# Number Theory

This broad category is a popular source for GMAT 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 k”*means*p*=*kq*+*r*; the integer*q*is called the quotient. For instance,*“The remainder is 1 when 7 is divided by 3”*means 7 = 3 x 2 + 1.

Example

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?- 2u + 5v = 4
- 2u – 5v = 2
- 4u + 5v = 2
- 4u – 5v = 2
- 3u – 5v = 2

Solution

Translating
Translating
Since both expressions equal
Rearranging and then combining like terms yields
The answer is (B).

*“When the integer n is divided by 2, the quotient is u and the remainder is 1”*into an equation gives*n*= 2

*u*+ 1

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

*v*+ 3

*n*, we can set them equal to each other:2

*u*+ 1 = 5*v*+ 32

*u*– 5*v*= 2

- A number
*n*is even if the remainder is zero when*n*is divided by 2:*n*= 2*z*+ 0, or*n*= 2*z*. - A number
*n*is odd if the remainder is one when*n*is divided by 2:*n*= 2*z*+ 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*

Example-2

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

*I*. (p + q)/p

*II. pq/3
III. q/p2*

- I only
- II only
- III only
- I and II only
- I and III only

Solution

For a fractional expression to be an integer, the denominator must divide evenly into the numerator.

Now, Statement I 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, cannot be an integer.

Next, Statement II can be an integer.

For example, if *p* = 2 and *q* = 3, then .

Finally, Statement III cannot be an integer. *p*^{2} = *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 *p*^{2}.

The answer is (E).

**C****onsecutive 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 × 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, and . Caution, the absolute value symbol acts only on what is inside the symbol, .**The product (quotient) of positive numbers is positive.****The product (quotient) of a positive number and a negative number is negative.****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 positives.**

is positive because there is an even number, 2, of positives.**The product (quotient) of an odd number of negative numbers is negative.****For example, is negative because there is an odd number, 3, of negatives. 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}=*π*^{4}≥ 0, and*x*^{2}≥ 0, and 0^{2}= 0 x 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.

II.

III.

*b*–*c*= 1II.

*abc*/3 is an integer.III.

*a*+*b*+*c*is even.- I only
- II only
- III only
- I and II only
- II and III only

Solution

Let
Plugging this into Statement I yields

Hence, Statement I is false.
As to Statement II, since
Hence,
As to Statement III, suppose
Then
Hence,
Thus, Statement III is not necessarily true.
The answer is (B).

*x*,*x*+ 1,*x*+ 2 stand for the consecutive integers*a*,*b*, and*c*, in that order.*b*–*c*= (*x*+ 1) – (*x*+ 2) = –1*a*,*b*, and*c*are three consecutive integers, one of them must be divisible by 3.*abc*/3 is an integer, and Statement II is true.*a*is even,*b*is odd, and*c*is even.*a*+*b*is odd since*even*+

*odd = odd*

*a*+

*b*+

*c*= (

*a*+ b) +

*c*= (

*odd*) +

*even*=

*odd*

Example-4

If both

*x*and*y*are prime numbers, which of the following CANNOT be the difference of*x*and*y*?- 1
- 3
- 9
- 15
- 23

Solution

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 , then

*x*=- –7
- –3
- 3
- 7
- 9

Solution

Working from the innermost parentheses out, we get
–
–
The answer is (C).

*x*= –(+3)*x*= –3*x*= 3