kmx = \frac{k}{m} is a rational number. Writing x=abx = \frac{a}{b} in lowest terms and substituting it back into x2+x=mx^2 + x = m shows that bb must divide a2a^2, forcing b=1b = 1. Thus, xx must be an integer, providing a definitive affirmative answer.

Adım Adım Çözüm

1
Evaluate Statement (1) independently.
Statement (1) is INSUFFICIENT.
If x=2x = 2, then x2+x=6x^2 + x = 6, which is an integer (YES). However, if x=−1+52x = \frac{-1 + \sqrt{5}}{2}, then x2+x−1=0  ⟹  x2+x=1x^2 + x - 1 = 0 \implies x^2 + x = 1, which is an integer, but xx is not an integer (NO). Thus, Statement (1) alone is insufficient.
2
Evaluate Statement (2) independently.
Statement (2) is INSUFFICIENT.
If x=2x = 2, then x3+x2=12x^3 + x^2 = 12, an integer (YES). If xx is the real root of x3+x2=1x^3 + x^2 = 1 (where x≈0.755x \approx 0.755), x3+x2x^3 + x^2 is an integer, but xx is not an integer (NO). Thus, Statement (2) alone is insufficient.
3
Evaluate Statements (1) and (2) together.
Statements (1) and (2) together are SUFFICIENT.
Let x2+x=mx^2 + x = m where m∈Zm \in \mathbb{Z}, and x3+x2=kx^3 + x^2 = k where k∈Zk \in \mathbb{Z}. Notice that x3+x2=x(x2+x)=x⋅m=kx^3 + x^2 = x(x^2 + x) = x \cdot m = k. If m=0m = 0, then x2+x=0  ⟹  x=0x^2 + x = 0 \implies x = 0 or x=−1x = -1, both of which are integers. If m≠0m \neq 0, then x=kmx = \frac{k}{m}, meaning xx must be a rational number. Express x=abx = \frac{a}{b} in lowest terms, where a,b∈Za, b \in \mathbb{Z}, b>0b > 0, and gcd⁡(a,b)=1\gcd(a, b) = 1. Substituting x=abx = \frac{a}{b} into x2+x=mx^2 + x = m yields a2+abb2=m  ⟹  a2+ab=mb2  ⟹  a2=b(mb−a)\frac{a^2 + ab}{b^2} = m \implies a^2 + ab = m b^2 \implies a^2 = b(mb - a). This implies that bb must divide a2a^2. Since gcd⁡(a,b)=1\gcd(a, b) = 1, bb can only divide a2a^2 if b=1b = 1. Therefore, x=a1=ax = \frac{a}{1} = a, which means xx MUST be an integer. The answer to the question is a definitive YES.

Anahtar Kavram

Testing real vs. integer constraints and applying rational root divisibility properties in Data Sufficiency.
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Number Properties and Integer Constraints in Data Sufficiency

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