Factors & Multiples: Level 2

Solution (E): We use remainder arithmetic. \(x \equiv 3 \pmod{7}\). We need \(2x^2 \pmod{7}\). \(x^2 \equiv 3^2 \equiv 9 \equiv 2 \pmod{7}\). \(2x^2 \equiv 2(2) \equiv 4 \pmod{7}\). *Correction:* Let’s re-check. \(x \equiv 3\). \(x^2 \equiv 9 \equiv 2\). \(2x^2 \equiv 2(2) = 4\). The remainder is 4. This is option D.

Solution (D): We can use remainders. We are given \(x \equiv 3 \pmod{7}\). We need to find the remainder of \(2x^2\). First, \(x^2 \equiv 3^2 \equiv 9\). The remainder of 9 div by 7 is 2. So \(x^2 \equiv 2 \pmod{7}\). Now, \(2x^2 \equiv 2(2) \equiv 4 \pmod{7}\). The remainder is 4.

Solution (C): If \(n\) is divisible by 6 and 8, it must be a multiple of their Least Common Multiple (LCM). Prime factors of 6 are \(2 \times 3\). Prime factors of 8 are \(2^3\). The LCM is \(2^3 \times 3 = 8 \times 3 = 24\). Therefore, \(n\) must be a multiple of 24.

Solution (B): 1. Find the prime factorization of 200: \(200 = 2 \times 100 = 2 \times 10^2 = 2 \times (2 \times 5)^2 = 2 \times 2^2 \times 5^2 = 2^3 \times 5^2\). 2. To find the number of factors, add 1 to each exponent and multiply: \((3+1) \times (2+1) = 4 \times 3 = 12\).

Solution (E): 1. Factor using difference of squares: \( (5^2 – 3^2)(5^2 + 3^2) \). 2. \( (25 – 9)(25 + 9) = (16)(34) \). 3. Find the prime factors: \( (2^4) \times (2 \times 17) = 2^5 \times 17 \). 4. The distinct prime factors are 2 and 17. The greatest is 17.

Solution (D): 1. First, evaluate the parenthesis: \((150 \# 12)\). 2. \(150 \div 12\). We know \(12 \times 12 = 144\). 3. \(150 – 144 = 6\). So the remainder is 6. 4. The expression becomes \((6) \# 5\). 5. \(6 \div 5\) has a remainder of 1.

Solution (B): The number \(abc,abc\) can be written as \(abc \times 1000 + abc\). This factors to \(abc \times (1000 + 1) = abc \times 1001\). The prime factorization of 1001 is \(7 \times 11 \times 13\). Therefore, the number must be divisible by 7, 11, 13, and 1001. It is not necessarily divisible by 2 (e.g., if \(abc = 123\), the number is 123,123, which is odd).

Solution (A): If \(n\) has a remainder of 7 when divided by 10, \(n\) must be a number ending in 7 (e.g., 7, 17, 27, …). The expression is \(n+8\). \((…7) + 8 = …15\). Any number ending in 15 is a multiple of 5, so the remainder when divided by 5 is 0.

Solution (C): If \(n\) is divisible by 3, 4, and 5, it must be a multiple of their LCM. \(LCM(3, 4, 5) = LCM(3, 2^2, 5) = 2^2 \times 3^1 \times 5^1 = 60\). The smallest possible value for \(n\) is 60. The number of factors for 60 is \((2+1)(1+1)(1+1) = 3 \times 2 \times 2 = 12\). *Rethink:* The question asks for the *smallest possible number of factors*. If \(n=60\), it has 12 factors. What if \(n=120\)? \(120 = 2^3 \times 3^1 \times 5^1\). Factors = \((3+1)(1+1)(1+1) = 4 \times 2 \times 2 = 16\). What if \(n=180\)? \(180 = 2^2 \times 3^2 \times 5^1\). Factors = \((2+1)(2+1)(1+1) = 3 \times 3 \times 2 = 18\). What if \(n=60^2\)? \(n = 2^4 \times 3^2 \times 5^2\). Factors = \((4+1)(2+1)(2+1) = 5 \times 3 \times 3 = 45\). It seems 12 is the smallest. Let me re-read the options. 12 is an option. Why did I select 16? Ah, I must have misread the question. The smallest possible value for \(n\) is 60, and 60 has 12 factors. The answer is D.

Solution (D): If \(n\) is divisible by 3, 4, and 5, it must be a multiple of their LCM. Prime factors: 3, \(2^2\), 5. LCM = \(2^2 \times 3^1 \times 5^1 = 60\). So, \(n\) must be a multiple of 60. The smallest possible value for \(n\) is 60. The number of factors for 60 is \((2+1)(1+1)(1+1) = 3 \times 2 \times 2 = 12\). Any other multiple of 60 (like 120, 180) will have at least as many, or more, factors.

Solution (D): 1. Find the prime factorization of 180: \(180 = 18 \times 10 = (2 \times 9) \times (2 \times 5) = (2 \times 3^2) \times (2 \times 5) = 2^2 \times 3^2 \times 5^1\). 2. We need \(180n\) to be a perfect square. The expression is \((2^2 \times 3^2 \times 5^1) \times n\). 3. For a perfect square, all exponents must be even. 4. \(2^2\) and \(3^2\) are fine. \(5^1\) is odd. We need to multiply by \(5^1\) to get \(5^2\). 5. The smallest \(n\) is 5.

Solution (E): 1. Remainder 3 when divided by 5: Numbers end in 3 or 8. {3, 8, 13, 18, 23, 28, 33, 38…} 2. Remainder 1 when divided by 4: Numbers are {1, 5, 9, 13, 17, 21, 25, 29, 33, 37…} 3. We look for a number on both lists. 4. 13 is on both lists. 5. 33 is on both lists. 6. Check the options. 13 is an option. 33 is an option. *Correction:* Let’s re-check 33. \(33 \div 5 = 6\) rem 3 (Correct). \(33 \div 4 = 8\) rem 1 (Correct). Let’s re-check 13. \(13 \div 5 = 2\) rem 3 (Correct). \(13 \div 4 = 3\) rem 1 (Correct). This is a “could be” question, so both A and E are valid. This is a bad question. I will fix the options. New options: A) 11, B) 21, C) 23, D) 31, E) 33

Solution (E): 1. \(n\) has a remainder of 1 when divided by 4. Let’s test the options for this first. (A) \(11 \div 4\) rem 3 (No) (B) \(21 \div 4\) rem 1 (Possible) (C) \(23 \div 4\) rem 3 (No) (D) \(31 \div 4\) rem 3 (No) (E) \(33 \div 4\) rem 1 (Possible) 2. Now we test the remaining options (B and E) for the second rule: remainder 3 when divided by 5. (B) \(21 \div 5\) rem 1 (No) (E) \(33 \div 5\) rem 3 (Yes) The only number that fits both conditions is 33.