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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?
  1. 2u + 5v = 4
  2. 2u – 5v = 2
  3. 4u + 5v = 2
  4. 4u – 5v = 2
  5. 3u – 5v = 2
Solution
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

 

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

  1. ​I only
  2. II only
  3. III only
  4. I and II only
  5. 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. 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.

 

The answer is (E).
 

  • 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 × 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, .
     
    For example, . 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 .
  • 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 = π4 ≥ 0, and x2 ≥ 0, and 02 = 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.    bc = 1
II.   abc/3 is an integer.
III.  a + b + c is even.
  1. I only
  2. II only
  3. III only
  4. I and II only
  5. II and III only
Solution
Let x, x + 1, x + 2 stand for the consecutive integers a, b, and c, in that order.
 
Plugging this into Statement I yields
bc = (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).
 
 
Example-4
If both x and y are prime numbers, which of the following CANNOT be the difference of x and y?
  1. 1
  2. 3
  3. 9
  4. 15
  5. 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 =
  1. –7
  2. –3
  3. 3
  4. 7
  5. 9
Solution
Working from the innermost parentheses out, we get
 
 
 
x = –(+3)
 
x = –3
 
x = 3
 
The answer is (C).
 




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