Number Theory Flashcards
If m = 44n/ 3,630 and m and n are integers, then what is the least possible value of n?
55 75 165 330 825
Answer = 165
Prime factorization of 44n and 3,630 help here
If the remainder when positive integer y is divided by 24 is 14, then which of the following must be a factor of y?
31 19 7 3 2
So if you say 24 goes into it once and then you add 14 more you have 38
So what are the factors of 38? = 1 * 38 and 2 * 19
Then you can try another number so what is 24 goes into the number twice and then add 14 so = 62 now what are the factors of 62 = 1 * 62 and 2 * 31 so since 2 is going into both questions, answer = 2
When a GRE question involves factors or multiples, what should you do?
Rewrite the numbers in the problem as the product of their primes
The remainder when 5 is divided by 9
9 divides into 5 zero times, so 5 is left over.
answer = 5
The remainder when 11 is divided by 3 is y. If 24 is divided by y, the remainder is?
11 divided by 3 is 3 with 2 left over, so y = 2. If 24 is divided by y (which y = 2) it divides evenly and there is no remainder so the answer = 0.
The remainder when 29 is divided by x is 1, and the remainder when 66 is divided by x is 3, so x = ?
Subtract the remainder from 29 and then consider the factors of the resulting number. So 29 - 1 = 28. The factors of 28 are 1, 2, 4, 7, 14 and 28.
Now do the same for 66. So, 66 - 3 = 63. The factors of 63 are 1, 3, 7, 9, 21, and 63.
The only number that is common in both and meets the criteria of the question is 7. So, the answer = 7
The remainder when positive integer x is divided by 6 is 4. If x > 4, then the value of x is ?
Find the value of x. An easy way to do that is to add the divisor (6) and the remainder (4) so 6 + 4 = 10 and x could = 10, which is one of the options. Since x is really 4 more than a multiple of 6, other multiples of 6 could also work. So, x could also equal 12 +4 = 16 or 18 + 4 = 22 (using different multiples of 6) However, since the question asks for the least value of x, the correct answer = 10.
Remainders are the most common number theory questions on the GRE !!!
If the remainder when integer x is divided by 4 is 3 and the remainder when integer y is divided by 4 is 2, what is the remainder when x + y is divided by 4?
x/4 Remainder 3, so x = 4 + 3 = (7)
y/4 Remainder: 2, so y = 4 + 2 = (6)
x + y = 13
(x + y) / 4: Remainder = 1
an easy way to calculate x and y is to add the remainder to the divisor so in this case the divisor is 4 and then you add each remainder to it.
7 / 3 = 2 remainder 1
in words “seven divided by three”
is asking?
how many times does 3 go into 7? The remainder is the whole number that is left over so the answer is = 2 remainder 1
fifteen divided by four ?
15 / 4 = 3 remainder 4
How many groups of 4 can you fit into 15? Then what is the whole number that is left over
25 / 7 = ?
= 3 with remainder 4
30 / 7 = ?
7 * 4 = 28
so answer = 4 remainder 2
41 / 6 = ?
Answer = 6 remainder 5
3771 is divided by 8 what is your remainder?
what you are dividing goes under the shape and what you are dividing by goes outside the shape
the number x “is divided by” y, which number goes on the outside?
y goes on the outside
If x and y are positive integers and 15x = sqrt(735y), what is the least possible value of x + y?
15x = sqrt(735y)
Start by re writing the numbers in the problem as the product of their prime factors. So, 15 = 3 * 5 and 735 = 3 * 5 * 7 * 7 and then rewrite the equation as (3 * 5)x = sqrt(( 3 * 5 * 7 * 7)y)
Next, because the question says that x and y are positive integers, in order for the square root of a number to be an integer, there must be an even number of each number under the radical sign. There are already two 7’s and since 7^2 = 49 and the sqrt(49) = 7, the radical can be rewritten, (3 * 5)x = 7sqrt((3 * 5)y) and then you know x = 7.
So, (3 * 5) = sqrt((3 * 5)y)
To solve for y, square both sides of the equation.
You get, 3^2 * 5^2 = 3 * 5y Then divide each side of the equation by 3*5 and you get y = 3 * 5
x = 7
y = 15
x + y = 22
In order for the square root of a number to be an integer, there must be an even number of each number under the radical sign
Even integers, when divided by 2 produce a remainder of what?
zero
Odd integers, when divided by 2 produce a remainder of what?
1
Which of the following numbers are both multiples of 3 and factors of 81?
3 6 9 18 27 81
To find if they are factors of 81, what you can do is prime factor 81 and also prime factor each of the answer options.
The prime factorization of 81 is (3 * 3 * 3 * 3)
prime factor or 3 is 3 - is there at least one 3 in the prime factors of 81? yes
prime factors of 6 is 3 * 2 is there a 3 in the pf of 81? yes , but is there a 2, no so 6 is not a factor of 81
prime factors of 9 yes, there are at least two threes in 81
prime factors of 18 are 2 * 3 * 3, in 81 there are two 3’s but no 2 so 18 is not a factor of 81
prime factor of 27 is 3^3 so yes
and 81 yes
Take any number that the divisor divides into evenly
So if you are looking for something that when divided by 4 gives you a remainder of 3 for example think about 8 + 3 = 11
11/ 4 = 2 remainder 3
remainders are the numerators in the fraction part of a mixed number
