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

11,13,15,17,19

I checked the RSM thingy

:)

Answer 2

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

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 = (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.