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Joined: 02 Jan 2020
Status:GRE Quant Tutor
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Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
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How to Solve: Even and Odd Numbers
[#permalink]
05 Jan 2024, 21:14
05 Jan 2024, 21:14
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Attachment:
How to Solve: Even and Odd Numbers
Hi All,
I have posted a video on YouTube to discuss Even and Odd Numbers
Attached pdf of this Article as SPOILER at the top! Happy learning!
Following is Covered in the Video
Theory of Even and Odd Numbers
- ⁍ What are Even Numbers ?
⁍ Even Number Problem ?
⁍ What are Odd Numbers ?
⁍ Odd Numbers Problem ?
⁍ Properties of Even and Odd Numbers ?
⁍ Solved Problems
What are Even Numbers
A number which gives 0 as remainder when divided by 2 is an even number.
⁍ Example: 18
⁍ Even numbers end with a units’ digit of 0, 2, 4, 6, 8
⁍ An even number “n” is represented as
n = 2k , where k is an integer
⁍ Example: Consecutive even numbers can be taken as
2k-4, 2k-2, 2k, 2k+2, 2k+4
Even Numbers Problem: Sum of three consecutive even numbers is 60. Find the numbers?
Show: ::
Sol: Let the three numbers be 2k-2, 2k, 2k+2
=> 2k-2 + 2k + 2k+2 = 60
=> 6k = 60
=> k = 10
=> Numbers are 18, 20, 22
=> 2k-2 + 2k + 2k+2 = 60
=> 6k = 60
=> k = 10
=> Numbers are 18, 20, 22
What are Odd Numbers
A number which gives 1 as remainder when divided by 2 is odd number.
⁍ Example: 19
⁍ Odd numbers end with a units’ digit of 1, 3, 5, 7, 9
⁍ An odd number “n” is represented as
n = 2k + 1 or 2k-1 , where k is an integer
⁍ Example: Consecutive odd numbers are
2k-5, 2k-3, 2k-1, 2k+1, 2k+3, 2k+5
Odd Numbers Problem: Sum of 4 consecutive odd numbers is 80. Find the numbers?
Show: ::
Sol: Let the three numbers be 2k-3, 2k-1, 2k+1, 2k+3
=> 2k-3 + 2k-1 + 2k+1 + 2k+3 = 80
=> 8k = 80
=> k = 10
=> Numbers are 17, 19, 21, 23
=> 2k-3 + 2k-1 + 2k+1 + 2k+3 = 80
=> 8k = 80
=> k = 10
=> Numbers are 17, 19, 21, 23
Properties of Even and Odd Numbers
| Addition and Subtraction | |
| -Addition E + E = E E + O = O O + E = O O + O = E Adding Odd number of Odds will give us O Adding even number of Odds will give us E where E -> Even, O -> Odd | -Subtraction E – E = E E – O = O O – E = O O – O = E Subtracting odd number of Odds will give us O Subtracting even number of Odds will give us E where E -> Even, O -> Odd |
| Division and Multiplication | |
| -Division E / E = E or F or O E / O = E or F O / E = F O / O = O or F where E -> Even, O -> Odd, F -> Fraction | -Multiplication E * E = E => \(E^{+ve Integer}\) = E E * O = E O * E = E O * O = O => \(O^{+ve Integer}\) = O where E -> Even, O -> Odd |
⁍ Product of numbers will be even when there is at least one number is Even.
Example: 3*3*2 = 18 = Even as there was one even number 2 on the left side
⁍ Product of numbers will be odd ONLY when all numbers are odd.
Example: 3*3*3 = 27 = Odd as all the numbers on the left side were odd
Solved Problems
Q1. If x, y, and z are integers and x + yz is odd, then which of the following must be true?
I. x + z is even
II. x + y is odd
III. y + z is odd
IV. xy is even
V. yz is even
VI. xz is odd
VII. xyz is even
Show: ::
Sol: x + yz = O
=> We will have four cases
Either x is E and yz is O.
=> Only one case possible
Case 1: x = E, y = O, z = O
Or x is O and yz is E
=> Three cases possible
Case 2: x = O, y = E, z = O
Case 3: x = O, y = O, z = E
Case 4: x = O, y = E, z = E
I. x + y is odd
We have to check for x + y in all the 4 possible cases
For x + y to be odd, one has to even and other has to be odd
But in 3rd case both x and y are odd. So, not possible
II. y + z is odd
Similar logic, in case 1 and case 4 its not possible
III. x + z is even
Similar logic, in case 1, 3,4 its not possible
IV. xy is even
At least one has to even. Not possible in case 3
V. yz is even
At least one has to even. Not possible in case 1
VI. xz is odd
Both have to odd. Not possible in case 1,3,4
VII. xyz is even
At least one has to even. Possible in all the cases
Answer VII
=> We will have four cases
Either x is E and yz is O.
=> Only one case possible
Case 1: x = E, y = O, z = O
Or x is O and yz is E
=> Three cases possible
Case 2: x = O, y = E, z = O
Case 3: x = O, y = O, z = E
Case 4: x = O, y = E, z = E
I. x + y is odd
We have to check for x + y in all the 4 possible cases
For x + y to be odd, one has to even and other has to be odd
But in 3rd case both x and y are odd. So, not possible
II. y + z is odd
Similar logic, in case 1 and case 4 its not possible
III. x + z is even
Similar logic, in case 1, 3,4 its not possible
IV. xy is even
At least one has to even. Not possible in case 3
V. yz is even
At least one has to even. Not possible in case 1
VI. xz is odd
Both have to odd. Not possible in case 1,3,4
VII. xyz is even
At least one has to even. Possible in all the cases
Answer VII
Q2.If x is even, y is odd, z is even, then whether the following are odd or even
I. x + yz
II. x + y + yz
III. xy + z
IV. (x+1)*(y+1)*(z+1)
V. xy*(z+1)
VI. \((x+1)^2*y*(z+1)^3\)
Show: ::
Sol:
I. E + O*E = E + E = E
II. E + O + O*E = E + O + E = O
III. E*O + E = E + E = E
IV. (E + O) * (O + O) * (E + O) = O ( E * O = E
V. E *O * (E + O) = E
VI. \((E+O)^2*O*(E+O)^3\) = O * O * O = O
I. E + O*E = E + E = E
II. E + O + O*E = E + O + E = O
III. E*O + E = E + E = E
IV. (E + O) * (O + O) * (E + O) = O ( E * O = E
V. E *O * (E + O) = E
VI. \((E+O)^2*O*(E+O)^3\) = O * O * O = O
Q3. Product of 4 consecutive numbers will be divisible by all of the following EXCEPT?
A. 6
B. 8
C. 12
D. 24
E. 48?
Show: ::
Sol: Let's take values to solve this. Lets take 4 numbers as 1, 2, 3, 4
Their product = 1*2*3*4 = 24 and will be divisible by all numbers except E
Answer E
Their product = 1*2*3*4 = 24 and will be divisible by all numbers except E
Answer E
Theory: Product of n consecutive numbers will always be divisible by n!
Q4. Sum of three consecutive even numbers is divisible by all of the following EXCEPT
A. 1
B. 2
C. 3
D. 4
E. 6
Show: ::
Sol: Let numbers be 2k-2, 2k, 2k+2
=> Sum = 6k
=> Will be divisible by 6 and all factors of 6
=> Divisible by 1, 2 , 3 and 6
Answer D
=> Sum = 6k
=> Will be divisible by 6 and all factors of 6
=> Divisible by 1, 2 , 3 and 6
Answer D
Q5. Sum of three consecutive odd numbers is divisible by which of the following
A. 2
B. 3
C. 4
D. 5
E. 6
Show: ::
Sol: Let numbers be 2k+1, 2k+3, 2k+5
=> Sum = 6k + 9 = 3*(2k + 3)
=> Will be divisible by 3
Answer B
=> Sum = 6k + 9 = 3*(2k + 3)
=> Will be divisible by 3
Answer B
Hope it helps!
gmatclubot
How to Solve: Even and Odd Numbers [#permalink]
05 Jan 2024, 21:14
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