<p>This isn't an SAT question but it's little bit tough. The number of integers between 1 and 1001 which are divisible by 2 and NOT divisible by 3 is equal to:
(A) 331
(B) 332
(C) 333
(D) 334
(E) 335</p>
<p>I will post much more harder questions later. Thank you!</p>
<p>Are you asking for the solution to this, or is this a thread to challenge students?</p>
<p>No DrSteve, I’m asking for the solution. I have more questions that I’m having trouble with but I will post them later after I understand the answer for this question.</p>
<p>is it “d”, 334??</p>
<p>@2200and beyond</p>
<p>That is correct. </p>
<p>To see how many such integers are divisible by 2 we can divide 1001 by 2 to get 500.5. Drop the “remainder” to see that there are 500 such integers divisible by 2. Of these, the ones that are also divisible by 3 are precisely the integers divisible by 6. Since 1001/6 ~ 166.8, there are 166 such integers. Finally the answer is 500-166 = 334, choice (D).</p>
<p>^ Yes… That was my method too… thanks :)</p>
<p>Thank you DrSteve, I understand the explanation but is there another method to solve this? It seems kind of hard to come with the idea of dividing 1001 by 6.</p>
<p>I agree it’s a bit tricky. Let me reformulate the problem in terms of set theory. Perhaps this will help you understand more.</p>
<p>Let the universe U consist of the integers between 1 and 1001. Let A be the set of integers divisible by 2 and let B be the set of integers divisible by 3. Let C be the intersection of A and B. Then C is the set of integers divisible by both 2 and 3, that is C is the set of integers divisible by 6.</p>
<p>If you were to draw this as a Venn diagram you would have a recatngle with 2 overlapping circles labelled A and B (and label the rectangle by U if you like).</p>
<p>Now the question is asking how many elements are in A - B, that is the set of elements in A but not in B. In the Venn diaram you would be shading the portion of the circle A that does not include the intersection. </p>
<p>Now note that as per my previous explanation there are 500 integers in set A and 166 integers in set C. So A - B has 500-166 = 334 integers.</p>
<p>Note that the set A - B is equal to the set A - C.</p>
<p>Hope this slight reformulation of the problem helps.</p>
<p>Well, I thought my method was extremely like DrSteve’s but maybe it’s a bit easier for you:</p>
<p>What it the last number before 1001 that is divisible by 2?
1000 which is equal to 2*500
so there are 500 numbers divisible by 2 between 1 and 1001</p>
<p>what is the last number before 1001 that is divisible by 6 (as in 2<em>3, because we want to subtract the number of integers divisible by both from the number of integers divisible by 2)?
996 which is equal to 6</em>166
so there are 166 numbers divisible by 6 between 1 and 1001.</p>
<p>the question asks for " number of integers between 1 and 1001 which are divisible by 2 and NOT divisible by 3" so 500-166= 334.</p>
<p>I found that this method would work between “number of integers between 1 and and (any number)”… (if I’m not mistaken 
)</p>
<p>hope that helps, and I haven’t confused you</p>
<p>It’s clear now. Thank you DrSteve and 2200andBeyond. Here is another problem: If m and n are real numbers such that m^2 = m+1, n^2 = n+1, and m <> n (m does not equal n), then m^5 + n^5 =
(A) 11
(B) 12
(C) 13
(D) 14
(E) 15</p>
<p>^ first of all, before i start solving… u sure u got the question right? </p>
<p>a real nb squared equals that same number plus one???</p>
<p>Yes, it’s like that! Obviously m and n are not integers or natural numbers. I tried to factor m and I think I solved it but I got stuck in n.</p>
<p>Well the answer is A (11), I used a CAS to solve that though, makes that m^2 -m-1 equals to either (-(root 5 -1) /2) or ((root5 +1)/2) take both of them to the power of 5 and you’ll get 11. I guess there’s an easier way that wouldn’t require actually finding out what m and n equals but just find out m^5 and n^5.</p>
<p>^ uugh! i realized how stupid my question ws and started solving… Then realized it needed a scientific calculator ! I’ve been googling casio fx 991 plus ever since!
anyway… yeah … i haven’t even found the calc… you giys know where to find an online virtual free one!!! really need that :/</p>
<p>any more questions? lol</p>
<p>Didn’t quite get it, but here is another problem. Any help will be appreciated.
Let a and b be positive integers such that 7 divides a^2 + b^2. When ab - 1 is divided by7, the remainder is
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6</p>
<p>Let’s do this by picking numbers. Let a=b=7. Then clearly a^2+b^2 is divisible by 7. Also, ab-1 = 49-1 = 48. When 48 is divided by 7 the remainder is 6.</p>
<p>So assuming the problem is sound, the answer is 6, choice (E).</p>
<p>As it turns out the problem is sound. I came up with the following proof without picking numbers. I first prove a preliminary lemma:</p>
<p>Lemma: If a^2+b^2 is divisible by 7, then both a and b are divisible by 7.</p>
<p>Proof: By the division algorithm we can write a=7k+j and b=7m+n where j,n are between 0 and 6 (inclusive). So a^2+b^2 =49k^2+14kj+j^2+49m^2+14mn+n^2 = 7(7k^2+2kj+7m^2+2mn)+m^2+n^2. So it suffices to show that m^2+n^2 is not divisible by 7 for m,n=1,2,3,4,5,6. You can show this by brute force. For example 1^2+1^2=2, 1^2+2^2=5, 1^2+3^2=10,…,6^2+6^2=72 (you must check all 21 cases).</p>
<p>It would be nice if someone can come up with a more elegant proof of this lemma (without having to do 21 computations).</p>
<p>Now to prove the theorem stated in the problem.</p>
<p>Proof: Suppose a^2+b^2 is divisible by. By the Lemma, a and b are divisible by 7. Thus, there are integers k and j such that a=7k and b=7j. So ab-1=(7k)(7j)-1=(7k)(7j)-7+6=7(7kj+1)+6. So ab-1=7z+6 for some integer z (z is the integer 7kj+1). Thus, ab-1 is divisible by 6.</p>
<p>Oh wow! I don’t even know why I posted this easy question! Thank you again DrSteve. These are another questions: The number of real roots of the equation
-8x^4 - 18x^2 - 11 + 1/x^4 = 0 is
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4</p>
<p>A bag contains 300 balls of 5 different colors, as follows: 90 red, 90 blue, 90 green, and the remaining balls are either black or white. If balls are drawn from the bag at random (without checking their colors before or after drawing them) and without replacement, then the smallest number of balls that must be drawn to ensure that 30 of them have the same color is
(A) 118
(B) 97
(C) 130
(D) 117
(E) 96</p>
<p>So here’s an intriguing question. For which primes p is the following theorem true?</p>
<p>Theorem: Suppose that a and b are integers with a^2+b^2 is divisible by p. Then a and b are both divisible by p.</p>
<p>For example, if p=7, I showed above that the theorem is true. If p=5, the theorem is false (let a=1, b=2).</p>
<p>To you number theory experts out there, give me a solution, or explain why it’s “too hard.” Or perhaps a partial solution? (maybe all primes of the form n^3-1 satisfy this theorem? Are there primes of this form other than 7? What’s the next one?)</p>