Dr. Steve's Level 6 SAT Math Problem Thread

<p>Hey CCers. For those of you going for an 800 in math on the SAT I thought I'd start a thread of "Level 6" SAT math problems. We all know that SAT problems only go up to Level 5, so these problems will be just a bit more difficult than anything that I've seen on an actual SAT. The problems I post will be just like SAT problems however. I will maintain the wording and style of actual SAT questions, and the same SAT strategies that I encourage students to use on the SAT can still be applied to these problems.</p>

<p>Please post your solutions. Let's see how many different solutions we can come up with for each problem. </p>

<p>Other SAT math experts can feel free to hijack this thread a bit and post your own questions. But let's try to make sure that they are similar to actual SAT problems (for example no Subject Test problems here), and let's try not to have more than a couple active problems at once. </p>

<p>Also, let's let students post their solutions before we give our own solutions.</p>

<p>Let me start off with an algebra question:</p>

<p>y=a-b+25
y=b-c-7
y=c-d+6
y=d-e+30
y=e-f-32
y=f-g+7
y=g-a-15</p>

<p>In the system of equations above, what is the value of y?</p>

<p>Add all the equations, 7y = 14, y = 2.</p>

<p>2011 AMC10 #16: Which of the following is equal to sqrt(9 - 6sqrt(2)) + sqrt(9 + 6sqrt(2))?</p>

<p>A: 3sqrt(2)
B: 2sqrt(6)
C: 7sqrt(2)/2
D: 3sqrt(3)
E: 6</p>

<p>@rspence: Is the answer B?
Let a = sqrt(9 - 6sqrt(2)) + sqrt(9 + 6sqrt(2))
=> a^2 = 18 + 2(81 - 2(36))^(1/2)
=> a^2 = 24
So, a = 2(6)^(1/2) or B.</p>

<p>Nice thread, DrSteve. :)</p>

<p>@yellowcat429 yep.</p>

<p>Let’s try a number theory problem:</p>

<p>The integer n is equal to k^3 for some integer k. Suppose that n is divisible by 45 and 400. The smallest possible value of n has the form ABC,DEF where A, B, C, D, E, and F are digits. What is the product of A and C?</p>

<p>216000
12</p>

<p>Just solved it. Not too difficult – you know that n is divisible by lcm(45,400)=3600 then find the smallest cube that is divisible by 3600 (which turns out to be 216000).</p>

<p>Another number theory problem:</p>

<p>How many ways can one choose two distinct numbers from the set {1,2,4,…,2^10} such that their product minus their sum is a multiple of 3? Assume that order doesn’t matter, i.e. {1,2} and {2,1} are considered the same.</p>

<p>its really simple
so 45,400 (3600)
factors 3^2,4,5^2
obviously the lowest one with an integer k^3 =n
obviously just
3^3 4^3 5^3=216000
AxC=12</p>

<p>next lowest 3^3 4^3 5^3 2^3
3^3 4^3 5^3 3^3
3^3 4^3 5^3 2^6</p>

<p>^Pretty much. Although you may want to write 2^4 instead of 4^2 to avoid possibly making a mistake. Because it’s the prime factorization that matters.</p>

<p>For rspence’s problem, take everything mod 3.
The set becomes {1,2,1,2,1,2,1,2,1,2,1} (note that the same elements are repeated; however we will continue to treat them as separate elements)</p>

<p>So we must have 2 elements a,b from the set where:
ab - a - b == 0 (mod 3)
Neither 1,1 nor 1,2 work. Only 2,2 works (observe that (2)(2) - 2 - 2 == 0 (mod 3)).</p>

<p>There are 5 2’s in this set, and since we see them as separate elements, we simply take:</p>

<p>5C2 = 10</p>

<p>@Kyrix1, mmhmm. :)</p>

<p>Here’s a probability problem:</p>

<p>The integers 1 through 7 are written on each of seven cards. The cards are shuffled and one card is drawn at random. That card is then replaced, the cards are shuffled again and another card is drawn at random. This procedure is repeated one more time (for a total of three times). What is the probability that the sum of the numbers on the three cards drawn was 18 or 19?</p>

<p>I see an easy solution that involves a little brute-forcing…</p>

<p>{7,7,4} → 3 ways
{7,6,5} → 6 ways
{6,6,6} → 1 way</p>

<p>{7,7,5} → 3 ways
{7,6,6} → 3 ways</p>

<p>Probability = (3+6+1+3+3)/(7^3) = 16/343</p>

<p>Here’s a fun one: Find the remainder when 55^12 is divided by 123.</p>

<p>This is too hard to be a level 6 problem ._.</p>

<p>But anyway, let x be our remainder, and then take it mod 123:</p>

<p>x == 55^12 <a href=“mod%20123”>i</a>*
x == (5^12)(11^12) <a href=“mod%20123”>i</a>*
x == ((5^3)^4)((11^2)^6) <a href=“mod%20123”>i</a>*
x == (2^4)((-2)^6) <a href=“mod%20123”>i</a>*
x == (2^10) <a href=“mod%20123”>i</a>*
x == 1024 <a href=“mod%20123”>i</a>*
x == 40 <a href=“mod%20123”>i</a>*
The remainder is thus 40.</p>

<p>Yep. I know, it’s a little more AMC level than SAT.</p>

<p>I posted this about eight years ago:</p>

<p>This classic problem would never appear on a SAT test because it does not have a fast and easy solution. While it is not an easy problem, it can be solved using SAT geometry concepts.</p>

<p>Here you go: </p>

<p>**Draw an isosceles triangle ABC with Side AB = Side AC. Draw a line from C to side AB and label that line CD. Now draw a line from B to side AC. Label that line BE. Let angle EBC = 60 degrees, angle BCD equal 70 degrees, angle ABE equal 20 degrees, and angle DCE equal 10 degrees. Now draw line DE. </p>

<p>Find what angle EDC is by using geometry only and no trigonometry. **</p>

<p>Hmm, okay I’ll take a look at it. :)</p>

<p>Darn, thought I had a solution until I realized I screwed up. Not a very easy problem – you have to find a way to get around those x = x statements (without using trig…). I’ve been playing around with cyclic quadrilaterals a bit.</p>

<p>Meh, don’t see a non-trig solution. It probably involves extending some segment or creating a new point which I apparently missed. If this was on an actual SAT I’d definitely use trig.</p>

<p>The only progress I made was that I proved that angle EDC is less than 60 degrees.</p>