Question

1. Find four consecutive even integers such that the sum of the first and third is 6 less than the largest

Answer 1

These are indeed quite a lot of exercises, but a lot of them are almost identical - only small computations will change. So, I'm glad to help you with one exercise from every cathegory, but I encourage you to solve the others on your own.

Exercise 1

You can call four consecutive integers as

$x,\ x+1,\ x+2,\ x+3$

So, the sum of the first and third is $x+(x+2) = 2x+2$

We want this quantity to be six less than the largest, i.e. $(x+3)-6 = x-3$

So, the equality is

$2x+2 = x-3$

Subtract x from both sides:

$x+2 = -3$

Subtract 2 from both sides:

$x = -5$

So, the consecutive integers are

$-5,\ -4,\ -3,\ -2$

In fact, the sum of the first and third is $-5-3 = -8$, which is indeed six less than the largest: $-8 = -2-6$

Exercise 2

If you call the first odd number $x$, the next consecutive odd numbers will be $x,\ x+2,\ x+4,\ x+6$

In fact, we have to count skipping two's, because we only want odd integers. From here, you go on like exercise 1: you write the largest (which is x+6), and set it to be two more than the sum of the other three (x, x+2 and x+4)

Exercise 3

By the same logic of exercise 2, two consecutive even integers are $x,\ x+2$, assuming that x is even.

So, you set the equation as usual: the smaller (which is x) is 26 less than three times the larger (which means 3(x+2)-26)

Exercise 4 to 8

These are all pretty identical to exercise 1: you start by listing three or four consecutive integers:

$x,\ x+1,\ x+2\quad\text{or}\quad x,\ x+1,\ x+2\ x+3$

and then you translate the request of each exercise accordingly. Remember that expressions like "three times the second number" means that you have to multiply: $3(x+1)$, while expression like "six more than the first" or "thirteen less than the first" imply adding/subtracting: $x+6$ or $x-13$.

Exercise 9

A multiple of 5 can be written as $5k$, for some integer k.

So, three consecutive multiples of 5 are

$5k, 5(k+1), 5(k+2) = 5k, 5k+5, 5k+10$

We want these three numbers to have a sum of 75. So, we have

$5k, 5k+5, 5k+10 = 75 \iff 15k+15 = 75 \iff 15k = 60 \iff k = 4$

So, the three numbers are

$5k, 5(k+1), 5(k+2) = 5\cdot 4, 5\cdot 5, 5\cdot 6 = 20, 25, 30$