<p>^^ Thanks everyone! Some of those studyhall explanations are hard to understand!</p>
<ol>
<li> It is asking what is the 50th digit form the left. The first numbers are all single digits, so they take up 9 spaces. What we have left for the double digit numbers is 50-9 = 41 spaces. Each of the next numbers takes up 2 spaces. So we divide 41 by 2 to get 20.5. Here comes the tricky part. We have to think of a simpler situation in order to find the pattern.<br></li>
</ol>
<p>Let's say that we want to find the 11th digit. So by subtracting the 9 single digits, we get 2. Then dividing by 2, we get 1. SO we can conclude that we must go 1 number over. We can easily see that 11th digits is the 0 from the number 10 (by writing the numbers out). So by this we learn that if we end up with an even number after dividing by 2, we get the 2nd digit of the number found by moving over as many spaces as what we get when dividing by 2. </p>
<p>So if it we got that it was 20 numbers to the right, we would get 9+20 = 29. Then we would pick the second digit, or 9. But we got 20.5 after dividing by 2, so we must go over an additional space, to the 3 digit of 30.</p>
<p>The answer is 3.</p>
<ol>
<li> The key thing to remember in this type of problem is that when making the equations, you must both subtract from the person who gives up something AND add to the person receiving that.</li>
</ol>
<p>SO the equations should look like this: (J = Jean's original amount; I = Irene's, and T = Todd's)</p>
<p>We know that J gives 10 to I, and I gives 6 to T.</p>
<p>So the preliminary equations would be:</p>
<p>J-10 = I+10-6 ___ = T +6 ___</p>
<p>But we are also told that after these transactions, Jean had 10 more than I still, and 20 more than T.<br>
So we fill that in to the equation as such
J - 10 = I + 10 + 10 = T + 6 + 20</p>
<p>Then we solve.</p>
<p>J-10 = I + 14= T + 26
Add 10 to all in order to get rid of the -10 on Jean's side.
J = I + 24 = J + 36</p>
<p>By this we can conclude that Jean originally had 24 more than Irene and 36 more than Todd. So the answer is E.</p>
<p>flipsta, you are awesome, thanks so much!</p>
<p>Any time. After all, this is probably beneficial to my SAT Math skills too. So ask away, my friend.</p>
<p>If you don't have the Big Red Book, heres the question:
1,234, . . . 1,920,21. . . ,484,950
The integer above is formed by writing the integers from 1 to 50, in order, next to each other. If the integer is read from left to right, what is the 50th digit from the left?
A. 0
B. 1
C. 2
D. 3
E. 9</p>
<p>I dont see how this question can be answered by simple reasoning and analyzing the answer choices. Id probably rule out just one: A.
50th digit and 0 go to well together for a random guessing.</p>
<p>The beginning is borrowed from Flipsta_G (post #2):
"It is asking what is the 50th digit form the left.
The first numbers are all single digits, so they take up 9 spaces":
123456789.
Let's throw them out.
If we drop the commas too, well have
10111213 . . . 484950.
We need to find the digit on the 50 - 9 = 41st place from the left.
41 : 2 = 20 R 1. This means that the first 20 consecutive two-digit number fit nicely into 40 places, and the following 21st number will have its first digit on place #41.</p>
<p>Let's start:
10111213...1819
(10, 11, 12,..., 19 first 10 two-digit numbers).
It'll be 10*2 = 20 digits.</p>
<p>Then:
20212221...2829
(20, 21, 22,..., 29 -- another 10 two-digit numbers).
10*2 = 20 more digits.
20 + 20 = 40.</p>
<p>On place 41 starts the number 30, so "3" is on place #41 in
101112...19202122...293303132....4950.
It means "3" is the 50th digit in the original number.</p>
<p>Answer D.</p>
<p>Comments.
1. Instead of writing out the first 20 consecutive two-digit numbers we could jump right to the last one on that list.
Whats the 20th term in the sequence 10, 11, 12,
?
You could use formula for arithmetic sequence
a{n} = a{1} + r<em>(n-1)
a{20) = 10 + 1</em>19 = 29
This formula really helps in BB p.653 q.5.
Also check out parallel post
<a href="http://talk.collegeconfidential.com/showthread.php?p=1128981&highlight=sequence#post1128981%5B/url%5D">http://talk.collegeconfidential.com/showthread.php?p=1128981&highlight=sequence#post1128981</a>.</p>
<p>Or you could reason:
the term #1 in this sequence is 10, which is greater than 1 by 9,
the term #2 in this sequence is 11, which is greater than 2 by 9,
the term #20 should be greater than 20 by 9, so it equals 29
(this is my preferred way of figuring it out; Flipsta_G seemed to use this route too).</p>
<ol>
<li><p>Finding a remainder can be very helpful in SAT questions.
See BB p.398 q.13.</p></li>
<li><p>Turning a sequence into a decimal number is a good example of misleading SAT questions.
See BB p.461 #9.</p></li>
</ol>
<p>RE: " Id probably rule out just one: A."
I looked back and felt that choice
E. 9
should be also ruled out - it stands out too much.
If you have run out of time or ideas, randomly choosing one out of
B. 1,
C. 2, and
D. 3
might bring you luck.</p>
<p>i don't get why you guys have to think so deeply... just write down on the paper 1 to 50. thats what i did. solved the question in like a minute.</p>
<p>JoeTM:
So they asked for the 50'th digit from the left, and you did it in a minute by writing out the entire number. Great. What if they'd asked for the 500'th?
Using reasoning (instead of brute force) to solve a problem is what the SAT is supposed to test, anyway...</p>
<p>Yes, reasoning is usually the best way. If you get completely stuck, however, and you have plenty of time left (i.e. you skipped it and then came back to it), then you can brute force it.</p>
<p>BB p.461 #9 (mentioned in post #5) is a proof to optimizerdad's point.
"What's the total number of 0's between the 98th and the 101st 1 in this decimal number?"
Joe, would you write out everything up to 101st 1 to answer it?</p>
<p>Two more arguments on #7:
Higher chance of a mistake: either when writing down a long sequence of numbers, or counting to the 50th digit.
Waste of time: on real test you would probalbly count twice to self-check.</p>
<h1>"Brute Force" vs. Reasoning.</h1>
<p>Flipsta_G gave a good example. It does happen - your mind goes blank on this one question. I'd leave it until I'm done with everything else, and then finish the bugger with any means necessary, Brute Force included.</p>
<p>BF might work sometimes as fast as reasoning.
BUT: if you practice BF exclusively when prepping for SAT, your BF does not improve, AND you are missing on shaping up your reasoning muscles.</p>
<p>What really helps build them up is trying to find many alternative solutions to your practice SAT math questions. You'll be churning out short-cuts on your real test like ...... - like I don't know what. You complete this sentence.</p>
<p>Without bragging - just to illustrate the workings of a trained mind:
let's use #7 again.
Start your stop-watch for this train of thoughts:
++++++++++++++++
1. 50th digit will be the first or the second digit of some TWO-digit number
2. 9 places out of 50 are used up by single-digit numbers =>
3. odd number of places is left for two-digit numbers =>
4. 50th digit will be the FIRST digit of a two-digit number - some number like twenty- or thirty- or fortysomething =>
5. 9 places for "singles"
+ 20 places for "teens" (total 29)
+ 20 places for twentysomethings (total 49),
are we there yet? no, we did not reach place #50 yet,
but thirtysomethings will cover it =>
the digit is 3.
STOP.
++++++++++++++++
Chances are it can be answered even faster.</p>
<p>Frankly, I think Joe skillfully plays a devil's advocate here.</p>
<p>Nice to have your predictions prove true.</p>
<p>I talked to a former student of mine (2400, multiple math competitions winner/runner-up), and he beat me on this one.</p>
<ol>
<li><p>Insert 0 in front of each single digit number:
010203...09 - nine of them altohether.
Now we are looking for 50+9=59th digit in this number:
010203...091011...484950.</p></li>
<li><p>We can consider all numbers in this sequence two-digit:
01,02,03,...,09,10,11,...,48,49,50.</p></li>
<li><p>59 : 2 = 29.5 - that means the first 59 places in
010203...091011...484950
are occupied by 29.5 consecitive two-digit numbers:
29 complete ones (01,02,03,...,29),
and the first "half" (digit) of number #30.</p></li>
</ol>
<p>The answer is 3.</p>
<p>I doubt it can be done any faster.</p>
<p>I'll pipe in with a slightly different approach. This problem is a simple problem that is made a bit more difficult because it uses two pattern in one sequence. However, the answer can be easily determined by writing a diagram that looks like this:</p>
<p>1 0 5 0 5 0 = pretty simple alternating pattern
2
3
4
5
6
7
8
9</p>
<h2>1 1 2 2 3 3 = pretty simple double pattern </h2>
<p>1 2 3 4 5 6</p>
<p>Since each column comprises ten digits, it takes no time to count to ... 5. I added a last row to show 1 through 6. :)</p>
<p>The solution is easily identified by looking at the FIRST row or tenth row. The first row shows 1st, 11th, 21th, etc, and the 51st digit is 0 - we just go back one for the 50th digit, and that is 3. A better option is to look directly at the 10th row, and count to five. Either way, it is three. </p>
<p>FWIW, you could find ANY numbered digit by simply expanding the pattern. As soon as you have a repeat pattern, the answer will be in front of you.</p>
<p>so far, BF has worked for me for the occasional question. I have yet to see a question on the SAT that obviously doesn't use BF (like that 100 -101 or whatever question) that isn't EASY to solve without BF anyways.</p>
<p>I was waiting for somebody to ask you to do this, but seems everybody understood it but me. :o</p>
<p>Would it work for an expanded question:
what's the 200th digit in
12,345,678,910,111,213, ... ,198,199,200 ?</p>
<p>Let's just look at your diagram under different angle:
instead of columns numbers follow each other in rows, wrapping to the next row.</p>
<hr>
<p>1,2,3,4,5,6,7,8,9,1 |
0,1 1,1 2,1 3,1 4,1 |
5,1 6,1 7,1 8,1 9,2 |
0,2 1,2 2,2 3,2 4,2 |
5,2 6,2 7,2 8,2 9,3 |
0,3 1,3 2,3 3,3 4,3 |
5,3 6,3 7,3 8,3 9,4 |
0,4 1,4 2,4 3,4 4,4 |
5,4 6,4 7,4 8,4 9,5 |
0</p>
<p>The first column shows that "11th, 21th,...,51st, etc., digit follow the pattern 0, 5, 0, 5, 0, etc. (1st digit 1 is not a part of the pattern though).
The tenth column shows that "10th, 20th,...,50th digit follow the pattern
1, 1, 2, 2, 3, etc.
That gives an answer 3.</p>
<p>xiggi, please correct me I misinterpreted your solution.</p>
<p>I am not sure this is an easy way of finding any numbered digit by expanding the pattern.</p>
<p>How would you use it in my proposed question (I understand it's not really an SAT type)?</p>
<p>What's the 200th digit in
12,345,678,910,111,213, ... ,198,199,200?</p>
<p>Jeffrey's (my student) universal method.
Let's convert all numbers in this sequence into three-digit numbers:</p>
<ol>
<li>
insert two zeroes in front of each one-digit number (9*2=18 altogether),
insert one zero in front of each two-digit number (90 altogether, since the first two-digit number is 10 and the last is 99).
Now we are looking for 200+18+90=308th digit in this number:
001002003...009010011012...098099100101...198199200.</li>
</ol>
<p>2.
308 : 3 = 102 2/3 - that means the first 102 places in
001002003...198199200
are occupied by 102 2/3 consecutive three-digit numbers:
102 complete ones (001,002,003,...,102),
and the first 2 digits (out of 3) of number #103, i.e.10,
therefore 308th digit in
001002003...009010011...198199200
is 0, which is also 200th digit in
12,345,678,910,111,213, ... ,198,199,200.</p>
<p>Done.</p>
<p>GCF, I just read your post and question. </p>
<p>First, the patterns are supposed the range of the questions. In the original question, we were asked to write "two digits" numbers. Thus, we look for patterns based on two numbers. When we expand the model beyond the original questions,we have to look for larger patterns. For instance, once we start writing THREE digits numbers, we need to identify the patterns for THREE numbers. </p>
<p>Let's look at the patterns, if column A represents the FIRST digits in your example, column B represents the 10th digits, and column C is given to follow the 10s easily. In this case, I started at 160. The 160th digit being 8 from 85. The 161th digit is thus 5. As we know, the patterns so far had been 0,5,0,5,0 for the first digits and 1,1,2,2,3,3 etc for the tenth digits.</p>
<p>Now, at 99 to 100, we need to change the analysis of patterns. As far as the answer to the question about the 200th digit, the answer would have been 0 - since after 99 you would have written 100. However, you could also have derived the answer by quickly filling the pattern after 95/99. Again, the 100 initiates a new pattern. </p>
<p>A B C </p>
<p>0 8 160
5 9 170
0 9 180
5 1 190 === change to different pattern.
0 0 200
3 6 210 === start new pattern for tenth
1 1 220
1 0 230
3 6 240 === start new pattern for first digits
1 1 250
1 1 260
3 6 270
1 1 280
2 2 290
3 6 300
1 1 310
3 3 320
3 6 330
1 1 340
4 4 350
3 6 360
1 1 370
5 5 380
3 6 390
1 1 400
6 6 410</p>
<p>Let's look at what we have, defining a couple of patterns (from several choices): </p>
<p>From A: 311, 311, 312, 313, 314, 315, 316 ...
From B: 610, 611, 612, 613, 614, 615, 616 ...</p>
<p>Following the above patterns, we can establish that </p>
<p>For instance, the 400th digit is a 1 or rge 380th is a 5.</p>
<p>5 ? 1,729 635,318,657</p>
<p>1000 999 972 954 936 918 900 ?</p>
<p>xiggi, I am always interested in alternative solutions. I did not get yours in post #16 yet, but I'll try.
I like sequences, but I am quite bad at them. Does not look like I can crack yours. Uncle!
Wait, I just discovered a pattern in #18!
Each term equals previous one minus a sum of its digits.
The "?" one is 900 - 9 = 891.</p>
<h1>18 is correct.</h1>
<p>The previous one is a very hard one. It is one of those problems that become easier ... after seeing the solution. There is something special about 1729, but unless you read about "taxicab" numbers, it won't mean that much. The sequences are part of the Nemesis test that originated in the Netherlands. I played with it in the summer of 2003, but did not find many answers, except for the numerical sequences. </p>