Divisibility and Prime

At the Canterbury Dog Fair, 1/4 of the poodles are also show dogs and 1/7 of the show dogs are poodles. What is the least possible number of dogs at the fair?

10. If 1/4 of the poodles are also show dogs, the number of poodles must be divisible by 4. (The number of dogs is necessarily an integer.) Since the least possible number is the goal, try an example with 4 poodles. If 1/7 of the show dogs are poodles, the number of show dogs must be divisible by 7. Since the least possible number is the goal, try an example with 7 show dogs. So far there are: 4 poodles, 1 of which is a show dog 7 show dogs, 1 of which is a poodle Note that the one poodle that is also a show dog is the same dog as the one show dog that is also a poodle! To get the total number of dogs, only count that dog once, not twice. In total: 3 poodles (non-show dogs) 1 dog that is both poodle and show dog 6 show dogs (non-poodles) This equals 10 dogs in total. This example met all the constraints of the question while using minimum values at each step, so this is the least possible number of dogs at the fair.

If p is divisible by 7 and q is divisible by 6, pq must have at least how many factors greater than 1? (A) 1 (B) 3 (C) 6 (D) 7 (E) 8

27. (D). This problem is most easily solved with an example. If p = 7 and q = 6, then pg = 42, which has the factors 1 & 42, 2 & 21, 3 & 14, and 6 & 7. That's 8 factors, but read carefully! The question asks how many factors greater than 1, so the answer is 7. Note that choosing the smallest possible examples (p = 7 and q = 6) was the right move here, since the question asks "at least how many factors...?" If testing p = 70 and q = 36, many, many more factors would have resulted. The question asks for the minimum.

How many factors greater than 1 do 120, 210, and 270 have in common? (A) 1 (B) 3 (C) 6 (D) 7 (E) 30

Find their prime factors 120: 2²*3*5 210:2*3*7*5 270:2*3³*5 Common Factors: 2*3*5 -> these are prime factors So now multiply them to get all the seven factors

n is divisible by 14 and 3. Which of the following statements must be true? Indicate all such statements. 12 is a factor of n 21 is a factor of n n is a multiple of 42

II and III only. Since n is divisible by 14 and 3, n contains the prime factors of both 14 and 3, which are 2, 7, and 3. Thus, any numbers that can be constructed using only these prime factors (no additional factors) are factors of n. Since 12 = 2 × 2 × 3, you CANNOT make 12 by multiplying the prime factors of n (you would need one more 2). However, you CAN construct 21 by multiplying two of the known prime factors of n (7 × 3 = 21), so the second statement is true. Finally, n must be at least 42 (= 2 × 7 × 3, the least common multiple of 14 and 3), so n is definitely a multiple of 42. That is, n can only be 42, 84, 126, etc...

New cars leave a car factory in a repeating pattern of red, blue, black, and gray cars. If the first car to exit the factory was red, what color is the 463rd car to exit the factory? (A) red (B) blue (C) black (D) gray (E) It cannot be determined from the information given.

Remember when we completely divide it's the last number 1,2,3,4 so 4th is the grey one similiarly 5,6,7,8 8 being the grey one and 9th will be red

The number of students who attend a school could be divided among 10, 12, or 16 buses, such that each bus transports an equal number of students. What is the minimum number of students that could attend the school?

The number of students must be divisible by 10, 12, and 16. So the question is really asking, "What is the least common multiple of 10, 12, and 16?" Since all of the answer choices end in 0, each is divisible by 10. Just use your calculator to test which choices are also divisible by 12 and 16. Because you are looking for the minimum, start by checking the smallest choices. Since and are not integers, the smallest choice that works is 240.

What is the remainder when 1317 + 1713 is divided by 10?

The remainder when dividing an integer by 10 always equals the units digit. You can also ignore all but the units digits, so the question can be rephrased as: What is the units digit of 317 + 713? The pattern for the units digits of 3 is [3, 9, 7, 1]. Every fourth term is the same. The 17th power is 1 past the end of the repeat: 17 - 16 = 1. Thus, 317 must end in 3. The pattern for the units digits of 7 is [7, 9, 3, 1]. Every fourth term is the same. The 13th power is 1 past the end of the repeat: 13 - 12 = 1. Thus, 713 must end in 7. The sum of these units digits is 3 + 7 = 10. Thus, the units digit is 0.

When x is divided by 10, the quotient is y with a remainder of 4. If x and y are both positive integers, what is the remainder when x is divided by 5? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4

This is a bit of a trick question—any number that yields remainder 4 when divided by 10 will also yield remainder 4 when divided by 5. This is because the remainder 4 is less than both divisors, and all multiples of 10 are also multiples of 5. For example, 14 yields remainder 4 when divided either by 10 or by 5. This also works for 24, 34, 44, 54, etc.