This one looks hard, but it isn't too bad once you think about it a bit.
Two odd numbers that would add up to 1088 and that would be consecutive (meaning if you were looking at the set of ALL odd numbers they would be right next to each other)
To solve these problems you divide the number you have by the number of "rooms."
So, 1088/2. The answer is 2 even numbers which will add to equal 1088.
544 + 544 = 1088
So we need to make them into two odd numbers that are consecutive and EQUAL to the above equation.
So, you can represent it like this:
544 + 544 + 1 - 1 = 1088.
All were doing is adding and subtracting the same number so that they would cancel out in the problem. So it still equals 1088, but now our two numbers are:
Our ANSWER is: Room #s: 545 and 543.
If you wanted to find 4 classroom numbers you would do something very similar.
1088 / 4 = 272
So,
272 + 272 + 272 + 272 = 1088
Now we need 4 numbers. To do that we need to add and subtract 4 different numbers. To keep it consecutive we'll need to do the lowest numbers. So once again we will use 1, and this time we will use 3 as well.
272 + 272 + 272 + 272 + 1 - 1 + 3 - 3 = 1088
So our numbers would be:
269, 271, 273, 275
