Find the Integers

[nycmathguy]

I want to know if the set up is correct.

There are three consecutive odd integers. Three times the largest is seven times the smallest. Find the integers.

Set Up:

Three consecutive odd integers:

2n + 1, 2n + 3, 2n + 5.

Let 2n + 5 = largest integer.

Let 2n + 1 = smallest integer.

3(2n + 5) = 7(2n + 1)

Yes?

 

[MathLover1]

correct

 

[nycmathguy]

correct

Ok. I wasn't too sure about this one. Not sure? After months playing with Precalculus, not sure?

 

[Country Boy]

You could have written the "three consecutive odd integers" as x, x+ 2, and x+ 4. 3(x+ 4)= 7x.
3x+ 12= 7x
Subtracting 3x from both sides:
4x= 12
Dividing both sides by 4:
x= 3.

The three consecutive odd integers are 3, 5, and 7.

Check: "three times the largest" is 3(7)= 21 and "seven times the smallest" is 7(3)= 21.

 

[nycmathguy]

You could have written the "three consecutive odd integers" as x, x+ 2, and x+ 4. 3(x+ 4)= 7x.
3x+ 12= 7x
Subtracting 3x from both sides:
4x= 12
Dividing both sides by 4:
x= 3.

The three consecutive odd integers are 3, 5, and 7.

Check: "three times the largest" is 3(7)= 21 and "seven times the smallest" is 7(3)= 21.

So, am I right or wrong? Let me guess: WRONG! SHAME ON ME!!

 

[MathLover1]

So, am I right or wrong? Let me guess: WRONG! SHAME ON ME!!

the general form of an odd integer can be expressed as n is n=2k+1, where k is an integer, and a pair of any consecutive odd integers are also 2 units apart
2k+3 is the second
2k+5 is the third odd integer

 

[MathLover1]

So, am I right or wrong? Let me guess: WRONG! SHAME ON ME!!

you are right

 

[nycmathguy]

you are right

Thank you. This Country Boy guy can't stop criticizing my effort to present correct calculations.
Remember: Precalculus on my days off only.

 

[Country Boy]

I pointed out a different way of doing the problem. That is NOT "criticism"!

If you write the three consecutive odd integers as n, n+ 2, and n+ 4 then "three times the largest is seven times the smallest is "3(n+ 4)= 7n". 3n+ 12= 7n. 4n= 12. n= 3.

Your formulation gave "3(2n + 5) = 7(2n + 1)". 6n+ 15= 14n+ 7. 15= 8n+ 7. 8n= 8. n= 1. But you had written the first number as 2n+ 1 so the first number is 2+ 1= 3, as before.

I would also encourage you (not intended as criticism!) to check the answer:
The three consecutive integers are 3, 5, and 7. "3 times the largest" is 3 times 7= 21. "7 times the smallest" is 7 times 3= 21.

 

[nycmathguy]

I pointed out a different way of doing the problem. That is NOT "criticism"!

If you write the three consecutive odd integers as n, n+ 2, and n+ 4 then "three times the largest is seven times the smallest is "3(n+ 4)= 7n". 3n+ 12= 7n. 4n= 12. n= 3.

Your formulation gave "3(2n + 5) = 7(2n + 1)". 6n+ 15= 14n+ 7. 15= 8n+ 7. 8n= 8. n= 1. But you had written the first number as 2n+ 1 so the first number is 2+ 1= 3, as before.

I would also encourage you (not intended as criticism!) to check the answer:
The three consecutive integers are 3, 5, and 7. "3 times the largest" is 3 times 7= 21. "7 times the smallest" is 7 times 3= 21.

A different way is ok by me.