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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.
 
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?
  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.
  • A common multiple is a multiple of two or more integers.
     
    For example, some common multiples of 2 and 5 are 0, 10, 20, 40, and 50.
  • The least common multiple (LCM) of two integers is the smallest positive integer that is a multiple of both.
     
    For example, the LCM of 4 and 10 is 20. The standard method of calculating the LCM is to prime factor the numbers and then form a product by selecting each factor the greatest number of times it occurs. For 4 and 10, we get
     
    4 = 22
    10 = 2 • 5
     
    In this case, select 22 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 22 • 5 = 4 • 5 = 20.
     
    For another example, let’s find the LCM of 8, 36, and 54. Prime factoring yields
    8 = 23
    36 = 22 • 32
    54 = 2 • 33
     
    In this case, select 23 because it has more factors of 2 than 22 or 2 itself, and select 33 because is has more factors of 3 than 32 does. Hence, the LCM is 23 • 33 = 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, .
     
    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 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.

  1. I only
  2. II
  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

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).
 

 
Example-4
If both x and y are prime numbers, which one 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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