<p>1) How many positive integers less than 1000 do not have 7 as any digit?
A)700
b)728
C)736
D)770
E)819</p>
<p>2) Points A, B, and X do not all lie on the same line. Point X is 5 units from A and 3 units from B. How many other points in the same plane as A, B, and X are also 5 units from A and 3 units from B ?</p>
<p>(A) None
(B) One
(C) Two
(D) Four
(E) More than four</p>
<ol>
<li><p>B</p></li>
<li><p>B, I think.</p></li>
</ol>
<p>lol please show how you got them</p>
<p>1.
The total number of positive integers less than 1000, is 10<em>10</em>10, the total number of positive integers less than 1000 that do not have 7 in it is (10-1)<em>(10-1)</em>(10-1). Which is 9^3, remember we are NOT counting 0 because it is asking for positive integers and 0 is not a positive integer so the answer is 9^3-1 or 728.
2. Very easy if you think about it conceptually, just draw A and B on a line and imagine where X could be if it was 5 away from A and 3 away from B, its easy, two possible positions, one on each side of the line.</p>
<ol>
<li>There are 19 sevens per every 100 numbers (7, 17, 27, 37, 47, 57, 67, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 87, 97) and there are 100 7’s between 700 and 799, so</li>
</ol>
<p>999 - [(19x9) + (100)] = 728</p>
<ol>
<li>On the second one, it is helpful to first picture the three points or to draw them. Then imagine the X being reflected or flipped over, like a mirror image. </li>
</ol>
<p>I don’t think I explained that very well…</p>
<pre><code> A b
x
</code></pre>
<p>Let’s pretend that x in this case is five away from A and three away from B. Naturally, you can flip it and it will still be the same space away</p>
<pre><code> X
A b
</code></pre>
<p>Hope this somewhat helped</p>
<p>clear for the first one. thanks
anybody have any idea for the second one, cuz it kinda obscure lol</p>
<p>X is 3 away from B, and 5 away from A. Since they’re not colinear, you know that it’s a triangle. It’s actually a 3,4,5, right triangle. No matter what point X is, if you flip it over the opposite side, you get it’s reflection. Thus, there is 1 other point that X could be, so B.</p>
<p>But Anomaly, it says the same plane. How do you define the same plane?</p>
<p>or anyone, please?</p>
<p>Why can’t points be overlapping? Can’t points all have the same coordinates?</p>
<p>Don’ tknow if this will help…but instead of triangles, picture circles. X is five units from point A. So it lies on the circle of radius 5, centered at A. But it is also 3 units from B. So it is on the circle of radius 3, centered at B. Those circles intersect at least once: at X
But they want to know how many other points could also be on the intersection of those two circles…Since two circles w/ different radii can only intersect at a maximum of two points, there can only be one other point that meets the requirements.</p>