Greatest Common Factor - SAT Math

Since v is divisible by 2, 3 and 15, v must be a multiple of 30. Any number that is divisible by both 2 and 15 must be divisible by their product, 30, since this is the least common multiple.

Out of all the answer choices, v + 30 is the only one that equals a multiple of 30.

The greatest common divisor of and is 10. This means that the prime factorizations of and must both contain a 2 and a 5.

The greatest common divisor of and is 9. This means that the prime factorizations of and must both contain two 3's.

The greatest common divisor of and is 8. This means that the prime factorizations of and must both contain three 2's.

Thus:

We substitute these equalities into the given expression and simplify.

Since and are two-digit integers (equal to and respectively), we must have and . Any other factor values for or will produce three-digit integers (or greater).

is equal to , so could be either 1 or 2.

Therefore:

or

Greatest common factor is a common factor shared by two or more numbers. Both numbers are even, so let's divide both numbers by two. We get . These are prime numbers (factors of one and itsef) in which we are done. Anytime we have two prime numbers or one prime and one composite number, we are finished. So the greatest common factor is .

Greatest common factor is a common factor shared by two or more numbers. is a multiple of , so let's divide for both numbers. We get . We are finished as these are the basic numbers. So the greatest common factor is .

Greatest common factor is a common factor shared by two or more numbers. is a prime number. is a composite number. Since we have a prime and composite number, the greatest common factor is .

Greatest common factor is a common factor shared by two or more numbers. Both numbers are even, so let's divide two for both numbers. We get . We have one prime and one composite number, so we are finished. The greatest common factor is .

Greatest common factor is a common factor shared by two or more numbers. If you know the divisibility rule of (sum of digits are divisible by ), then the answer is just as the quotient is . We have a prime and composite number. However, if you don't and only know the divisibility rule of , then we can divide both numbers by to get . We do it once more to get . Since we divided twice by , we multiply these factors and this is our greatest common factor of . Our answer is .

Greatest common factor is a common factor shared by two or more numbers. If you know divisibility rule of , then this is the answer. However, this isn't easy to spot, so we will do process of elimination. The numbers are odd and if we have even factors, we never generate odd numbers so is wrong. Next, check divisibility rule of . The digits of add to which isn't divisible by so is wrong. Next, let's divide into . We get a decimal value and that's wrong since if we consider to be a multiple of , it should be a whole number and not a decimal. Finally, by dividing and , it's also . This is our answer. To find out if is divisible by , just add the outside digits and match the middle one. Since it does, is divisible by .

These two numbers are definitely divisible by . When we divide both numbers by , we get and remaining. Since we have a combination of a prime and composite number, then we can't find any more factors. Our answer is .

Let's do some divisibility rules. For , the sum of the digits must be divisible by .

We have:

They are both good so when we divide both numbers by , we get . Lastly they are both divisible by . So we multiply both factors to get an answer of .

Greatest common factor involves all the numbers in the set. Even though three of the numbers are divisble by , isn't. The only factor that satisfies all the numbers is .

Although the first three numbers are divisible by , doesn't divide evenly into . The next best factor is . The remainder will be . This won't go any further as most of the numbers are even except . Our final answer is just .

We know all of the numbers are divisible by so when we divide all the numbers by , we have . Next, we can divide al of them by , because the sum of the digits of all numbers are divisible by . So we get . This is as best as we can go so now we multiply the factors to get as an answer.

Because they are all even and divisible by , we can divide each number to get . Next, let's divide by to get . We are finished as we have a mixture of prime and composite numbers. We multiply the factors to get .

The key here is to find the prime factorizations of both numbers and multiplying the common prime factors together:

Both prime factorizations have in common, so is our answer.