Among the integers 1 to 30 any 2 intergents are taken such that the second one is bigger than first one and their sum is divisible by 2 .

Question

How many possible ways to do this? 

Answer 1

1, 2, 3, 4, ..., 27, 28, 29, 30

15 are odd integers and 15 are even integers

working with the odd integers...

1,3,5,7,...,23,25,27,29

an odd integer plus an odd integer gives you an even integer

an even integer plus an even integer gives you an even integer

all even integers are divisible by 2

1 can be paired with 14 other larger integers (1+3, 1+5, 1+7,...1+29)

3 can be paired with 13 other larger integers (3+5, 3+7, 3+9,...3+29)

5 can be paired with 12 other larger integers

7 can be paired with 11 other larger integers

do you see the pattern ?

9 with 10 others

11 with 9 others

and so on until you come to 25 paired with 2 others and 27 paired with 1 other

so this amounts to adding 14+13+12+11+10+9+8+7+6+5+4+3+2+1=105 sums of odd integers

do the same for the even integers

2,4,6,8,...24,26,28,30

2 is paired with 14 other larger integers

4 is paired with 13 other larger integers

6 is paired with 12 other larger integers

and so on

you end up adding 14+13+12+...+3+2+1=105 sums of even integers

105+105=210 sums altogether

Answer 2

There are 210 such pairs.

Method 1 (computer program):
   Count = 0
   FOR A = 1 TO 30
     FOR B = (A+1) TO 30 /* B is greater than A; both 1 to 30
       IF MOD( (A+B), 2) = 0 THEN /* Divisible by 2
         COUNT = COUNT+1
     NEXT B
   NEXT A
   OUTPUT Count

Method 2 (analysis and math):
  The sum of two odd numbers is even (that is, divisible by 2).
  The sum of two even numbers is even (that is, divisible by 2).
  The sum of an even number and an odd number is odd (that is, not divisible by 2).
 
  From 1 to 30 the odd numbers are 1, 3, 5, 7, … 29.
    For each of them (as x), there can be associated (29-x)/2 odd numbers.
    That’s 14, 13, 12, 11, 10, …, 1
  From 1 to 30, the even numbers are 2, 4, 6, … 28.
    For each of them, there can be associated (30-x)/2 even numbers.
    That’s 14, 13, 12, 11, 10, …, 1

 Reversing the order:,
  That makes (1+2+…+14)+(1+2+…+14) potential larger values.

 Gauss’ famous equation for the sum of the numbers from 1 to n is (n)(n+1)/2.
  So, the sum of 1+2+…+14) is 14*15/2 = 105.
  And, twice that sum is 210.

Answer 3

It's a good question, worthy of thought. Here are some hints for you to build your own thoughts around.

This type of problem can only benefit a student who gets involved; as Thomas Edison said, you learn so much by so much failure (of experiments).

 

First, I will assume that you meant to talk about INTEGERS (proper spelling). Next, if you choose two numbers with the proviso that the second one chosen is > the first one chosen, this means that the first choice may have any result (call it x) from 1 to 29.  The second choice (call it y) is restricted to numbers ranging from x+1 to 30.

 

The overall choice is restricted by the requirement that x+y must be EVEN--either both x & y are odd, or both are even, therefore.

 

Work with these ideas and develop a pattern to represent this problem. You might make charts showing all possibilties, and ultimately compute/count the number of ways in which this can be done.