How do i attack these kind of math problems?

<p>When i took the last SAT test in June, i noticed there were a certain type of math questions that at some point i gave up finding the answer, thus no points and less time left. For my first SAT test, i got 640 on the math section, which is not bad. If i can get these problems right, i believe i can get at least 720. </p>

<p>Q1) In writing all of the integers from 1 to 300, how many times is the digit 1 used?</p>

<p>For this i tried to write down every integer with the digit 1, but after a few mins into it i got tired; i gave up knowing of all the possibilities. </p>

<p>So is there an easy to to solve this problem???</p>

<p>Here's another one.</p>

<p>Q2) How many integers between 1 and 1000 are the product of two consecutive integers?</p>

<p>2x3=6
3x4= 12
4x5=20
5x6=30</p>

<p>.........i find this tedious and get's me nowhere except waste my time.</p>

<p>Any help?</p>

<p>If you guys can find more questions like these, please post them here.</p>

<p>Thanks.</p>

<p>1) I’m thinking 100-199 is 100
then 10 1’s from 1-99 and 10 more from 201-299
100+10+10=120
2)greatest product of consecutive integers that is less than 1000 is 32x33
so I’m thinking this one is 33</p>

<p>For the first question, it could be solved by finding all the integers within the 1-100 range: (1,11,12,13,14,15,16,17,18,19,21,31,41,51,61,71,81,91) = 18 integers.</p>

<p>18 values, plus 100 (since there is always a 1 in the hundreds place from 100-200), plus another 18 (since its the same integers as 1-100, but with a different hundreds place value)</p>

<p>18 + 100 + 18 = 136</p>

<p>1) Look at each digit place.</p>

<p>The ones digit: in each 100 terms there are ten 1s (e.g. 1, 11, 21, 31, …, 91) so 10x3 = 30
The tens digit: in each 100 terms there are ten 1s (e.g. 210, 211, 212, 213…, 219) so 10x3 = 30
The hundreds digit: 100-199 = 100 more 1s</p>

<p>100+30+30 = 160</p>

<h1>2 - 1x2 = 2, 2x3 = 6, etc. Notice this counts as “two” integers even though it ends with 2x3.</h1>

<p>The highest we can go is 31x32 = 992. By the pattern above, this would yield 31 integers.</p>

<p>The key with these: Look for patterns.</p>

<p>The question is slightly ambiguous to me. Does it mean how many times the digit one appears so that 111 counts as three, or does it mean how many numbers have the digit one, in which case 111 would count as one? Depending the meaning of the question either JBIG is correct of zephyr is correct…</p>

<p>JBIG is correct since it says, “How many times is the digit 1 used?” In order to use zephyr’s method, it would have to say, “How many integers contain the digit 1?”</p>