Chapter 5.3, Problem 43PPS

To determine

Find the sets of all four consecutive odd positive integers with a sum less than 42.

Answer to Problem 43PPS

The set of 3 positive odd numbers are (1,3,5,7) , (3,6,7,9,) , (5,7,9,11) , (7,9,11,13)

Explanation of Solution

Given: Find the sets of all four consecutive odd positive integers with a sum less than 42

Concept Used:

Difference of consecutive even integers is 2.

Condition: Sum of all four consecutive odd positive integers less than 42.

Let the least positive odd number be x, the other three are: $x+2;x+4;x+6$

Their sum: $x+x+2+x+4+x+6$

Equation: $x+x+2+x+4+x+6<42$

Calculation:

Equation: $x+x+2+x+4+x+6<42$

Solve for x, the least number of the set.

  $\begin{array}{l}x+x+2+x+4+x+6<42\\ 4x+12<42\\ 4x+12-12<42-12\text{[}\text{Subtract}12\text{from}\text{both}\text{sides}\text{]}\\ 4x<30\\ \frac{4x}{4}<\frac{30}{4}\text{[}\text{Divide}\text{both}\text{sides}\text{by}4\text{]}\\ x<7.5\end{array}$

So, the least positive odd number be no more than 7.5.

All possible set of four odd numbers whose sum is less than 42:

First set = (1,3,5,7) Sum =   $1+3+5+7=16<42$ Second set = ( 3 , 5 , 7 , 9 ) Sum =   $3+5+7\text{ + 9}=24<42$ Third set = ( 5 , 7 , 9 , 11 ) Sum =   $5+7\text{ + 9 +11}=32<42$ Fourth set = ( 7 , 9 , 11 , 13 ) Sum =   $7\text{ + 9 +11 +13}=40<42$

Where   $x=1$ (the first positive odd integers less than 7.5. Any number less that 7.5 would satisfy the condition. )   $x=3$   $x=5$   $x=7$

So, there are possible 4 odd integers whose sum is less than 42 are:

(1,3,5,7) , (3,6,7,9,) , (5,7,9,11) , (7,9,11,13)

Thus, the set of 3 positive odd numbers are (1,3,5,7) , (3,6,7,9,) , (5,7,9,11) , (7,9,11,13)