Number Range Manipulation Problems

Test Preparation SAT Preparation
1

<p>I know that number range problems are a common pitfall for most SAT test-takers.</p>

<p>Example from the 2009 PSAT:</p>

<p>How many positive integers less than 1,000 do not have 7 as any digit?
(A) 700
(B) 728
(C) 736
(D) 770
(E) 819 </p>

<p>Here's College Board's official solution: (skim if you wish)
The number of positive integers less than 1,000 that do not have 7 as any digit is equal to the number of one-digit positive integers other than 7 plus the number of two-digit positive integers without 7 as any digit plus the number of three-digit positive integers without 7 as any digit. There are 8 positive one-digit integers other than 7. If a two-digit positive integer does not have 7 as any digit, there are 8 possibilities for the tens digit and 9 possibilities for the ones digit; there are thus a total of (8)(9) = 72 such two-digit positive integers. If a three-digit positive integer does not have 7 as any digit, there are 8 possibilities for the hundreds digit, 9 possibilities for the tens digit and 9 possibilities for the ones digit; there are thus a total of (8)(9)(9) = 648 such three-digit positive integers. Therefore, there are a total of 8 + 72 + 648 = 728 positive integers less than 1,000 that do not have 7 as any digit.</p>

<p>Looks rather lengthy. Doesn't it?</p>

<p>I'm sure there are mathematical methods for quickly manipulating number range problems such as these. (I've been exposed to these types of problems frequently in my AP Computer Science courses)</p>

<p>Here's an example (4 methods for adding the numbers 1 to 100):
Techniques</a> for adding the numbers 1 to 100 | BetterExplained</p>

<p>PLEASE Post your own methods/techniques (or someone else's) for solving number range problems.</p>

2

<p>I think in this case, the solution given by collegeboard is the easiest and the most straightforward, really.</p>

3

<p>I could give you another solution; it may be simpler to understand. </p>

<p>First, we know we are looking at numbers 1 - 999, excluding 1000. </p>

<p>I will first find the number of numbers with a digit 7.</p>

<p>Now, we know that 07,17,27,37,47,57,67,77,87,97. Yields 10 numbers with a digit 7. However we have to remember that 7 is apparent in 70,71,72,73,74,75,76,77,78,79. Since we have already counted it as 1 , we must subtract 1 from 10. So far we have 9 numbers per 99 excluding the 70-79. Now 70-79 has 10 numbers with digit 7. Now we have 19 numbers per interval of 99 numbers. </p>

<p>Now we may multiply this 19 by 10 since 1-999 has 10 intervals of 99. 19 X 10=190 numbers with digit 7.</p>

<p>Once again, we have to remember that 700-799 all have a digit 7. Therefore there are another 100 numbers with digit 7 in 700-799. Now we must also subtract 19 from 190 since we double counted 19 of those 100 numbers in 700-799. </p>

<p>We now get 190-19+100 = 271 numbers with digit 7 from 1-999.</p>

<p>Since there are 999 numbers total. We subtract 271 from 999 to yield 728 numbers WITHOUT a digit 7.</p>

<p>This method may be longer to a sense because i did not really type it up in a concise manner.</p>

4

<p>Hmm Im surprised that no one has replied with something close to mine. Well I sort of think of it as a probablility problem. The positive integers numbers less than 1000 are 000 to 999 (ok 000 doesn’t count but go along with it). There are 10 possible numbers in the hundreds, tens, and ones place (0,1,2,3,4,5,6,7,8,9). So its 10x10x10=1000 possible numbers from 0-999. Now if we take out the 7, its 9 possible outcomes for the hundreds,tens, and ones place. 9x9x9=729 numbers without any 7 from 000-999. Subtract 1 (because 000 isnt a positive integer) and you get the answer 728.</p>

5

<p>I just did 9 x 9 x 9, because there were 9 possibilities for each spot and got 729. Then I subtracted 1 because 000 isn’t an integer. So I ended up using the same method as idontthinkiremem.</p>

6

<p>a number < 1000 can be represented by XYZ, X ranging from 0-9, Y=0-9, Z=0-9. That makes it 10x10x10 = 1000 numbers from 0-999. if digit 7 is not to be counted, then we have 9x9x9 = 729 numbers. Because 0 is not to be counted, we have 728. This is the easiest to count. Though the straightforward method of counting is easy, it consumes lot of time with risk of making errors in addition.</p>

<p>SealPoint Academy
[S&lt;/a&gt; P A R K &nbsp;](<a href=“http://www.SealPointAcademy.com%5DS”>http://www.SealPointAcademy.com)</p>