GMAT Logic: Level 1

Solution (B): The 10 consecutive integers are: -5, -4, -3, -2, -1, 0, 1, 2, 3, 4. The positive integers in the list are {1, 2, 3, 4}. Their sum is \(1 + 2 + 3 + 4 = 10\).

Solution (D): Let the integers be \(n, n+1, n+2, n+3\). Sum \(x = 4n + 6 = 2(2n + 3)\). Since \(2n+3\) is always odd, the sum \(x\) is 2 times an odd number. This is always an even number, but it is never divisible by 4. Example: \(1+2+3+4 = 10\). \(2+3+4+5 = 14\).

Solution (A): 1. \(ab > 0\) means \(a\) and \(b\) have the same sign (both positive or both negative). 2. \(bc < 0\) means \(b\) and \(c\) have opposite signs. Case 1: If \(a, b\) are (+), then \(c\) must be (-). This makes \(a\) (+) and \(c\) (-). Case 2: If \(a, b\) are (-), then \(c\) must be (+). This makes \(a\) (-) and \(c\) (+). In both cases, \(a\) and \(c\) have opposite signs, so their product \(ac\) must be negative.

Solution (B): 1. Find the prime factorization of 72: \(72 = 8 \times 9 = 2^3 \times 3^2\). 2. The expression is \( (2^3 \times 3^2) \times n \). 3. For a number to be a perfect square, all exponents in its prime factorization must be even. 4. We have \(3^2\) (even) but \(2^3\) (odd). We need one more factor of 2 to make the exponent \(2^4\). 5. Therefore, the smallest positive integer \(n\) is 2.

Solution (E): 1. Find the prime factorization of 440: \(440 = 44 \times 10 = (4 \times 11) \times (2 \times 5) = (2^2 \times 11^1) \times (2^1 \times 5^1) = 2^3 \times 5^1 \times 11^1\). 2. To make all exponents even, we need one more 2 (to get \(2^4\)), one more 5 (to get \(5^2\)), and one more 11 (to get \(11^2\)). 3. The smallest integer \(y\) must be \(2 \times 5 \times 11 = 110\).

Solution (C): 1. Find the prime factorization of 54: \(54 = 2 \times 27 = 2^1 \times 3^3\). 2. The expression is \((2^1 \times 3^3) \times x = n^3\). 3. For a number to be a perfect cube, all exponents must be multiples of 3. 4. We have \(3^3\) (multiple of 3) but \(2^1\). We need \(2^3\). 5. We must multiply by \(2^2\) to get \(2^1 \times 2^2 = 2^3\). 6. Therefore, the smallest \(x\) is \(2^2 = 4\).

Solution (E): A) \( \text{Even} + \text{Odd} + \text{Odd} = \text{Even} + \text{Even} = \text{Even} \) B) \( \text{Even} \times \text{Odd} = \text{Even} \) C) \( \text{Even} + 2(\text{Odd}) = \text{Even} + \text{Even} = \text{Even} \) D) \( \text{Odd} + 1 = \text{Even} \) E) \( \text{Even} + \text{Odd} = \text{Odd} \). This must be true.

Solution (C): Let the integers be \(n, n+1, n+2\). Sum: \(n + (n+1) + (n+2) = 18\) \(3n + 3 = 18 \implies 3n = 15 \implies n = 5\). The integers are 5, 6, and 7. Their product is \(5 \times 6 \times 7 = 30 \times 7 = 210\).

Solution (C): \(x = 10k = 2 \times 5 \times k\). I. Since \(x\) has a factor of 2, it must be divisible by 2. (True) II. Since \(x\) has a factor of 5, it must be divisible by 5. (True) III. If \(k=1\), \(x=10\). 10 is not divisible by 20. (Not must be true) Therefore, only I and II must be true.

Solution (D): We need to test the options. \(a\) = Negative Even (e.g., -2, -4, -6…) \(b\) = Positive Odd (e.g., 1, 3, 5…) The result \(a/b\) must be negative. This eliminates A, B, and E. (C) -2.5: To get -2.5, we could have \(-5 / 2\). Here, \(a=-5\), which is not even. (D) -4: To get -4, we could have \(a = -4\) and \(b = 1\). This fits the conditions.