GMAT Math: Logic & Integers (Level 3)

Solution (E): The number \(n\) is \(p^1 q^1 r^1\). Therefore, \(n^2 = (p^1 q^1 r^1)^2 = p^2 q^2 r^2\). The formula for the number of factors is \((e_1+1)(e_2+1)(e_3+1)\), where \(e\) are the exponents. Number of factors = \((2+1)(2+1)(2+1) = 3 \times 3 \times 3 = 27\).

Solution (B): 1. If \(bc\) is odd, then both \(b\) and \(c\) must be odd. 2. If \(ab\) is even and we know \(b\) is odd, then \(a\) must be even. 3. So: \(a = \text{Even}\), \(b = \text{Odd}\), \(c = \text{Odd}\). 4. Check the options: \(a+c = \text{Even} + \text{Odd} = \text{Odd}\). This must be true.

Solution (C): \(k\) leaves a remainder of 4 when divided by 6. We can write this as \(k = 6n + 4\) for some integer \(n \ge 0\). We need to find the remainder of \(3k\). \(3k = 3(6n + 4) = 18n + 12\). Now, divide \(18n + 12\) by 9. 1. \(18n\) is always divisible by 9, so the remainder is 0. 2. \(12 \div 9\) leaves a remainder of 3. The total remainder is \(0 + 3 = 3\).

Solution (D): If the product is 0, one of the integers must be 0. The sum of 5 consecutive integers is \(5 \times \text{median}\). Since the median of 5 consecutive integers is always the 3rd integer, the median is an integer. Therefore, the sum must be a multiple of 5. Examples: \{-2, -1, 0, 1, 2\}, Sum = 0 (a multiple of 5) \{-4, -3, -2, -1, 0\}, Sum = -10 (a multiple of 5) \{0, 1, 2, 3, 4\}, Sum = 10 (a multiple of 5)

Solution (A): Any prime number \(p > 3\) must be of the form \(6k \pm 1\). Case 1: \(p = 6k + 1\). \(p^2 = (6k+1)^2 = 36k^2 + 12k + 1 = 12(3k^2 + k) + 1\). The remainder is 1. Case 2: \(p = 6k – 1\). \(p^2 = (6k-1)^2 = 36k^2 – 12k + 1 = 12(3k^2 – k) + 1\). The remainder is 1. In both cases, the remainder is 1. (e.g., \(7^2=49 \to 49 \div 12\) rem 1. \(11^2=121 \to 121 \div 12\) rem 1).

Solution (B): If \(a\) and \(b\) are both odd: \(a^2 = (\text{Odd})^2 = \text{Odd}\). \(b^2 = (\text{Odd})^2 = \text{Odd}\). \(c^2 = a^2 + b^2 = \text{Odd} + \text{Odd} = \text{Even}\). This means \(c\) must be even. An odd number squared is of the form \(4k+1\). So \(a^2 = 4m+1\) and \(b^2 = 4n+1\). \(c^2 = (4m+1) + (4n+1) = 4(m+n) + 2\). A perfect square cannot have a remainder of 2 when divided by 4 (it must be 0 or 1). Thus, two odd squares can never sum to a perfect square.

Solution (A): 1. Multiples of 3: Smallest is 102, Largest is 300. \(n = \frac{(300 – 102)}{3} + 1 = \frac{198}{3} + 1 = 66 + 1 = 67\) terms. *Correction:* 102 is 3*34. 300 is 3*100. Count = 100 – 34 + 1 = 67. 2. Multiples of 9: Smallest is 108, Largest is 297. \(n = \frac{(297 – 108)}{9} + 1 = \frac{189}{9} + 1 = 21 + 1 = 22\) terms. 3. Result: (Total multiples of 3) – (Total multiples of 9) = \(67 – 22 = 45\). *Wait, let me re-check Q1.* 300 is inclusive. 100 is inclusive. Multiples of 3: 102, 105, …, 300. Count is \( (300-102)/3 + 1 = 198/3 + 1 = 66+1 = 67 \). Multiples of 9: 108, 117, …, 297. Count is \( (297-108)/9 + 1 = 189/9 + 1 = 21+1 = 22 \). Result = \( 67 – 22 = 45 \). This is option B. *Let me try 100 to 299.* Mult of 3: 102…297. \( (297-102)/3 + 1 = 195/3 + 1 = 65+1 = 66 \). Mult of 9: 108…297. \( (297-108)/9 + 1 = 189/9 + 1 = 21+1 = 22 \). Result = \( 66 – 22 = 44 \). This is option A. The question says “between 100 and 300, inclusive”, so 300 is included. My first calculation (45) is correct.

Solution (B): 1. Find all multiples of 3 from 100 to 300. The smallest is 102 (3×34) and the largest is 300 (3×100). The count is \(100 – 34 + 1 = 67\) numbers. 2. Find all multiples of 9 from 100 to 300. The smallest is 108 (9×12) and the largest is 297 (9×33). The count is \(33 – 12 + 1 = 22\) numbers. 3. Every multiple of 9 is also a multiple of 3. To find the numbers that are multiples of 3 *but not* 9, we subtract: \( (\text{Total multiples of 3}) – (\text{Total multiples of 9}) = 67 – 22 = 45 \).

Solution (A): We use modular arithmetic. We are given \(x \equiv 3 \pmod{8}\). Substitute this into the expression: \(x^2 \equiv 3^2 \equiv 9 \equiv 1 \pmod{8}\) \(5x \equiv 5(3) \equiv 15 \equiv 7 \pmod{8}\) \(6 \equiv 6 \pmod{8}\) Now sum the remainders: \(1 + 7 + 6 = 14\). Find the remainder of the sum: \(14 \div 8\) leaves a remainder of 6.

Solution (C): If \(k+4\) is divisible by 5, then \(k+4 \equiv 0 \pmod{5}\). Subtracting 4 from both sides: \(k \equiv -4 \pmod{5}\). A remainder of -4 is the same as a remainder of 1 (since \(-4 + 5 = 1\)). So, \(k \equiv 1 \pmod{5}\). Now we evaluate \(k^2 + 1\): \(k^2 + 1 \equiv (1)^2 + 1 \equiv 1 + 1 \equiv 2 \pmod{5}\). The remainder is 2.

Solution (E): For the product to be 0, one of the integers must be 0. Since they are all even, 0 must be in the set. We list the possible sets of 4 consecutive even integers containing 0: Set 1: \(-6, -4, -2, 0\). Sum = -12. Set 2: \(-4, -2, 0, 2\). Sum = -4. Set 3: \(-2, 0, 2, 4\). Sum = 4. Set 4: \(0, 2, 4, 6\). Sum = 12. The largest possible sum from these options is 12.