Factors Flashcards
The quick method for counting the number of factors of a perfect square remains the same, with a slight twist:
1) Test divisibility of the integer by all integers from 1 to the square root of the integer. Use the rules of divisibility to ask “Is this Integer divisible by 1? by 2? by 3?” etc. If you don’t know the rules of divisibility yet, don’t worry about it - we’ll see those later on.
2) Record the small factors (the numbers that the original integer IS divisible by) in the left side of the table.
3) Count the number of small factors and multiply by 2.
4) Subtract 1 from the total - the square root cannot be counted twice. The result is the number of factors.
If n is a positive integer, is √n an integer?
(1) n has exactly three positive divisors, including 1 and n.
(2) n has an odd number of positive divisors, including 1 and n.
Stat.(1): Recall the factor chart - factors come in pairs, except for Perfect squares which have an odd number of factors, since the last factor is counted only once. Therefore, if n has 3 factors, it must be a perfect square, and √n must be an integer. Stat.(1)->Yes->S->AD.
Alternative method: Plug in numbers for n, and see which ones satisfy stat. (1).
n=cannot be 2, 3, 5, or 7 - prime numbers have only 2 factors: 1 and themselves.
n CAN be 4: factors of 4 are 1, 2 and 4.
n cannot be 6 or 8 - those will have an even number of factors.
n CAN be 9: factors of 9 are 1, 3 and 9.
…Get the pattern?
Stat.(2): Again, recall the factor chart - factors come in pairs, except for Perfect squares which have an odd number of factors, since the last factor is counted only once. therefore, if n has 3 factors, it must be a perfect square, and √n must be an integer. Hence, Stat.(2)->Yes->S->D.
What is the smallest integer that is greater than 500 and is divisible by 8, 30 and 45?
Correct.
To find the LCM (Least Common Multiple) of 2 or more numbers -
- Break down each integer into its “building blocks” using the factor tree.
- Build a list comprised of the least number of prime “building blocks” required to “build” each of the integers.
- Multiply the building blocks in the list to find the LCM.
Remember that each prime number is included the minimum number of times required to build the LCM.
Found the LCM of the three numbers? Good. Now find the first multiple over 500 of that LCM, and you got it.
Factor the three integers:
Note that 2 is counted 3 times , because 8 requires 3 “2”s to make. These three “2’s already include the 2 needed to make 30.
The same goes for the two “3”s needed to make a 45 - they already include the single 3 needed to make a 30.
The LCM is 2×2×2×3×3×5=360.
This isn’t the answer, though, since you’re looking for a multiple that is over 500, thus the next multiple of 360 –> 720 is the multiple you’re looking for.
If a2=60n, where a and n are integers, then n MUST be divisible by which of the following:
A 12
B 15
C 20
D 30
E 60
We move now to some of the tougher questions the integer subject has to offer. Take a look at the following problem:
If a2=60n, where a and n are integers, then n MUST be divisible by which of the following:
A 12
B 15
C 20
D 30
E 60
First, note that the problem supplies an equation. As with all equations, whatever is on one side must equal the other.
Let’s analyze right side. Find the prime factors of 60 using the factor tree - 60=2×2×3×5. The equation now looks like this:
a2=22×31×51×n
It’s clear that a must be an integer with at least 2, 3, and 5 as its prime factors. But note an even more interesting fact: since the left side presents a2, every factor of a must appear twice on the right side. If, for example, a=2×3×5, then a2=a×a=(2×3×5)⋅(2×3×5) = 22×32×52 - twice for each factor. If the left side had a3, the right would need to include every factor three times; a4 on the left = four times on the right, etc.
Back to our equation: a2=22×31×51×n - the right side has 22, but is still missing a 3 and 5 to complete everything to the second power. These must be “supplied” by n, so n MUST include 31×51.
If x and y are integers, and 6^x·3^y=2^6·9^4, what is the value of y?
–> 6x·3y=26·94
2 and 3 are the common factors of all the bases in this equation.
–> (2·3)x·3y=26·(32)4
On the left-hand side, we have a multiplication of two powers with the same base - add the exponents:
- -> 2x·3x·3y=26·38
- -> 2x·3x+y=26·38
Now the bases are the same on both sides, so we can equate the exponents:
- -> x=6,
- -> x+y=8
Plug in the first equation into the second:
- -> 6+y=8
- -> y=2
If a^2=60^n, where a and n are integers, which of the following is the greatest integer that n MUST be divisible by?
A 3
B 5
C 15
D 30
E 60
Sure.
If the question hadn’t included the “greatest”, we’d have 3 right answer choices. Since n MUST include at least 31×51, n must be divisible by 3 (answer choice A), by 5 (answer choice B), by 15=31×51 (answer choice C). The demand for the “greatest number that must divide n” serves to eliminate A, and B and keep C as the only right answer.
To sum up:
Solving questions asking for the greatest number that n must be divisible by:
1) Use the information in the question to find the minimum list of building blocks that you know for sure n MUST be divisible by. Eliminate all answer choices greater than that list. Eliminate any answer choices NOT on that list, if there are any.
2) From the remaining answer choices, eliminate the smaller numbers and choose the greatest.
If x,y and z are positive integers, and 30x=35y=42z, then which of the following must be divisible by 3?
I) x II) y
III) z
Correct.
a. x = 35y/30 = 7y/6.
Since x is an integer, it follows that 7y/6 is an integer. This means that the 6 in the denominator is canceled out; since 7 is not divisible by 6, it follows that y must be divisible by 6, and therefore must also be divisible by 3. Do some constructive POE - since y must be divisible by 3, the correct answer choice must include II. POE A, C and D, since they do not include II.
b. 35y=42z
You already know that y is divisible by 3, so there’s no use to put y over 42 and show that it must be an integer. Doing the opposite y = 42z/35 = 6z/5 proves that z is divisible by 5, but not that it is divisible by 3. z may be divisible by 3, but you cannot prove that it MUST be divisible by 3.
c. 30x=42z
Isolate x: x = 42z/30 = 7z/5. Again, the fact that x is an integer means that z must be divisible by 5, but not that it must be divisible by 3.
Isolate z: z = 30x/42 = 5x/7 - the same. x must be divisible by 7, but you don’t know that it must be divisible by 3.
Therefore, y is the only variable that you can say for sure must b divisible by 3, so the correct answer is II.
If a and b are integers, is ab/8 an integer?
(1) 2a=b
(2) ab/4 is an integer.
Correct.
Stat. (1): Plug in good numbers for a and b, such as a=2 and b=4. These numbers satisfy stat. (1), but a⋅b=8 is divisible by 8, so the answer is “Yes”. However, is this always true, for any number? Try to find values of a and b that satisfy the equation and yield an answer of “No”. If a=1, then b=2, and a⋅b=2, which is not divisible by 8, yielding an answer of “No”. No definite answer, so Stat.(1)->Maybe->IS->BCE.
Stat. (2): If a·b/4 is an integer, then a·b is divisible by 4. Plug in for a·b as a whole : if a·b=4, then a·b is not divisible by 8 and the answer is “No”; however, if a·b=8, then a·b is divisible by 8 and the answer is “Yes”. No definite answer, so Stat.(2)->Maybe->IS->CE.
Stat. (1+2): from stat. (2), a·b is divisible by 4. This eliminates the possibility that a=1 and b=2, as 1·2 is not divisible by 4. Under test conditions, after plugging in several more values of a and b that satisfy both statements, you can reach the conclusion that a·b is divisible by 4 and the answer is “Yes”. Stat. (1+2)->Yes->S->C.
If you’re not sure, take a step back and work the problem algebraically: Plug in b=2a into stat. (2):
ab/4 = a·2a/4 = 2a2/4 = a2/2 is an integer. This means that a2 is divisible by 2. Since a is an integer, a cannot be √2, so a2 must be divisible by 4 and a must be divisible by 2.
Plug this information into stat. (1): Since b=2a, if a is divisible by 2, then b must be divisible by 2·2, and a·b must be divisible by 2·4=8.
If a is an integer and (a^2)/(123) is odd, which of the following must be an odd integer?
a/4
a/12
a/27
a/36
a/72
Think about what value a could be. There are two things we know about a2, which we can use to infer about a:
1) a2 is divisible by 123, so it must contain all of the prime factors of 123 (at least) in order to be able to reduce the top of the fraction with the bottom and come up with an odd integer result.
2) a2 must be a perfect square, i.e. must have an integer square root.
Break down 123 : (4⋅3)3 = 43⋅33 = (22)3⋅33 = 26⋅33.
So a2 should at least equal 26⋅33. Can we take the square root of 26⋅33 and get an integer result? Taking the square root is the equivalent of multiplying the powers by 1/2, so if we take the square root of a2 = 26⋅33 we get a=23⋅33/2 - which is not an integer.
square root of 26 is √26 = 26/2 = 23. That’s fine - the problem is with the 33 - the odd power doesn’t allow an integer square root. This means that a2 must include at least an extra 3, to make a2 = 26⋅34, leaving a as √(26⋅34) = 23⋅32= 8⋅9 = 72. Which works, since in this case (a2)/(123) = 26⋅34 / 26⋅33 = 3, which is indeed an odd integer, as the question requires.
All this leads us to the conclusion that a COULD equal 72 - it’s not the only value of a, but it’s a valid plug in. According to the question stem, one of the answer choices must always result in an odd integer for every value of a. Work with what you have: plug in a=72 into the answer choices and see if you get an odd result. If you don’t, then eliminate the answer choice, as we’ve found a single counter example of a that does not give an odd result for that answer choice.
72/12 = 6 -> not odd. POE this answer choice because a/12 is not necessarily an odd integer.
A 72/4 = 18 - not odd
B 72/12 = 6 - not odd.
C 72/27 = that’s not even an integer.
D 72/36 = 2 - not odd.
E 72/72 = 1 - the only odd answer. Since this is enough to eliminate answer choice A, B, C and D, E remains as the only right answer.
