Math - Number Amount Problems

<p>I do decently well on the SAT math (I almost always score 700+), but I realized that I am having trouble with a specific type of problem.</p>

<p>Here are some examples of this specific type:</p>

<p>How many numbers between 1-1000 do not have 8 as a digit?</p>

<p>How many positive integers less than 1000 are multiples of 10 and are equal to 4 times an even integer?</p>

<p>How many multiples of 5 and 6 are less than 2000?</p>

<p>What trips me up is not that these problems are necessarily difficult but that we are required to complete them with great rapidity. Does anyone have any formulaic or rapid approaches to this specific problem type? What do you all do when you see something like this?</p>

<p>No idea, bro. Personally, I just write out all the numbers haha</p>

<p>How many numbers between 1-1000 do not have 8 as a digit?</p>

<p>The best approach is to just count the #s that do have 8, because there are more of those. </p>

<p>8, 18, 28, 38, 48, 58, 68, 78, 80~89, 98.</p>

<p>Up to 100 there are 19. So 19 * 9 = the amount of #'s with 8 as a digit within the intervals of 1-799 and 900-1000. Now add on the 100 extra because of all the #'s in the 800-899 range.</p>

<p>You should come up to approximately 271. </p>

<p>The question states how many don’t have 8 as a digit, so you do 1000 - 271, which is equal to 729, which oddly enough is 27^2 and 9^3. </p>

<p>My answer might be off be 1 or 2, but if so just choose the closest answer.</p>

<p>How many multiples of 5 and 6 are less than 2000?</p>

<p>2000 / 5 = 400. Because they want less than, there are 399 multiples of 5 less than 2000.
2000 / 6 = 333. </p>

<p>You would think that there are 400 + 333 (733) multiples of numbers less than 2000 of 5 and 6, but there is a catch…</p>

<p>You must now find how many of these 733 are the same number (both multiples of 5 and 6). To do so, you must use the Least Common Multiple of 5 and 6 which is 30.</p>

<p>2000 / 30 = 66.</p>

<p>733 - 66 = 667. </p>

<p>667 should be the answer.</p>

<p>How many positive integers less than 1000 are multiples of 10 and are equal to 4 times an even integer?</p>

<p>So basically this is saying how many integers less than 1000 are multiples of 80 (because 4x2=8[2 is the even integer] and 8x10=80). That should be much easier now 80,160,240,320,400,480,560,640,720,800,880,960. So the answer is 12.</p>

<p>Woops my calculations were wrong. I screwed that up.</p>

<p>To answer the question," How many positive integers less than 1000 are multiples of 10 and are equal to 4 times an even integer?" I think you want to look for multiples of 40, rather than 80. For example, consider 40 itself. It meets the criterion.</p>

<p>2000/5 = 400. That means that you’ve got 400 multiples of 5 ranging from 0 to 2000?
By the way, the technical correct answer is an infinite number of values, unless they say “between 0 and 2000”.
Also, the question isn’t clear. Does it ask for common multiples of 5 and 6 less than 2k or just multiples of each added?</p>

<p>How many numbers between 1-1000 do not have 8 as a digit?</p>

<p>This is another problem that can be solved with the counting principle: multiply the number of choices you have for each option (in this case, each digit).</p>

<p>Think of it as one of those bicycle locks that has 3 digits that you can set. For each of them, you usually have 10 digits to choose from and thus 10x10x10 possible combinations.</p>

<p>Now supposed you could not use the digit 8. Then you would only have 9 choices for each digit and thus 9x9x9=729 possible combinations. But that would represent numbers from 0 to 999. We don’t want to count 0 but we did want to count 1000 so it is still 729.</p>