SAT Factors, Divisibility, and Remainders

Correct Answer: A

Solution 1:

Tip: Replace variables with numbers.
We first find an example of a positive integer $n$ satisfying the given conditions. An example of an integer that leaves a remainder of $3$ upon division by $5$ is

$n=(5 \times 1)+3= 5 + 3 = 8.$

Then

$n+3 = 8 + 3 = 11.$

Now, we divide $11$ by $5$ to obtain $11 = (5 \times 2) + 1,$ which shows $11$ leaves a remainder of $1$ upon division by $5.$ This shows (A) is the correct answer.

Solution 2:

Since $n$ leaves a remainder of $3$ upon division by $5$ , $n$ can be written as $5k + 3$ for some non-negative number $k$ . Then

$n+3 = (5k + 3) + 3 = 5k + 6 = 5(k+1) + 1.$

Therefore, dividing $n+3$ by $5$ leaves a remainder of $1$ , which is answer (A).

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Incorrect Choices:

(B)
If the remainder of $n+3$ upon division of $5$ is $2$ , then we can express $n+3$ as $n+3=5a+2$ for some non-negative integer $a$ . Subtracting $3$ from both sides, we get $n=5a-1$ , which can be rewritten as $n=5a-5+4=5(a-1)+4$ . This shows when $n$ is divided by $5$ , it leaves a remainder of $4$ . Since this contradicts the condition in the problem statement, Choice (B) may be eliminated.

(C)
If the remainder of $n+3$ upon division of $5$ is $3$ , then we can express $n+3$ as $n+3=5a+3$ , for some non-negative integer $a$ . Subtracting $3$ from both sides, we get $n=5a$ . This shows when $n$ is divided by $5$ , it leaves a remainder of $0$ . Since this contradicts the condition in the problem statement, Choice (C) may be eliminated.

(D)
If the remainder of $n+3$ upon division of $5$ is $4$ , then we can express $n+3$ as $n+3=5a+4$ , for some non-negative integer $a$ . Subtracting $3$ from both sides, we get $n=5a+1$ . This shows when $n$ is divided by $5$ , it leaves a remainder of $1$ . Since this contradicts the condition in the problem statement, Choice (D) may be eliminated.

(E)
Dividing any integer by $5$ leaves a remainder of $0,1,2,3,$ or $4$ . Since dividing an integer by $5$ cannot leave a remainder of $5$ , choice (E) may be eliminated.

n = 5 k + 3 , k = 1 , 2 , 3 , ........

n + 3 = 5 k + 3 + 3 = 5(k + 1) + 1

the answer is 1