# Number Theory

This broad category is a popular source for test 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 — 2 + 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?*

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

*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

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.****A***common multiple*is a multiple of two or more integers.**The***least**common multiple*(LCM) of two integers is the smallest positive integer that is a multiple of both.^{2}

10 = 2 • 5^{2}instead of 2 because it has the greater number of factors of 2, and select 5 by default since there are no other factors of 5. Hence, the LCM is 2^{2}• 5 = 4 • 5 = 20.

8 = 2^{3}

36 = 2^{2}• 3^{2}

54 = 2 • 3^{3}^{3}because it has more factors of 2 than 2^{2}or 2 itself, and select 3^{3}because is has more factors of 3 than 3^{2}does. Hence, the LCM is 2^{3}• 3^{3}= 8 • 27 = 216.

A shortcut for finding the LCM is to just keep adding the largest number to itself until the other numbers divide into it evenly. For 4 and 10, we would add 10 to itself: 10 + 10 = 20. Since 4 divides evenly in 20, the LCM is 20. For 8, 36, and 54, we would add 54 to itself: 54 + 54 + 54 + 54 = 216. Since both 8 and 36 divide evenly into 216, the LCM is 216.**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 — 0 = 0 ≥ 0.

*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.*

*I only**II**III only**I and II only**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, abc/3 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).

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

- 1
*3**9**15**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).

*If , then x =*

- –7
*–3**3**7**9*

Working from the innermost parentheses out, we get

–x = –(+3)

–x = –3

x = 3

The answer is (C).