The integers m and p are such that 2 , and (m is not a factor of p. If r...

Understanding the Question

We're asked whether \(\mathrm{r} > 1\), where r is the remainder when integer p is divided by integer m.

Given Information

• m and p are integers
• \(2 < \mathrm{m} < \mathrm{p}\)
• m is NOT a factor of p (so p isn't divisible by m)
• r is the remainder when p is divided by m

What We Need to Determine

Since m doesn't divide p evenly, we know there WILL be a remainder. By the division algorithm, when we divide p by m, we get: \(\mathrm{p} = \mathrm{mq} + \mathrm{r}\), where \(0 < \mathrm{r} < \mathrm{m}\).

The question asks: Is this remainder greater than 1? In other words, is r at least 2?

Key Insight

Since r must be between 1 and (m-1), and we're asking if r > 1, we're really asking: Can r be exactly 1, or must it be 2 or more?

Analyzing Statement 1

Statement 1: The greatest common factor of m and p is 2.

What Statement 1 Tells Us

If \(\mathrm{gcd}(\mathrm{m}, \mathrm{p}) = 2\), then both m and p are even numbers (they both have 2 as a factor).

The Strategic Insight

Here's the key realization: When we divide one even number by another even number, what happens to the remainder?

Think about it this way:

• p is even (let's say \(\mathrm{p} = 2\mathrm{k}\) for some integer k)
• m is even (let's say \(\mathrm{m} = 2\mathrm{j}\) for some integer j)
• When we divide p by m: \(\mathrm{p} = \mathrm{mq} + \mathrm{r}\)

This means: \(2\mathrm{k} = 2\mathrm{j} \times \mathrm{q} + \mathrm{r}\)

Rearranging: \(\mathrm{r} = 2\mathrm{k} - 2\mathrm{jq} = 2(\mathrm{k} - \mathrm{jq})\)

Since r equals 2 times some integer, r must be even!

Conclusion

The remainder r must be:

1. Positive (since m doesn't divide p)
2. Less than m
3. Even (as we just proved)

The smallest positive even number is 2. Therefore, \(\mathrm{r} \geq 2\), which means \(\mathrm{r} > 1\).

[STOP - Statement 1 is SUFFICIENT!]

This eliminates choices B, C, and E.

Analyzing Statement 2

Now let's forget Statement 1 completely and analyze Statement 2 independently.

Statement 2: The least common multiple of m and p is 30.

What Statement 2 Provides

If \(\mathrm{lcm}(\mathrm{m}, \mathrm{p}) = 30\), then both m and p must be divisors of 30.

The divisors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30

Given that \(2 < \mathrm{m} < \mathrm{p}\), the possible values for m are: 3, 5, 6, 10, 15

Testing Different Scenarios

Let's examine two specific cases to see if we always get the same answer:

Case 1: \(\mathrm{m} = 3\)

• For \(\mathrm{lcm}(3, \mathrm{p}) = 30\), we need p to be a multiple of 10
• Using the relationship: \(\mathrm{lcm}(3, \mathrm{p}) \times \mathrm{gcd}(3, \mathrm{p}) = 3\mathrm{p}\)
• We get: \(30 \times \mathrm{gcd}(3, \mathrm{p}) = 3\mathrm{p}\)
• This gives us \(\mathrm{p} = 10 \times \mathrm{gcd}(3, \mathrm{p})\)
• Since \(\mathrm{gcd}(3, 10) = 1\), we have \(\mathrm{p} = 10\)
• Dividing: \(10 = 3 \times 3 + 1\), so \(\mathrm{r} = 1\) ✗

Case 2: \(\mathrm{m} = 6\)

• For \(\mathrm{lcm}(6, \mathrm{p}) = 30\), using similar logic: \(30 \times \mathrm{gcd}(6, \mathrm{p}) = 6\mathrm{p}\)
• So \(\mathrm{p} = 5 \times \mathrm{gcd}(6, \mathrm{p})\)
• We need p > 6, and checking \(\mathrm{gcd}(6, 10) = 2\), we get \(\mathrm{p} = 10\)
• Dividing: \(10 = 6 \times 1 + 4\), so \(\mathrm{r} = 4\) ✓

Conclusion

We found:

• When \((\mathrm{m}, \mathrm{p}) = (3, 10)\): \(\mathrm{r} = 1\) (not greater than 1)
• When \((\mathrm{m}, \mathrm{p}) = (6, 10)\): \(\mathrm{r} = 4\) (greater than 1)

Since r can be either equal to 1 or greater than 1, we cannot determine whether \(\mathrm{r} > 1\).

Statement 2 is NOT sufficient.

This eliminates choices B and D.

The Answer: A

Statement 1 alone tells us that both m and p are even, which forces the remainder to be even (and therefore at least 2). Statement 2 alone allows for remainders of 1 or greater.

Answer Choice A: Statement 1 alone is sufficient, but Statement 2 alone is not sufficient.