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Number Theory
 

This broad category is a popular source for GRE questions. At first, students often struggle with these problems since they have forgotten many of the basic properties of arithmetic. So before we begin solving these problems, let’s review some of these basic properties.
 

“The remainder is r when p is divided by q” means p = qz + r; the integer z is called the quotient. For instance, “The remainder is 1 when 7 is divided by 3” means 7 = 3•2 + 1.
 

Example 1:

When the integer n is divided by 2, the quotient is u and the remainder is 1. When the integer n is divided by 5, the quotient is v and the remainder is 3. Which one of the following must be true?

 

(A) 2u + 5v = 4
(B) 2u – 5v = 2
(C) 4u + 5v = 2
(D) 4u – 5v = 2
(E) 3u – 5v = 2

Translating "When the integer n is divided by 2, the quotient is u and the remainder is 1" into an equation gives
n = 2u + 1

Translating “ When the integer n is divided by 5, the quotient is v and the remainder is 3” into an equation gives
n = 5v + 3

Since both expressions equal n, we can set them equal to each other:
2u + 1 = 5v + 3

Rearranging and then combining like terms yields
2u – 5v = 2

The answer is (B).

 

  • A number n is even if the remainder is zero when n is divided by 2: n= 2z + 0, or n= 2z
  • A number n is odd if the remainder is one when n is divided by 2: n= 2z + 1.
  • The following properties for odd and even numbers are very useful—you should memorize them:

even x even = even
odd x odd = odd
even x odd = even

even + even = even
odd + odd = even
even + odd = odd


[Select One or More Answer Choices]

Example 2:

Suppose p is even and q is odd. Then which of the following CANNOT be an integer?

 

A. p + q p
B.
p q3 
C.
q p2


For a fractional expression to be an integer, the denominator must divide evenly into the numerator. Now, choice A cannot be an integer. Since q is odd and p is even, p + q is odd. Further, since p + q is odd, it cannot be divided evenly by the even number p. Hence,
p + q p cannot be an integer. Next, Choice B can be an integer. For example, if p = 2 and q = 3, then p q 3 = 2 · 3 3 = 2. Finally, Choice C cannot be an integer. p2 = p•p is even since it is the product of two even numbers. Further, since q is odd, it cannot be divided evenly by the even integer p2. Thus, the answer consists of choices A and C.
 

  • Consecutive integers are written as x, x + 1, x + 2, . . .
  • Consecutive even or odd integers are written as x, x + 2, x + 4, . . .
  • The integer zero is neither positive nor negative, but it is even: 0 = 2 x 0.
  • A prime number is a positive integer that is divisible only by itself and 1.
    The prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, . . .
  • A number is divisible by 3 if the sum of its digits is divisible by 3.
    For example, 135 is divisible by 3 because the sum of its digits (1 + 3 + 5 = 9) is divisible by 3.
  • The absolute value of a number, | |, is always positive. In other words, the absolute value symbol eliminates negative signs.
    For example, |-7| = 7 and |-π| = π . Caution, the absolute value symbol acts only on what is inside the symbol. For example, -|-(7 - π)| = -(7 - π) . Here, only the negative sign inside the absolute value symbol but outside the parentheses is eliminated.
  • The product (quotient) of positive numbers is positive.
  • The product (quotient) of a positive number and a negative number is negative.
    For example, –5(3) = –15 and
    6-3=-2
  • The product (quotient) of an even number of negative numbers is positive.
    For example, (–5)(–3)(–2)(–1) = 30 is positive because there is an even number, 4, of negatives.

    -9-2=92is positive because there is an even number, 2, of negatives.
  • The product (quotient) of an odd number of negative numbers is negative.
    For example, (-2)(-π )(-√ 3) = -2π√ 3is negative because there is an odd number, 3, of negatives. 
    (-2)(-9)(-6)(-12)(-182)=-1 is negative because there is an odd number, 5, of negatives.
  • The sum of negative numbers is negative.
    For example, –3 – 5 = –8. Some students have trouble recognizing this structure as a sum because there is no plus symbol, +. But recall that subtraction is defined as negative addition. So –3 – 5 = –3 + (–5).
  • A number raised to an even exponent is greater than or equal to zero.
    For example, (–π)4 = π4 ≥ 0, and x2 ≥0, and 0 x 2 = 0•0 = 0≥0.

 

Example 3:

If a, b, and c are consecutive integers and a < b < c, which of the following must be true?

 

I. b – c = 1
II.
abc/3 is an integer.
III. a + b + c is even.

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only

 

Let x, x + 1, x + 2 stand for the consecutive integers a, b, and c, in that order. Plugging this into Statement I yields

b – c = ( x + 1) – ( x + 2) = –1


Hence, Statement I is false.


As to Statement II,
since a, b, and c are three consecutive integers, one of them must be divisible by 3.


Hence,
abc3 is an integer, and Statement II is true.


As to Statement
III, suppose a is even, b is odd, and c is even. Then a + b is odd since

even + odd = odd


Hence,

a + b + c = ( a + b) + c = (odd) + even = odd


Thus, Statement III is not necessarily true. The answer is (B).

Example 4:

 

If both x and y are prime numbers, which of the following CANNOT be the difference of x and y?

 

(A) 1
(B) 3
(C) 9
(D) 15
(E) 23

 

Both 3 and 2 are prime, and 3 – 2 = 1. This eliminates (A). Next, both 5 and 2 are prime, and 5 – 2 = 3. This eliminates (B). Next, both 11 and 2 are prime, and 11 – 2 = 9. This eliminates (C).

 

Next, both 17 and 2 are prime, and 17 – 2 = 15. This eliminates (D). Hence, by process of elimination, the answer is (E).

 

Example 5:


If -x = -|-(-2 + 5)|, then x =

 

(A) –7
(B) –3
(C) 3
(D) 7
(E) 9

 

Working from the innermost parentheses out, we get

The answer is (C).

 

Example 6:

 

Which two of the following numbers have a product less than –28 ?

 

(A) –1
(B) –2
(C) –5
(D) 14

Since the product must be negative, the only possible products are (–1)(14) = –14, (–2)(14) = –28, and (–5)(14) = –70. Only -70 is less than –28, so the answer consists of Choice (C) and Choice (D).





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