<p>Q: a certain number y has a remainder of 3 when divided by 5. That same number has a remainder of 2 when divided by 4. If y is an integer between 10 and 40, what is one possible value of y?</p>

<p>The book answer is to list out the numbers between 10 and 40 that have a remainder of3 when divided by 5. then, divide the numbers by 4 to see which give a remainder of 2.</p>

<h2>is there a shortcut to this problem?</h2>

<p>what about longer problems like this one (i made up the variables):

a certain number y has a remainder of 3 when divided by 5, a remainder of 2 when divided by 4, a remainder of x when divided by z, and a remainder of w when divided by v. what is y?</p>

<p>This isn't exactly a "short cut" but think about it this way:

Any number that will have a remainder of 3 when divided by 5 will either end in 3 or 8 (since multiples of 5 end in 0 and 5).

If it has a remainder of 2 when divided by 4, then it must be an even number, so this narrows it down to numbers that end in 8.

If you're looking for a number between 10 and 40, then you now have only 3 options (18, 28, 38).

You should know off the top of your head that 4x4 = 16 and 4x9=36, so 18 and 38 are possible answers. Or you could eliminate 28, because you should also know off the top of your head that 7x4=28. Either way you do it, it shouldn't take too long.</p>

<p>ILoveBrown's approach is probably the fastest and most direct solution.</p>

<p>An equivalent and more formal approach would be:</p>

<p>Since Y divided by 5 leaves 3, we know Y is equal to a multiple of 5 plus 3, so</p>

<p>*Y = 5a + 3, for some arbitrary a*</p>

<p>Likewise Y divided by 4 leaves 2, we know Y is equal to a multiple of 4 plus 2, so</p>

<p>*Y = 4b + 2, for some arbitrary b*</p>

<p>Equating and doing arithmetic</p>

<p>*5a + 3 = 4b + 2,*</p>

<p>*5a + 1 = 4b*</p>

<p>This type of equation, requiring integer solutions, is a linear Diophantine equation in two variables. At this point , we see there exist solutions for a = 1 (Y = 18), a = 7 (Y=38), a= 11 (Y = 58) and so on. For discussion of Diophantine equations, see <a href="http://mathworld.wolfram.com/DiophantineEquation.html%5B/url%5D">http://mathworld.wolfram.com/DiophantineEquation.html</a></p>

<p>p.s. the initial problem is succinctly stated in terms of modular equivalences:

*Y is equivalent to 3 modulo 5*

Y is equivalent to 2 modulo 2</p>