When you encounter a problem asking for numbers that satisfy multiple conditions simultaneously, you need to find the intersection of those conditions. Here, you're looking for positive integers that are both factors of 720 and multiples of 18.

Start by finding the prime factorizations. For 720: $$720 = 2^4 \times 3^2 \times 5^1$$. For 18: $$18 = 2^1 \times 3^2$$.

Since $$n$$ must be a multiple of 18, it must contain at least $$2^1 \times 3^2$$ in its prime factorization. Since $$n$$ must also be a factor of 720, it cannot contain prime powers greater than those in 720's factorization.

Therefore, $$n$$ must have the form $$n = 2^a \times 3^b \times 5^c$$ where:

• $$1 \leq a \leq 4$$ (at least $$2^1$$ from 18, at most $$2^4$$ from 720)
• $$b = 2$$ (exactly $$3^2$$ from both conditions)
• $$0 \leq c \leq 1$$ (at most $$5^1$$ from 720)

This gives you $$4 \times 1 \times 2 = 8$$ possibilities.

Choice A) 6 likely comes from miscounting the exponent possibilities. Choice C) 10 might result from incorrectly allowing $$b$$ to vary from 0 to 2, giving $$4 \times 3 \times 2 = 24$$, then making calculation errors. Choice D) 12 could come from forgetting the multiple of 18 constraint and just counting factors of 720 that are multiples of some smaller number.

The answer is B) 8.

Strategy tip: For problems involving both "factor of" and "multiple of" conditions, always find the prime factorizations first, then determine the constraints each condition places on the exponents.

When you encounter remainder problems involving different divisors, you need to understand how remainders relate to each other through the division algorithm.

Since $$n$$ divided by 15 gives remainder 7, we can write $$n = 15k + 7$$ for some integer $$k$$. This means $$n$$ could be 7, 22, 37, 52, 67, and so on.

To find what remainder $$n$$ gives when divided by 5, let's rewrite our expression. Since $$15 = 3 \times 5$$, we have:

$$n = 15k + 7 = 3 \times 5k + 7 = 5(3k) + 7$$

Now we need to find what remainder 7 gives when divided by 5. Since $$7 = 5 \times 1 + 2$$, we get:

$$n = 5(3k) + 5(1) + 2 = 5(3k + 1) + 2$$

This shows that $$n$$ always gives remainder 2 when divided by 5.

Let's verify with examples: $$7 ÷ 5 = 1$$ remainder $$2$$, $$22 ÷ 5 = 4$$ remainder $$2$$, $$37 ÷ 5 = 7$$ remainder $$2$$.

Choice A) 1 is incorrect because no number of the form $$15k + 7$$ gives remainder 1 when divided by 5.

Choice C) 3 is incorrect for the same reason—the remainder is always 2, never 3.

Choice D) 4 is incorrect because the mathematical relationship forces the remainder to be 2.

Choice B) 2 is correct, as our algebra demonstrates.

Strategy tip: When working with remainder problems involving related divisors, express the number using the first condition, then manipulate that expression to find what happens with the second divisor. Look for common factors between the divisors.

When you see an expression like $$2^6 - 1$$, recognize this as a difference of powers that can often be factored using algebraic identities. This question tests your ability to factor completely and identify prime factors.

First, let's calculate and factor $$2^6 - 1$$. Since $$2^6 = 64$$, we have $$2^6 - 1 = 64 - 1 = 63$$. Now we need to find the prime factorization of 63.

Start by testing small prime divisors: $$63 ÷ 3 = 21$$, and $$21 ÷ 3 = 7$$. Since 7 is prime, we have $$63 = 3^2 × 7$$. The prime factors are 3 and 7, so the largest prime factor is 7.

Alternatively, you can use the difference of squares formula. Since $$2^6 = (2^3)^2 = 8^2$$, we have $$2^6 - 1 = 8^2 - 1 = (8-1)(8+1) = 7 × 9 = 7 × 3^2$$, confirming our factorization.

Looking at the wrong answers: (A) 3 is indeed a prime factor of 63, but it's not the largest one. (B) 5 is not a factor of 63 at all—this might tempt students who miscalculate $$2^6$$ or make arithmetic errors. (D) 9 is a factor of 63, but it's not prime since $$9 = 3^2$$.

The answer is (C) 7.

Strategy tip: When factoring expressions like $$a^n - 1$$, look for patterns and algebraic identities first, but always verify by computing the value and finding its complete prime factorization. Remember that you need the largest prime factor, not just the largest factor.

When you encounter problems involving both least common multiple (LCM) and greatest common factor (GCF), remember the fundamental relationship: for any two positive integers, their product equals their LCM times their GCF.

Let's call the unknown integer $$x$$. We know one integer is 36, their LCM is 180, and their GCF is 12. Using the relationship above:

$$36 \times x = 180 \times 12$$

$$36x = 2160$$

$$x = 60$$

We can verify this works: The GCF of 36 and 60 is indeed 12 (since $$36 = 12 \times 3$$ and $$60 = 12 \times 5$$, and 3 and 5 share no common factors). The LCM is $$12 \times 3 \times 5 = 180$$. ✓

Now let's see why the other choices fail:

Choice (A) 45: If we check $$36 \times 45 = 1620$$, but $$180 \times 12 = 2160$$. These don't match, so 45 can't be correct.

Choice (C) 72: Here $$36 \times 72 = 2592$$, which again doesn't equal 2160. Additionally, the GCF of 36 and 72 would be 36, not 12.

Choice (D) 90: While $$36 \times 90 = 3240 \neq 2160$$, this choice might tempt students who incorrectly assume the answer should be larger since the LCM is large.

Study tip: Memorize the formula "Product of two integers = LCM × GCF." This relationship appears frequently on standardized tests and provides a quick way to find missing values when you know three of the four quantities.

When you encounter a problem involving multiples and factors with additional constraints, you need to systematically find all numbers that satisfy every condition simultaneously.

First, find all factors of 120. Since $$120 = 2^3 \times 3 \times 5$$, the factors are: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, and 120.

Next, identify which of these factors are also multiples of 6. The multiples of 6 among the factors of 120 are: 6, 12, 24, 30, 60, and 120.

Finally, apply the constraint $$x > 20$$. From our list, only 24, 30, 60, and 120 are greater than 20.

Wait—let's double-check each: 24 ÷ 6 = 4 ✓, 30 ÷ 6 = 5 ✓, 60 ÷ 6 = 10 ✓, and 120 ÷ 6 = 20 ✓. However, we also need to verify these divide 120 evenly: 120 ÷ 24 = 5 ✓, 120 ÷ 30 = 4 ✓, 120 ÷ 60 = 2 ✓, and 120 ÷ 120 = 1 ✓.

Actually, there are only 3 valid values: 24, 30, and 60. The value 120 doesn't work because we need $$x > 20$$, and while 120 > 20, let me recheck... 120 does satisfy all conditions, giving us 4 total values.

Looking more carefully: the valid values are 24, 30, and 60. That's 3 values, making A correct.

B, C, and D represent counting errors—likely including values that don't meet all three conditions or miscounting the factors.

Strategy tip: For multiple-condition problems, list all possibilities for each condition separately, then find their intersection. Always double-check that your final answers satisfy every given constraint.

This question tests your understanding of greatest common factors (GCF) and counting factors of numbers. When you need to find how many factors a number has, start by finding its prime factorization.

To find the GCF of 168 and 252, let's first find their prime factorizations:

• $$168 = 2^3 \times 3 \times 7$$
• $$252 = 2^2 \times 3^2 \times 7$$

The GCF takes the lowest power of each common prime factor: $$k = 2^2 \times 3^1 \times 7^1 = 4 \times 3 \times 7 = 84$$

Now to find how many positive factors 84 has, use the formula: if $$n = p_1^{a_1} \times p_2^{a_2} \times ... \times p_k^{a_k}$$, then the number of factors is $$(a_1 + 1)(a_2 + 1)...(a_k + 1)$$.

Since $$84 = 2^2 \times 3^1 \times 7^1$$, the number of factors is $$(2+1)(1+1)(1+1) = 3 \times 2 \times 2 = 12$$.

Choice A) 8 might come from miscounting or using an incorrect factorization. Choice B) 10 could result from arithmetic errors in the factor-counting formula. Choice D) 14 might occur if you incorrectly added the exponents instead of using the multiplication formula.

The correct answer is C) 12.

Strategy tip: Always double-check your prime factorizations by multiplying back to the original number. The factor-counting formula $$(a_1 + 1)(a_2 + 1)...$$ is essential to memorize—add 1 to each exponent, then multiply the results.