FInd all sets of three consective odd integers whose sum is between 20 and 30?

Answer 1

Let n be the smallest in which tripplet:

20 <= n + (n+2) + (n+4) <= 30

20 <= 3n + 6 <= 30 // -6

14 <= 3n <= 24 // divide by 3

4 2/3 <= n <= 8

The Smallest n to satisfy 4 2/3 <= n is 5
The greatest which satisfies n<=8 is 7

5 + 7 + 9 = 21
7 + 9 + 11 = 27

Answer 2

If a, b and c are consecutive odd integers then a=b-2 and c=b+2
So a+b+c = 3b

So for a+b+c to be between 20 and 30 we actually just need to find values for b such that 3b is between 20 and 30, where b is odd

3x7 = 21 => (5+7+9) = 21 is a solution
3x9 = 27 => (7+9+11) = 27 is a solution

No other odd numbers which meet the criteria 20 <=3b <= 30 and b is odd

Answer 3

x+(x+2)+(x+4)= Anything between 20 and 30

x can equal 5 or 7. That's all

So, it could be 5, 7, and 9
Or 7, 9, 11

If you look, x has to be an odd integer. So, the next odd integer would have to be x+2. For example, 3 , 3+2=5

After that, the next odd integer would be x+4, then x+6, so on


If you had to solve without guessing, you would do this:

x+(x+2)+(x+4)= 20,21,22,23,24,25,26,27,28,29,or 30 or

x+x+2+x+4= 20,21,22,23,24,25,26,27,28,29,or 30 or
combine like terms(x with x, numbers with numbers)

3x+6= 20,21,22,23,24,25,26,27,28,29,30

Check and see which ones are true, which are:

3x+6= 27 or
3x+(6-6)=(27-6) subtract 6 from both sides
3x=21
x=7

also

3x+6= 21 or
3x+(6-6)=(21-6) subtract 6 from both sides
3x=25
x=5

Answer 4

5,7,9
7,9,11

Note that since the sum is 3 times the middle number, this is equivalent to finding all odd numbers than lie between 20/3 = 6 2/3 and 30/3 = 10.

Answer 5

OK. Let the 3 integers be 2n+1, 2n+2, and 2n+5. Now, 3 odd integers will be an odd sum, so solve
(2n+1)+(2n+3)+(2n+5) = k for k={21, 23, 25, 27, 29} and you're all done.

Doug

Answer 6

2n-1 + 2n + 1 + 2n + 3 = 6n + 3

20<=6n+3<=30

Substracting 3:

17 <= 6n <= 27

17 is not a multiple of 6, so, 17 <= 17 acutally means 18 <= 6n, so 3<= n

6n<= 27 means that 6n <= 24, so n <= 4

So:

3<= n <= 4, and 3 and 4 are the only possible n. Now plug both values of n in 2n-1, 2n+1 and 2n+3 and you are done.

Ana