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Difficulty: HardNumber Properties and Integer Constraints in Data Sufficiency

If nn is a positive integer, is nn a multiple of 18?

(1) n2n^2 is divisible by 108.
(2) n3n^3 is divisible by 243.

  1. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.Answer
  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. D
    EACH statement ALONE is sufficient.
  5. E
    Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (1) alone is sufficient because prime factor analysis of n2n^2 being divisible by 108=22×33108 = 2^2 \times 3^3 requires the exponent of 3 in n2n^2 to be at least 4 (since nn is an integer and exponents in n2n^2 must be even). This guarantees nn is divisible by 21×32=182^1 \times 3^2 = 18. Statement (2) alone is not sufficient because n3n^3 being divisible by 243=35243 = 3^5 only guarantees nn is a multiple of 9, allowing n=9n = 9 (not a multiple of 18) or n=18n = 18 (a multiple of 18).

Step-by-Step Solution

1
Rephrase the target question using prime factorizations.
The target asks if nn is a multiple of 18=21×3218 = 2^1 \times 3^2. For nn to be a multiple of 18, its prime factorization must contain at least one factor of 2 and at least two factors of 3.
Decomposing into prime factors simplifies the condition required for sufficiency.
2
Evaluate Statement (1): n2n^2 is divisible by 108=22×33108 = 2^2 \times 3^3.
Since nn is a positive integer, any prime factor in the prime factorization of n2n^2 must have an even exponent. For n2n^2 to be divisible by 22×332^2 \times 3^3, the power of 2 in n2n^2 must be at least 2, so nn has at least one factor of 2. The power of 3 in n2n^2 must be an even integer greater than or equal to 3, meaning it must be at least 4. Thus, nn must contain at least two factors of 3 (32=93^2 = 9). Consequently, nn must be divisible by 21×32=182^1 \times 3^2 = 18. Statement (1) alone yields a definitive 'Yes'.
An integer's square must have even exponents for all its prime factors.
3
Evaluate Statement (2): n3n^3 is divisible by 243=35243 = 3^5.
For n3n^3 to be divisible by 353^5, the exponent of 3 in n3n^3 must be a multiple of 3 greater than or equal to 5, which means it must be at least 6. Therefore, nn must contain at least two factors of 3 (32=93^2 = 9). However, statement (2) places no restriction on whether nn is even. If n=9n = 9, n3=729=243×3n^3 = 729 = 243 \times 3 (divisible by 243), but 9 is NOT a multiple of 18. If n=18n = 18, n3=5832=243×24n^3 = 5832 = 243 \times 24 (divisible by 243), and 18 IS a multiple of 18. Since nn can yield both 'No' and 'Yes', Statement (2) alone is insufficient.
Testing specific values demonstrates that nn may or may not be even.

Key Concept

Prime Factor Exponents in Integer Powers and Data Sufficiency Sufficiency Logic
Estimated Time:2m 0s
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Topics

Number Properties and Integer Constraints in Data Sufficiency

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