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# Integers: Remainder Problems

If $$m$$ is a two-digit number, what is the remainder when $$m$$ is divided by 3? >(1) $$m+1$$ is divisible by 3. >(2) $$m$$ is positive, and the sum of its two digits is 8.
Incorrect. [[snippet]] Think about some ways you could narrow down the possible remainders of $$\frac{m}{3}$$. Here are some hints: Stat. (1): All remainders are positive, even when the divisor or dividend is negative. Stat. (2): You can manually test all possible values of $$m$$ to see if this statement is sufficient.
Correct. [[snippet]] For stat. (1), If $$m+1$$ is divisible by 3, then $$m-2$$ is also divisible by 3. In this case, dividing $$m$$ by 3 can only leave a remainder of 2. You could also use __Plugging In__: >If $$m+1 = 12$$, then $$m = 11$$, which has a remainder of 2 when divided by 3. >If $$m+1 = -12$$, then $$m= -13$$, which also has a remainder of 2 when divided by 3 since $$-13 = -5\times 3+2$$ All such $$m$$'s leave a remainder of 2 when divided by 3. (Keep in mind that all remainders are positive, even when the divisor or dividend is negative.) Therefore, **Stat.(1) → S → AD**. For stat. (2), mark $$m$$'s digits as $$XY$$, so that $$X+Y = 8$$. According to the __Rule of Divisibility by 3__, $$m$$ isn't divisible by 3, but $$m+1$$ is divisible by 3 because the sum of its digits is $$X + (Y+1) = 9$$, which is divisible by 3. Now, as you've established, if $$m+1$$ is divisible by 3, then $$m-2$$ is also divisible by 3, and thus $$m \div 3$$ must have a remainder of 2. You could also test all cases. There aren't that many: >List all two-digit numbers for which the sum of the figures is 8, such as 17, 26, 35, 44, 53, 62, 71, and 80. The nearest multiples of 3 are 15, 24, 33, 42, 51, 60, 69, and 78, respectively. All of the numbers you listed as $$m$$ are 2 more than the nearest multiple of 3, which means they all leave a remainder of 2 when divided by 3. **Stat.(2) → S → D**.
Incorrect. [[snippet]] In this question, either one of the statements is __Sufficient__. Think about why. For stat. (1), If $$m+1$$ is divisible by 3, then $$m-2$$ is also divisible by 3. In this case, dividing $$m$$ by 3 can only leave a remainder of 2. Therefore, **Stat.(1) → S → AD**. For stat. (2), mark $$m$$'s digits as $$XY$$, so that $$X+Y = 8$$. According to the __Rule of Divisibility by 3__, $$m$$ isn't divisible by 3, but $$m+1$$ is divisible by 3 because the sum of its digits is $$X + (Y+1) = 9$$, which is divisible by 3. Now, as you've established, if $$m+1$$ is divisible by 3, then $$m-2$$ is also divisible by 3, and thus $$m \div 3$$ must have a remainder of 2. **Stat.(2) → S → D**.
Incorrect. [[snippet]] Yes, Stat. (2) is __Sufficient__, but what about Stat. (1)? Note that remainders are always positive, even when the divisor or the dividend is negative.
Incorrect. [[snippet]] Yes, Stat. (1) is __Sufficient__, but what about Stat. (2)? If you're not sure, manually test all possible $$m$$'s. There aren't that many.
Statement (1) ALONE is sufficient, but Statement (2) alone is not sufficient to answer the question asked.
Statement (2) ALONE is sufficient, but Statement (1) alone is not sufficient to answer the question asked.
BOTH Statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.