Question

There are five consecutive odd integers. The sum of the three smallest is 3 more than the sum of the two largest. Find the integers.

Answer 1

Answer:

The answer is "$\bold{11, \ 13, \ 15, \ 17, \ 19}$".

Step-by-step explanation:

In word problem, it is solved by 2k+1, among the most common forms of even an, unlike number.

Let another odd integer become 2k+1.....(a)

(2K+1)+(2k+3)+(2k+5)=(2k+7)+(2k+9)+3(to take the left side 3 extra)  

Quality for the k:

$\to 6k+ 9 = 4k +19 \\\\\ \to 6k-4k= 19+9 \\\\ \to 2k = 10 \\\\ \to k = \frac{10}{2}\\\\ \to K = 5$

put the value of k, in the equation (a):

$\to 2k+1 \\ \to 2(5)+ 1 \\ \to 10 + 1\\\to 11$

There are also 11, 13, 15, 17, and 19 episodes.  

Let's test it! Let's test it. 11+13+15=39  

17+ 19 = 36.  

So we're right, the first 3 integer numbers are 3 more than the last 2.

Answer 2

Answer:

11,13,15,17,19

I checked the RSM thingy

:)