What are four consecutive odd integers that the sum of the second and third integers is 48 less than the sum of all 4

Question

The Sum of all four integers must be 48 less than the second and third integers. 

Integers must be consecutive

integers must be odd numbers

Answer 1

Hello Kaylyn,

 

Let the first odd integer be x.

The second odd integer would be x+2, since odd numbers are 2 apart.

The third odd integer would be x+2+2, or x+4.

The fourth would be x+2+2+2, or x+6.

 

Using your criteria,

 

x+2 (2nd term) + x+4 (third term) = x+x+2+x+4+x+6 (sum of all digits) - 48.  Solve for x

 

2x+6 = 4x-36

x=21

 

The first odd digit is 21.  The next is 23, then 25, 27.

 

Checking 23 + 25 = 48.  The sum of all is 21+23+25+27 = 96.  Subtract 48, yields 48, so our digits are correct.

 

Hope this helps.

 

Frank

Answer 2

X= 1st odd number

x+2= 2nd odd number

x+4= 3rd odd number

x+6= 4th odd number

 

Your equation will be:

x+2+x+4= (x+x+2+x+4+x+6)-48  (the sum of the 2nd and 3rd integer is 48 less than the sum of all 4 integers

2x+6= 4x+12-48  (combine like terms)

2x+6=4x-36    (combine like terms)

6=2x-36    (subtract 2x from both sides)

42=2x  (add 36 to both sides)

21=x  (divide both sides by 2)

 

check:  23+25=(21+23+25+27)-48

simplify  48=96-48

simplify  48=48