GMAT Logic: Level 3

Solution (D): Let the integers be \(n, n+1, n+2, n+3\). The sum is \(4n + 6\). We are given \(4n + 6\) is a multiple of 6. \(4n + 6 = 6k\) for some integer \(k\). \(4n = 6k – 6 = 6(k-1)\). This means \(4n\) is a multiple of 6. Let’s test the options for \(a\) (which is \(n\)). (A) \(n=1\): \(4(1) = 4\). Not a multiple of 6. (B) \(n=2\): \(4(2) = 8\). Not a multiple of 6. (C) \(n=3\): \(4(3) = 12\). This is a multiple of 6. (Sum = \(3+4+5+6 = 18\)). (D) \(n=4\): \(4(4) = 16\). Not a multiple of 6. *Correction:* Let’s re-read. Oh, “could be”. My test for C worked. Let’s re-test D. \(n=4\). Sum = \(4+5+6+7 = 22\). Not a multiple of 6. My test for C: \(n=3\). Sum = \(3+4+5+6 = 18\). This *is* a multiple of 6. So C is a possible answer. Let me re-check my algebra. \(4n+6 = 6k \implies 4n = 6(k-1)\). \(2n = 3(k-1)\). This means \(2n\) must be a multiple of 3. Since 2 is not, \(n\) must be a multiple of 3. The only option that is a multiple of 3 is C.

Solution (C): Let the integers be \(n, n+1, n+2, n+3\). The sum is \(4n + 6\). We need \(4n + 6\) to be a multiple of 6. \(4n + 6 = 6k\) for some integer \(k\). \(4n = 6k – 6 = 6(k-1)\). This means \(4n\) is a multiple of 6. \(2n = 3(k-1)\). This means \(2n\) must be a multiple of 3. Since 2 is not, \(n\) must be a multiple of 3. The only option for \(a\) (which is \(n\)) that is a multiple of 3 is 3. (Check: If \(a=3\), the integers are 3, 4, 5, 6. Sum = 18, which is a multiple of 6.)

Solution (E): Let the integers be \(n, n+2, n+4, n+6, n+8, n+10\). The sum \(x = 6n + 30 = 6(n + 5)\). Since \(n\) is an even integer, \(n+5 = \text{Even} + \text{Odd} = \text{Odd}\). So \(x = 6 \times (\text{Odd})\). Let’s test this: \(2+4+6+8+10+12 = 42\). \(42 = 6 \times 7\). Test again: \(4+6+8+10+12+14 = 54\). \(54 = 6 \times 9\). The sum \(x\) is always a multiple of 6, but not necessarily 12. *Rethink:* The middle two terms are \(n+4\) and \(n+6\). The average is \(n+5\). Sum = \(n \times \text{Avg} = 6 \times (n+5)\). If \(n\) is even, \(n+5\) is odd. The sum is \(6 \times \text{Odd}\). This is divisible by 3 and 2, but not 4 or 12. This means I, but not II or III. This is option A. Let’s re-read Q2 from the image. “six consecutive integers”. My question is “six consecutive *even* integers”. Let’s test my answer A. Sum = 42. Divisible by 3 (Yes). Divisible by 4 (No). Divisible by 12 (No). This fits option A. What if my algebra is wrong? Let the integers be \(2k, 2k+2, 2k+4, 2k+6, 2k+8, 2k+10\). Sum = \(12k + 30 = 6(2k + 5)\). Since \(2k\) is even, \(2k+5\) is odd. The sum is \(6 \times \text{Odd}\). This is always divisible by 3 and 2. It is *not* always divisible by 4 or 12. This confirms the answer is I only. *Let’s check the provided image Q2 again*. “six consecutive integers” \(n, n+1, n+2, n+3, n+4, n+5\). Sum = \(6n + 15 = 3(2n + 5)\). Since \(2n+5\) is odd, the sum is \(3 \times \text{Odd}\). This is always divisible by 3, but never by 2 or 6. So the answer to the image Q2 is “I only”. My question is different and the answer is also “I only”. *Wait, I see my error.* My sum: \(x = 6(n+5)\). \(n\) is the *first even integer*. So \(n = 2k\). \(x = 6(2k+5)\). \(2k+5\) is always odd. \(x = 6 \times \text{Odd}\). This means \(x\) is divisible by 3. (Yes, I is true). This means \(x\) is divisible by 2. Is \(x\) divisible by 4? No, because \(6 \times \text{Odd} = (2 \times 3) \times \text{Odd}\). It only has one factor of 2. (II is false). Is \(x\) divisible by 12? No. (III is false). The answer is A. *Let me try a different set of 6 consecutive evens.* \(-4, -2, 0, 2, 4, 6\). Sum = 6. 6 is div by 3 (Yes). 6 is not div by 4 (No). 6 is not div by 12 (No). This confirms the answer is “I only”.

Solution (A): Let the integers be \(2k, 2k+2, 2k+4, 2k+6, 2k+8, 2k+10\). The sum \(x = 12k + 30 = 6(2k + 5)\). Since \(2k\) is even, \(2k+5\) is always odd. So, \(x = 6 \times (\text{Odd number})\). Test the properties: I. Is \(6 \times \text{Odd}\) divisible by 3? Yes, always. II. Is \(6 \times \text{Odd}\) divisible by 4? No. It can be written as \(2 \times 3 \times \text{Odd}\), which only has one factor of 2. III. Is it divisible by 12? No, same reason. Therefore, only I must be true.

Solution (A): We are given \(a\) is negative, \(b\) and \(c\) are positive, and \(b < c\). A) \((b-c) / a\): Since \(b < c\), \(b-c\) is negative. \(a\) is negative. \( \text{Negative} / \text{Negative} = \text{Positive} \). This must be true. B) \(a / b\): \( \text{Negative} / \text{Positive} = \text{Negative} \). C) \(a(b-c)\): \(a\) is negative. \(b-c\) is negative. \( (\text{Neg}) \times (\text{Neg}) = \text{Positive} \). This also must be true. *Rethink:* My analysis shows A and C are both "must be positive". Let me re-read. A) \((b-c) / a \to \text{Neg} / \text{Neg} \to \text{Pos}\). Correct. C) \(a(b-c) \to (\text{Neg}) \times (\text{Neg}) \to \text{Pos}\). Correct. This is a flawed question. I will fix it by changing C. New C: \(a(c-b)\). \(a\) is negative. \(c-b\) is positive. \( (\text{Neg}) \times (\text{Pos}) = \text{Negative} \). New D: \(c(a-b)\). \(c\) is positive. \(a-b\) is negative. \( (\text{Pos}) \times (\text{Neg}) = \text{Negative} \). Let's use the first option.

Solution (A): We are given \(a\) is negative, \(b\) and \(c\) are positive, and \(b < c\). A) \((b-c) / a\): Since \(b < c\), the term \((b-c)\) is negative. \(a\) is also negative. \( \text{Negative} / \text{Negative} = \text{Positive} \). This must be true. B) \(a / c\): \( \text{Negative} / \text{Positive} = \text{Negative} \). C) \(a(c-b)\): \(a\) is negative. Since \(c > b\), \((c-b)\) is positive. \( (\text{Neg}) \times (\text{Pos}) = \text{Negative} \). D) \(c(a+b)\): \(c\) is positive. \((a+b)\) could be positive (e.g., -1+5) or negative (e.g., -5+1). Not “must be”. E) \(abc\): \( (\text{Neg}) \times (\text{Pos}) \times (\text{Pos}) = \text{Negative} \).

Solution (D): For \(n\) to be a perfect square, all exponents in its prime factorization must be even. Current exponents: \(2^4\) (Even), \(3^1\) (Odd), \(5^3\) (Odd). We need to multiply by at least \(3^1 \times 5^1\), so \(p\) must be a multiple of \(3 \times 5 = 15\). This eliminates A and B. Let \(k\) be an integer. \(p\) must be of the form \(15 \times k^2\). Let’s test the remaining options: (C) \(p = 15\). This works. \(n = 2^4 \times 3^2 \times 5^4\). (D) \(p = 75\). \(75 = 3 \times 25 = 3 \times 5^2\). This also works. \(n = 2^4 \times (3^1 \times 3^1) \times (5^3 \times 5^2) = 2^4 \times 3^2 \times 5^5\). *Correction:* \(n = (2^4 \times 3^1 \times 5^3) \times (3 \times 5^2) = 2^4 \times 3^2 \times 5^5\). This is NOT a perfect square. Let’s re-read the explanation. \(p = 15 \times k^2\). (C) \(p=15\). Is \(15 = 15 \times k^2\)? Yes, if \(k=1\). (D) \(p=75\). Is \(75 = 15 \times k^2\)? \(75/15 = 5\). \(k^2=5\). No. (E) \(p=105\). Is \(105 = 15 \times k^2\)? \(105/15 = 7\). \(k^2=7\). No. This means the only possible answer is C. The question is “which of the following *could be*”. Why did I create D=75? Let me check my own logic. \(n = 2^4 \times 3^1 \times 5^3 \times p\) We need \(p\) to supply \(3^1\) and \(5^1\) (to make \(3^2\) and \(5^4\)). So, \(p\) must have a factor of \(3 \times 5 = 15\). The *remaining* factors of \(p\) must be a perfect square. Let \(p = 15 \times (\text{perfect square})\). Test options: (C) \(p = 15\). \(15 = 15 \times 1\). 1 is a perfect square. This is possible. (D) \(p = 75\). \(75 = 15 \times 5\). 5 is NOT a perfect square. This is not possible. (E) \(p = 105\). \(105 = 15 \times 7\). 7 is NOT a perfect square. This is not possible. This is a bad question, only one answer is possible. Let me fix (D). I want (D) to be possible. Let \(p = 15 \times 4 = 60\). New (D) = 60. Let’s try again. What if \(p = 75\)? \(n = 2^4 \times 3^1 \times 5^3 \times (3 \times 5^2) = 2^4 \times 3^2 \times 5^5\). Exponent 5 is odd. Not a square. What if \(p = 135\)? \(135 = 15 \times 9\). \(k^2 = 9\). This works. I will rewrite the question. If \(n = 2^3 \times 3^2 \times 5^1 \times p\)… We need \(2^1 \times 5^1 \times k^2\). So \(p\) must be a multiple of 10 times a square. \(p = 10 \times k^2\). Options: A) 10 (\(k=1\)), B) 20 (\(10 \times 2\)), C) 30 (\(10 \times 3\)), D) 40 (\(10 \times 4\), \(k=2\)), E) 50 (\(10 \times 5\)). Answers A and D are possible. This is a good “could be” question.

Solution (D): For \(n\) to be a perfect square, all exponents must be even. Current exponents: \(2^3\) (Odd), \(3^2\) (Even), \(5^1\) (Odd). We need to multiply by at least \(2^1\) and \(5^1\) to get \(2^4\) and \(5^2\). So, \(p\) must have a factor of \(2 \times 5 = 10\). The remaining factors of \(p\) must be a perfect square. So, \(p\) must be of the form \(10 \times k^2\) (where \(k\) is an integer). Test options: (A) \(p=15\). Not a multiple of 10. (B) \(p=20\). \(20 = 10 \times 2\). 2 is not a perfect square. (C) \(p=30\). \(30 = 10 \times 3\). 3 is not a perfect square. (D) \(p=40\). \(40 = 10 \times 4\). 4 is a perfect square (\(k=2\)). This is possible. (E) \(p=50\). \(50 = 10 \times 5\). 5 is not a perfect square.

Solution (B): 1. Find the prime factorization of 432. \(432 = 2 \times 216 = 2 \times 6^3 = 2 \times (2 \times 3)^3 = 2 \times 2^3 \times 3^3 = 2^4 \times 3^3\). 2. We have \((2^4 \times 3^3) \times y = n^3\). 3. For a perfect cube, all exponents must be multiples of 3. 4. \(3^3\) is fine. \(2^4\) is not. The next multiple of 3 after 4 is 6. 5. We need to multiply \(2^4\) by \(2^2\) to get \(2^6\). 6. Therefore, the smallest \(y\) is \(2^2 = 4\). *Correction:* Let me re-check. \(2^4 \times 3^3\). We need exponents to be 0, 3, 6, 9… \(3^3\) is fine. For \(2^4\), we need to get to \(2^6\). We must multiply by \(2^2\). \(y = 2^2 = 4\). This is not an option. Let’s re-factor 432. \(432 / 2 = 216 = 6^3 = 2^3 \times 3^3\). So \(432 = 2^4 \times 3^3\). My factorization is correct. My logic \(y=4\) is correct. The options are wrong. Let’s re-read the Q7 from image. \(450y = n^3\). \(450 = 45 \times 10 = (9 \times 5) \times (2 \times 5) = 2^1 \times 3^2 \times 5^2\). We need \(2^2 \times 3^1 \times 5^1\). \(y = 4 \times 3 \times 5 = 60\). Let me fix my Q5 options. \(y=4\) should be an answer. New options: A) 2, B) 4, C) 6, D) 12, E) 18

Solution (B): 1. Find the prime factorization of 432. \(432 = 216 \times 2 = 6^3 \times 2 = (2 \times 3)^3 \times 2 = (2^3 \times 3^3) \times 2^1 = 2^4 \times 3^3\). 2. We have \((2^4 \times 3^3) \times y = n^3\). 3. For a perfect cube, all exponents must be multiples of 3 (e.g., 0, 3, 6, 9…). 4. The exponent \(3^3\) is fine. The exponent \(2^4\) is not. The next multiple of 3 after 4 is 6. 5. We must multiply \(2^4\) by \(2^2\) to get \(2^6\). 6. Therefore, the smallest integer \(y\) must be \(2^2 = 4\).

Solution (E): 1. The list has 21 integers and starts at -12. 2. The largest integer is \(-12 + (21-1) = -12 + 20 = 8\). 3. The list is \(-12, -11, …, 0, …, 7, 8\). 4. The positive integers in the list are {1, 2, 3, 4, 5, 6, 7, 8}. 5. The range of *these* positive integers is \(\text{Largest} – \text{Smallest} = 8 – 1 = 7\).

Solution (D): 1. Let the integers be \(n, n+1, n+2, n+3\). 2. \(a=n\), \(b=n+1\), \(c=n+2\), \(d=n+3\). 3. We are given the equation \(d = a+b\). 4. Substitute the expressions: \(n+3 = n + (n+1)\). 5. \(n+3 = 2n + 1\). 6. \(2 = n\). 7. The integers are 2, 3, 4, 5. 8. The value of \(c\) is \(n+2 = 2+2 = 4\).

Solution (B): Let’s determine the sign and parity of the result. \(a^2 = (\text{Neg Odd})^2 = \text{Positive Odd}\). \(a^2 b = (\text{Pos Odd}) \times (\text{Neg Even}) = \text{Negative Even}\). \(c = \text{Positive Odd}\). Expression = \( \frac{\text{Negative Even}}{\text{Positive Odd}} \). The result must be Negative. This eliminates C, D, E. The result must also be Even (since \(\text{Even} / \text{Odd}\) is Even, if it’s an integer). Let’s test A: -12. Can we get -12? Let \(a=-1, b=-12, c=1\). \(a^2 b = (1)(-12) = -12\). \( \frac{-12}{1} = -12 \). This works. \(a=-1\) (Neg Odd), \(b=-12\) (Neg Even), \(c=1\) (Pos Odd). Let’s test B: -4. Can we get -4? Let \(a=-1, b=-4, c=1\). \(a^2 b = (1)(-4) = -4\). \( \frac{-4}{1} = -4 \). This also works. *Rethink:* This is a “possible value” question. Both A and B are possible. Let me check the provided image Q8. “a possible value”. This implies only one option works. Let’s re-read my Q8. “a possible value”. This is fine. Both A and B are possible. Maybe I should make it “must be”. No, the image question is “a possible value”. I will stick with my question. I will select B as the answer, but A is also correct. I’ll fix the options to make only one correct. New options: A) -13, B) -4, C) 4, D) 6, E) 1.5

Solution (B): \(a = \text{Neg Odd}\) (e.g., -1, -3) \(b = \text{Neg Even}\) (e.g., -2, -4) \(c = \text{Pos Odd}\) (e.g., 1, 3) Let’s determine the sign: \(a^2 = (\text{Neg Odd})^2 = \text{Positive Odd}\). \(a^2 b = (\text{Pos Odd}) \times (\text{Neg Even}) = \text{Negative Even}\). \(\frac{a^2 b}{c} = \frac{\text{Negative Even}}{\text{Positive Odd}}\). The result must be Negative. This eliminates A, C, D. The result must be an Even number divided by an Odd number. If it is an integer, it must be Even. Test B: -4. Let \(a=-1, b=-4, c=1\). \(\frac{(-1)^2(-4)}{1} = \frac{1(-4)}{1} = -4\). This is possible. Test E: -5.5. This is not an integer. Can we get it? \(\frac{a^2 b}{c} = -5.5 = -11/2\). \(c\) would have to be 2, but \(c\) must be odd. Not possible.

Solution (C): 1. Find prime factors of 490: \(490 = 49 \times 10 = (7^2) \times (2 \times 5)\). List P = {2, 5, 7}. 2. Find prime factors of 180: \(180 = 18 \times 10 = (2 \times 9) \times (2 \times 5) = (2 \times 3^2) \times (2 \times 5) = 2^2 \times 3^2 \times 5\). List Q = {2, 3, 5}. 3. The union of P and Q is the set of all unique elements from both lists: {2, 3, 5, 7}. *Correction:* Re-check. P={2, 5, 7}. Q={2, 3, 5}. Union = {2, 3, 5, 7}. The count of unique numbers is 4. This is option B.

Solution (B): 1. Find prime factors of 490: \(490 = 49 \times 10 = (7^2) \times (2 \times 5)\). List P = {2, 5, 7}. 2. Find prime factors of 180: \(180 = 18 \times 10 = (2 \times 9) \times (2 \times 5) = (2 \times 3^2) \times (2 \times 5) = 2^2 \times 3^2 \times 5\). List Q = {2, 3, 5}. 3. The union of P and Q is the set of all unique elements from both lists: {2, 3, 5, 7}. 4. The count of these unique numbers is 4.

Solution (D): 1. Find the prime factorization of 500: \(500 = 5 \times 100 = 5 \times 10^2 = 5 \times (2 \times 5)^2 = 5 \times 2^2 \times 5^2 = 2^2 \times 5^3\). 2. We have \(x^2 y = 2^2 \times 5^3\). 3. \(x^2\) must have even exponents. \(y\) must contain all the “leftover” odd-exponent factors. 4. Case 1: \(x^2 = 2^2\). Then \(x=2\). This leaves \(y = 5^3 = 125\). (This works: \(x=2, y=125\)). 5. Case 2: \(x^2 = 2^2 \times 5^2\). Then \(x = 2 \times 5 = 10\). This leaves \(y = 5^1 = 5\). (This works: \(x=10, y=5\)). 6. Let’s check the options based on these two possibilities: (A) \(y\) is even? No, \(y=125\) and \(y=5\) are both odd. (B) \(x\) is odd? No, \(x=2\) and \(x=10\) are both even. (D) \(y\) is a multiple of 5? Yes, \(y=125\) and \(y=5\) are both multiples of 5. This must be true. (E) \(x\) is a multiple of 5? No, \(x=2\) is not.