<p>p.839</p>
<li>Which of the following could be the remainders when4 consecutive positive integers are each divided by 3?</li>
</ol>
<p>How would I go about this problem?</p>
<p>Let's choose numbers:</p>
<p>Let's just find the remainders of 10 numbers if divided by 3. This is to see a pattern.</p>
<p>The remainders would be (If starting with number 3 and moving towards 13):</p>
<p>0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1</p>
<p>We see a pattern. This same pattern is consistent with D</p>
<p>Ans: D</p>
<p>The answer would be either</p>
<p>0,1,2,0
1,2,0,1
2,0,1,2</p>
<p>Try out 4 consecutive positive integers and see for yourself!</p>
<p>Let's say your first integer is 4. Okay, so when you do 4 /3, your remainder is 1.</p>
<p>Now, let's do 5. </p>
<p>5 = 3(1) + 2. So the remainder is 2.</p>
<p>4 = 3(1) + 1
5 = 3(1) + 2
6 = 3(2) + 0
7 = 3(2) + 1
8 = 3(2) + 2
9 = 3(3) + 0.</p>
<p>See the pattern?</p>
<p>The remainder from division by 3 can not be greater than 2.
(D) is the only answer which satisfies that condition.</p>
<p>For example
23 = 6x3 + 5 -> 5 "contains" one more 3.
23 = 7x3 + 2 -> 2 is the correct remainder.</p>