SAT math problem with calculator help

<p>x = the sum of all odd integers from 1 to 49, inclusive
y = the sum of all even integers from 2 to 50, inclusive
z = the sum of all integers from 1 to 48, inclusive
Based on the definition of x, y, and z above, what is the value of x + y - z?</p>

<p>Can somebody show me how to do this with a calculator and without a calculator?</p>

<p>An odd integer can be expressed in the form 2n - 1, where n is an integer. If we start from the number 1, the sum of the first n elements is n^2. Since x is this very sum from 1 to 25 (since 2*25 - 1 = 49), x = 25^2.</p>

<p>But the sum of 2n - 1 = the sum of 2n - the sum of 1 (spanning from 1 to n, of course). So the sum of 2n, which is how we represent the even integers is given by the formula [(sum of 2n - 1) + (sum of 1)] from 1 to n, which equals n^2 + n. In this case, we're taking the sum from 1 to 25, so y = 25^2 + 25.</p>

<p>The sum of the first n integers is n(n+1)/2, and n = 48 in this case, so z = 24*49.</p>

<p>Add them up and you got your answer. It's all in recognition of formulas, and it's pretty easy to see the formulas yourself if you write out the terms.</p>

<p>I can't help you with the calculator part unless you use a TI-89, because that's the only calculator I've ever used, but I'd definitely say that this is a calculator problem. They can't expect you to figure out these formulas during the test and they sure as hell can't assume you've already memorized them.</p>


<p>Edit: Yeah, even if you were trying to find x + y + z instead of x + y - z, you could just use 2*n(n+1)/2 where n = 48, and add that to 49 + 50.</p>

<p>Here's the easy way.
You know that if you combine x and y together, you get all the integers from 1 to 50 inclusive.
And since z includes only the integers from 1 to 48 inclusive, when you subtract z from x + y you are left with the numbers 49 and 50. Find their sum, which is 99 and you have your answer.</p>

<p>Note that calculations of x, y, and z's actual values were not needed.</p>

<p>Haha. Yeah, I misread it as x + y + z. His method is obviously the best one.</p>

<p>x + y= sum of all integers from 1-50.
z = sum of all integers from 1-48.
x+ y > z, by 49 + 50.
49 + 50 = 99</p>

<p>Thank you guys! You are just great!!!</p>

<p>you could get this answer really quickly if you just reason it out-recognize that (x+y) is simply the sum of all integers between 1 and 50 (inclusive). given that z is the sum of all integers between 1 and 48 inclusive, every term within z is also in (x+y) except for 49 and 50. subtract the terms in z from (x+y) and you end up with 49 and 50 in (x+y) and 49+50 = 99.</p>