Divisibility questions are presented below with clear and detailed solutions. When an integer \( n \) is divided by a nonzero integer \( m \), the result can be written in the form
\[ n = qm + r \]where \( n \) is the dividend, \( m \) is the divisor, \( q \) is the quotient, and \( r \) is the remainder.
If a positive integer \( n \) is divided by 5, the remainder is 3. Which of the following expressions yields a remainder of 0 when divided by 5?
A) \( n + 3 \) B) \( n + 2 \) C) \( n - 1 \) D) \( n - 2 \) E) \( n + 1 \)
If \( n \) leaves a remainder of 3 when divided by 5, then
\[ n = 5k + 3 \]for some integer \( k \). Adding 2 to both sides gives
\[ n + 2 = 5k + 5 = 5(k + 1) \]Thus, \( n + 2 \) is divisible by 5 and leaves a remainder of 0.
Answer: B
If an integer \( n \) is divisible by 3, 5, and 12, what is the next larger integer divisible by all three numbers?
A) \( n + 3 \) B) \( n + 5 \) C) \( n + 12 \) D) \( n + 60 \) E) \( n + 15 \)
If \( n \) is divisible by 3, 5, and 12, then it must be a multiple of their least common multiple:
\[ \mathrm{lcm}(3,5,12) = 60 \]Thus,
\[ n = 60k \]The next larger integer divisible by all three is
\[ n + 60 = 60(k + 1) \]Answer: D
What is the smallest positive integer that is a multiple of 5, 7, and 20?
A) 70 B) 35 C) 200 D) 280 E) 140
The smallest number divisible by 5, 7, and 20 is their least common multiple:
\[ \mathrm{lcm}(5,7,20) = 140 \]Answer: E
When the integer \( n \) is divided by 8, the remainder is 3. What is the remainder when \( 6n \) is divided by 8?
A) 0 B) 1 C) 2 D) 3 E) 4
Since \( n \) leaves a remainder of 3 when divided by 8,
\[ n = 8k + 3 \]Multiplying both sides by 6:
\[ 6n = 6(8k + 3) = 48k + 18 \]Rewrite 18 as \( 16 + 2 \):
\[ 6n = 8(6k + 2) + 2 \]Therefore, the remainder is 2.
Answer: C
If \( n \) is an integer, when \( (2n + 2)^2 \) is divided by 4, the remainder is
A) 0 B) 1 C) 2 D) 3 E) 4
First expand the expression:
\[ (2n + 2)^2 = 4n^2 + 8n + 4 \]Factor out 4:
\[ (2n + 2)^2 = 4(n^2 + 2n + 1) \]Thus, the expression is divisible by 4 and the remainder is 0.
Answer: A
What is the smallest positive two-digit integer divisible by 3 such that the sum of its digits is 9?
A) 27 B) 33 C) 72 D) 18 E) 90
Let the two-digit number be \( 10x + y \), where \( x \) and \( y \) are digits. Divisibility by 3 implies
\[ 10x + y = 3k \]The sum of the digits is 9:
\[ x + y = 9 \]Solving for \( y \) gives \( y = 9 - x \). Substitute:
\[ 10x + 9 - x = 3k \] \[ 9x + 9 = 3k \] \[ x = \frac{k - 3}{3} \]Trying integer values of \( k \), the smallest valid digit is \( x = 1 \), giving \( y = 8 \). The number is 18.
Answer: D
Which of the following numbers is not divisible by 3?
A) 339 B) 342 C) 552 D) 1111 E) 672
A number is divisible by 3 if the sum of its digits is divisible by 3.
A) \( 3 + 3 + 9 = 15 \) (divisible by 3)
B) \( 3 + 4 + 2 = 9 \) (divisible by 3)
C) \( 5 + 5 + 2 = 12 \) (divisible by 3)
D) \( 1 + 1 + 1 + 1 = 4 \) (not divisible by 3)
Thus, 1111 is not divisible by 3.
Answer: D
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