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# Integers: Remainder Problems

If \$\$m\$\$ and \$\$n\$\$ are prime numbers, is the remainder when \$\$3m\$\$ is divided by \$\$n\$\$ greater than 5? >(1) \$\$m \lt 10\$\$ >(2) \$\$n \lt 6\$\$
Incorrect. [[snippet]] According to Stat. (2), \$\$n \lt 6\$\$, so \$\$n\$\$ can be 2, 3, or 5, and \$\$m\$\$ can be any prime number. It is a general rule that if you divide two numbers, the remainder must be strictly _less_ than the divisor (i.e. the number you are dividing by). For example, when dividing by 5, the greatest possible remainder is 4. For Stat. (2), no matter what the value of \$\$m\$\$, the maximum possible value of \$\$n\$\$ is 5. Hence, the remainder _cannot_ be greater than 5. This is a definite answer, so **Stat.(2) → Yes → S → BD**.
According to Stat. (2), \$\$n \lt 6\$\$, so \$\$n\$\$ can be 2, 3, or 5, and \$\$m\$\$ can be any prime number. It is a general rule that if you divide two numbers, the remainder must be strictly _less_ than the divisor (i.e. the number you are dividing by). For example, when dividing by 5, the greatest possible remainder is 4. For Stat. (2), no matter what the value of \$\$m\$\$, the maximum possible value of \$\$n\$\$ is 5. Hence, the remainder _cannot_ be greater than 5. This is a definite answer, so **Stat.(2) → Yes → S → B**.
Incorrect. [[snippet]] According to Stat. (1), \$\$m \lt 10\$\$, so \$\$m\$\$ can be 2, 3, 5, or 7, and \$\$n\$\$ can be any prime number. First plug in \$\$m = 3\$\$ and \$\$n = 5\$\$. Then the remainder when \$\$3m\$\$ is divided by \$\$n\$\$ is 4, which is less than 5. Now, how could the remainder be greater than 5? Since there's no limitation on \$\$n\$\$, the easiest way is to plug in a value for \$\$n\$\$ that is greater than \$\$3m\$\$. If you plug in \$\$m = 5\$\$ and \$\$n = 17\$\$, then the remainder when \$\$3m\$\$ is divided by \$\$n\$\$ is 15, which is greater than 5. Thus, there is no definite answer, so **Stat.(1) → No → IS → BCE**.
Incorrect. [[snippet]] According to Stat. (1), \$\$m \lt 10\$\$, so \$\$m\$\$ can be 2, 3, 5, or 7, and \$\$n\$\$ can be any prime number. First plug in \$\$m = 3\$\$ and \$\$n = 5\$\$. Then the remainder when \$\$3m\$\$ is divided by \$\$n\$\$ is 4, which is less than 5. Now, how could the remainder be greater than 5? Since there's no limitation on \$\$n\$\$, the easiest way is to plug in a value for \$\$n\$\$ that is greater than \$\$3m\$\$. If you plug in \$\$m = 5\$\$ and \$\$n = 17\$\$, then the remainder when \$\$3m\$\$ is divided by \$\$n\$\$ is 15, which is greater than 5. Thus, there is no definite answer, so **Stat.(1) → No → IS → BCE**.
Correct. [[snippet]] According to Stat. (1), \$\$m \lt 10\$\$, so \$\$m\$\$ can be 2, 3, 5, or 7, and \$\$n\$\$ can be any prime number. First plug in \$\$m = 3\$\$ and \$\$n = 5\$\$. Then the remainder when \$\$3m\$\$ is divided by \$\$n\$\$ is 4, which is less than 5. Now, how could the remainder be greater than 5? Since there's no limitation on \$\$n\$\$, the easiest way is to plug in a value for \$\$n\$\$ that is greater than \$\$3m\$\$. If you plug in \$\$m = 5\$\$ and \$\$n = 17\$\$, then the remainder when \$\$3m\$\$ is divided by \$\$n\$\$ is 15, which is greater than 5. Thus, there is no definite answer, so **Stat.(1) → No → IS → BCE**.
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.