Factors & Multiples: Level 3
1.
When positive integer \(x\) is divided by 8, the remainder is 3. What is the remainder when \( (x+1)(x+3) \) is divided by 8?
Solution (D): We are given \(x \equiv 3 \pmod{8}\).
Substitute this into the expression:
\((x+1) \equiv (3+1) \equiv 4 \pmod{8}\)
\((x+3) \equiv (3+3) \equiv 6 \pmod{8}\)
The expression becomes \( (4) \times (6) = 24 \).
Now, find the remainder of 24 when divided by 8.
\(24 \div 8 = 3\) with a remainder of 0.
*Correction:* Let’s re-check. \(x=3\). \((3+1)(3+3) = 4 \times 6 = 24\). \(24 \div 8\) rem 0.
Let’s try \(x=11\). \((11+1)(11+3) = 12 \times 14\).
\(12 \equiv 4 \pmod 8\). \(14 \equiv 6 \pmod 8\).
\(4 \times 6 = 24\). \(24 \equiv 0 \pmod 8\).
The answer is 0.
1.
When positive integer \(x\) is divided by 8, the remainder is 3. What is the remainder when \( (x+1)(x+3) \) is divided by 8?
Solution (A): We use remainder arithmetic. We are given \(x \equiv 3 \pmod{8}\).
Substitute this into the expression \((x+1)(x+3)\):
The first term: \((x+1) \equiv (3+1) \equiv 4 \pmod{8}\)
The second term: \((x+3) \equiv (3+3) \equiv 6 \pmod{8}\)
Multiply the remainders: \( (4) \times (6) = 24 \).
Now, find the remainder of 24 when divided by 8.
\(24 \div 8 = 3\) with a remainder of 0.
2.
If \(n\) is divisible by 4, 9, and 10, then \(n\) must be a multiple of which of the following?
Solution (D): If \(n\) is divisible by 4, 9, and 10, it must be a multiple of their Least Common Multiple (LCM).
Prime factors:
4 = \(2^2\)
9 = \(3^2\)
10 = \(2 \times 5\)
The LCM must have the highest power of each prime: \(2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 = 36 \times 5 = 180\).
Therefore, \(n\) must be a multiple of 180.
3.
A positive integer \(n\) leaves a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, and a remainder of 3 when divided by 4. What is the remainder when \(n\) is divided by 12?
Solution (E): In each case, the remainder is 1 less than the divisor.
\(n \equiv 1 \pmod{2} \implies n \equiv -1 \pmod{2}\)
\(n \equiv 2 \pmod{3} \implies n \equiv -1 \pmod{3}\)
\(n \equiv 3 \pmod{4} \implies n \equiv -1 \pmod{4}\)
This means \(n+1\) is divisible by 2, 3, and 4.
\(n+1\) must be a multiple of the \(LCM(2, 3, 4)\), which is 12.
So, \(n+1 = 12k\).
\(n = 12k – 1\). This means \(n\) has a remainder of 11 when divided by 12.
4.
When \(x\) is divided by \(y\), the remainder is 4. When \(a\) is divided by \(b\), the remainder is 7. What is the smallest possible value for \(y+b\)?
Solution (D): The remainder must be less than the divisor.
1. \(x\) div by \(y\) rem 4 \(\implies y > 4\). The smallest integer value for \(y\) is 5.
2. \(a\) div by \(b\) rem 7 \(\implies b > 7\). The smallest integer value for \(b\) is 8.
3. The smallest possible value for \(y+b\) is \(5 + 8 = 13\).
5.
If \(x \# y\) represents the remainder when \(x\) is divided by \(y\), what is the value of \((90 \# 21) \# (50 \# 7)\)?
Solution (A):
1. Evaluate \((90 \# 21)\). \(21 \times 4 = 84\). \(90 – 84 = 6\). The remainder is 6.
2. Evaluate \((50 \# 7)\). \(7 \times 7 = 49\). \(50 – 49 = 1\). The remainder is 1.
3. The expression becomes \(6 \# 1\).
4. \(6 \div 1 = 6\) with a remainder of 0.
6.
How many positive *even* factors does the number 360 have?
Solution (E):
1. Prime factorization of 360: \(360 = 36 \times 10 = (4 \times 9) \times (2 \times 5) = (2^2 \times 3^2) \times (2 \times 5) = 2^3 \times 3^2 \times 5^1\).
2. Total factors = \((3+1)(2+1)(1+1) = 4 \times 3 \times 2 = 24\).
3. Odd factors are formed using only the odd primes (\(3^2 \times 5^1\)).
Number of odd factors = \((2+1)(1+1) = 3 \times 2 = 6\).
4. Even factors = Total factors – Odd factors = \(24 – 6 = 18\).
7.
When \(x\) is divided by 7, the quotient is \(y\) and the remainder is 2. When \(x\) is divided by 4, the quotient is \(z\) and the remainder is 1. Which of the following expresses \(z\) in terms of \(y\)?
Solution (D):
1. Write the two equations from the given info:
\(x = 7y + 2\)
\(x = 4z + 1\)
2. Set the expressions for \(x\) equal to each other: \(7y + 2 = 4z + 1\).
3. Solve for \(z\):
\(7y + 1 = 4z\)
\(z = \frac{7y+1}{4}\)
8.
If \(12^n\) is a divisor of \(31,104\), what is the largest possible integer value of \(n\)?
Solution (C):
1. We need to find the prime factors of \(12^n\) and 31,104.
2. \(12^n = (2^2 \times 3)^n = 2^{2n} \times 3^n\).
3. \(31,104 = 4 \times 7776 = 2^2 \times (1296 \times 6) = 2^2 \times (6^4 \times 6) = 2^2 \times 6^5\).
4. \(31,104 = 2^2 \times (2 \times 3)^5 = 2^2 \times 2^5 \times 3^5 = 2^7 \times 3^5\).
5. For \(12^n\) to be a divisor, its prime factors must be “contained” in \(2^7 \times 3^5\).
6. We need \(2n \le 7\) AND \(n \le 5\).
7. From \(2n \le 7\), we get \(n \le 3.5\). From \(n \le 5\), we get \(n \le 5\).
8. Both must be true. The limiting condition is \(n \le 3.5\). The largest integer \(n\) is 3.
9.
A group of \(n\) students can be divided into equal groups of 4 with 1 student left over, or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of \(n\)?
Solution (C):
1. \(n \equiv 1 \pmod{4}\). Possible values: {1, 5, 9, 13, 17, 21, 25, 29, 33, …}
2. \(n \equiv 3 \pmod{5}\). Possible values: {3, 8, 13, 18, 23, 28, 33, …}
3. The smallest value on both lists is 13.
4. The next smallest value on both lists is 33.
5. The sum of the two smallest values is \(13 + 33 = 46\).
10.
The greatest common factor (GCF) of 16 and \(n\) is 4. The GCF of \(n\) and 45 is 3. Which of the following could be the value of \(n\)?
Solution (D):
1. GCF(16, \(n\)) = 4. This means \(n\) must be a multiple of 4, but *not* a multiple of 8 (otherwise the GCF would be 8 or 16). \(n = 4 \times (\text{Odd})\).
2. GCF(\(n\), 45) = 3. This means \(n\) must be a multiple of 3.
3. Combine: \(n\) is a multiple of 4 and 3, so \(n\) is a multiple of 12.
4. And \(n/4\) must be odd.
5. Test multiples of 12:
\(n=12 \implies 12/4 = 3\) (Odd). This is possible.
\(n=24 \implies 24/4 = 6\) (Even). Not possible.
\(n=36 \implies 36/4 = 9\) (Odd). This is possible.
\(n=48 \implies 48/4 = 12\) (Even). Not possible.
\(n=60 \implies 60/4 = 15\) (Odd). This is possible.
6. From the options, 24 is not possible. 36 and 60 are possible. The question asks what *could be* the value. 36 is an option.
Score: 0 / 10