<p>1) A large solid cube is assembled together by gluing together identical unpainted small cubic blocks. All six faces of the large cube are then painted red. If exactly 27 of the small cubic blocks that make up the large cube have no red paint on them, how many small cubic blocks make up the large cube?</p>
<p>2) The sum of the positive odd integers less than 100 is subtracted from the sum of even integers less than or equal to 100. What is the resulting difference?</p>
<p>This one was easy but time consuming. So, I was wondering if there is a faster way to solve it. </p>
<p>3) If x and y are two different integers and the product 35xy is the square of an integer, which of the following could be equal to xy?</p>
<p>(A) 5
(B) 70
(C) 105
(D) 140
(E) 350</p>
<p>I didn't put the answers so you guys can practice and help me at the same time</p>
<ol>
<li>125
If 27 blocks are unpainted, that means 3<em>3</em>3 is the inner cube. You can then picture the entire cube to be 5<em>5</em>5, 125.</li>
</ol>
<ol>
<li>I would do up to 10 instead of 100.
2-1=1
4-3=1
6-5=1
8-7=1
10-9=1
Each pair adds 1 to the sum. 50 pairs up to 100 = 100/2 answer is 50.</li>
</ol>
<p>That was easy. Thank you</p>
<p>huh, i never thought of adding up to 10. but can you solve it algebraically? and you got them both right.</p>
<p>1) You can think of the unpainted cubes as the “inner” cube, and the painted cubes as the “outer” one. If the inner cube has 27 little cubes, the inner cube must be 3 by 3 by 3. Since the outer cube has to completely surround the inner cube, it means the whole cube has to be 5 by 5 by 5, so the large cube would be made up of 125 little cubes.</p>
<p>2) This problem is basically :
(2+4+6…100) - (1+3+5+…99) = ?
right?
so it can be rearranged as :
(2-1)+(4-3)+(6-5)…+(100-99).
If arranged this way, it would equal in 50 pairs that would all equal 1, so the answer would be 50 times 1 or 1.</p>
<p>3) if 35 xy is the square of a number, it would mean that all its factors need to be in pairs (ex. 2 sevens, 2 fives, etc). 35’s factors are 7 and 5, so it means that xy needs to consist of one 7, one 5, and maybe other numbers that are already squares. The least possible xy would have been 35, but since that is not one of the choices, multiply it by the next smallest square, 4. The product is 140, which is oneof your choices, D.</p>
<p>I hope this helped,… I hope I made everything clear!
p.s. it’s 3 annoying math problems not 2 haha</p>
<p>2) Just picture it like this</p>
<pre><code>(2 ~ 4 ~ 6 … 100)
</code></pre>
<ul>
<li> (1 ~ 3 ~ 5 … 99)</li>
</ul>
<p>= 1 ~ 1 ~ 1 … 1</p>
<p>50 pairs total so 1 x 50 =50. There’s the difference.</p>
<ol>
<li>- D</li>
</ol>
<p>I just did 35 x (sub in each of those answer choices), and then see which one came out to be a perfect square.</p>
<ol>
<li>35xy = 5<em>7</em>x<em>y
Since perfect sq * perfect sq = perfect sq, we need to find an answer choice that has perfect square * 5</em>7.
A. 5 = 1<em>5 NO
B. 70 = 5</em>7<em>2 = it has 5 and 7 but 2 it not a perfect sq
C. 105 = 5</em>7<em>3 = same reason
D. 140 = 5</em>7<em>2</em>2 = meets all criteria 2<em>2=4 is a perfect sq.
E. 350 = 2</em>5<em>5</em>7 = 10 is not a perfect sq.</li>
</ol>
<p>I hate typing numbers!! This is why I never answer the math questions on CC haha. these three problems look like they’re time consuming but in reality, they only took me like one minute for all? I just couldn’t think of good ways to explain since most of my works are done in my head. These are like the brain teasers in middle school textbooks. Once you understand what they are really asking for, it’s easy to see the shortcuts. You just have to truly understand why College Board chooses these sets of numbers rather than others random ones. I don’t think SAT has any real time consuming questions… I suggest you take few more seconds thinking about the question than to start jumping in to solve it.</p>
<p>Thanks, everyone.
@JefferyJung I solved it the same way you did but I thought it was time consuming cause I had to go through all the choices</p>
<p>and I also hate the backsolve method, I only resort to it when there is no other choice</p>
<p>For 3, just divide out 35 from the answer choices and it literally takes 10 seconds maybe…</p>