<p>I seem to be having a serious problem with this particular kind of math question.
I would really appreciate it if you could help me out...</p>
<ol>
<li><p>The integer m is between 40 and 100. When m is divided by 3, the remainder is 2. When m is divided by 7, the remainder is 1. What is one possible value of m?</p></li>
<li><p>When a positive integer is divided by 5, the remainder is 4. Which of the following expressions will yield a remainder of 2 when it is divided by 5?</p></li>
</ol>
<p>2) The answer is C. If you take 9, for example, and divide by 5, you get 1 with a remainder of 4. Then, if you add 3 to 9, you get 12. 12 divided by 5 = 2 with a remainder of 2.</p>
<h1>1 on the other hand, I have no idea how to do that without guessing and checking, which isnt incredibly time efficient when you’re doing it for 58 integers.</h1>
<p>thank you…
but I actually wanted to know how to solve it without the guessing and checking method.
For example:
When a positive number is divded by 4 the remainder is 1…when it is divided by 5 the remainder is 2… when it is divided by 6 the remainder is 3. What is the smallest 3 digit number?</p>
<p>you would solve it like this :
A=4p+1 A=5q+2 A=6r+3
4(p+1)-3 5(q+1)-3 6(r+1)-3
so the answer is multiples of 60 minus 3… </p>
<p>can I do something similar with this question?</p>
<p>I just figured out a way to do #1. On my TI-84, I entered the equation y=7x+1 into the y= screen, and then pressed 2nd and Graph, which brought me to the table for that equation. I then recorded all of the y values between 40 and 100. Then I did the same thing for y=3x+2. Finally, I compared the y values for the two equations until I found one that was in both, which is 50. 50/7 = 7 and 1/7, and 50/3 = 16 and 2/3. I’m interested to see how other people would solve that, but I see the way I did it as the easiest method.</p>
<p>x and y are both integers, so the difference must be an integer. y > x, so the difference will be a negative integer. However, for the difference to be less than -1, our mysterious value would need to be greater than 8 - not one of the allowed choices. Therefore, y is one more than x (x+1), and…</p>
<p>i think it’s been impossible for you to give up trying again and again and checking since the question asks you to give only one possible answer ,which means there are so many answers, and you’ll not list them all ,right??</p>