<p>The value of k is an integer between 50 and 90 and is a multiple of 4. When k is divided by 5, the remainder is 3. When k is divided by 3, the remainder is 2. What is the value of k?</p>
<p>This is a grid in question. Please describe a method to do this without listing out all the values. </p>
<p>Thanks a lot</p>
<p>52, 56, 60, 64, 68, 72, 76, 80, 84, 88</p>
<p>Just from looking it has to be either 68 or 88… because their 3 more than any multiple of 5… 65, 85.</p>
<p>68/3 = 22 + 2/3 </p>
<p>Bingo</p>
<p>Key is to be able to do the divided by 5 part real fast first.</p>
<p>EDIT: I just saw you said not listing the values… but listing would still be the fastest way imo.</p>
<p>Damn, i didnt realize how few numbers r between 50 - 90 that r multiples of 4</p>
<p>thanks</p>
<p>There are probably algebraic ways to solve this. Number theory anyone? I’ve seen a bunch of AIME problems like this… except the range is 100 times as big so you can’t handpick.</p>
<p>^Ha, when I saw this problem, I thought about the Modular Artihmetic stuff I read about in AoPS</p>
<p>The first criteria I looked at was “when k is divided by 5, the remainder is 3.” What does that tell us? </p>
<p>Well, we know that all numbers divisible by 5 end in either 0 or 5. Thus, if when a number is divided by 5 and the remainder is 3, then the number must end in 8 or 3 (5+3,0+3).</p>
<p>Next I looked at “When k is divided by 3, the remainder is 2.” This eliminates all numbers that end in 3. Why? because any number that ends in 3 and gives a remainder of two MUST end in a 1 which is NEVER divisible by 5. For example, when 83 is divided by 3, the answer is 27 (27 x 3 = 81) with a remainder of 2. However, 81 is not divisible by 5.</p>
<p>So now we have eliminated everything except those numbers which end in 8, namely:
58,68,78 and 88
cross out 58 and 78 because they arent divisible by 4.
next cross out 88 because it doesn’t give the remainder 2 when divided by 3.
That leaves us with (drumroll): 68!</p>
<p>Quite elementary, yeah?</p>
<p>
</p>
<p>Yes, but is your second step the correct one? It is not circular to use the "“When k is divided by 3, the remainder is 2” twice (in step 2 and 4).</p>
<p>Let’s look at a slightly different POE:</p>
<p>You start with 1:</p>
<p>
</p>
<p>But why look at “When k is divided by 3, the remainder is 2” now? Why not use the “divisibility by four” here? Can a number ending in 3 be divisible by four? Nope … use this POE step because it is very simple!</p>
<p>As you said, now “we have eliminated everything except those numbers which end in 8, namely: 58,68,78 and 88. Cross out 58 and 78 because they arent divisible by 4.”</p>
<p>And, now you can use the last clue to eliminate 88!</p>