<PRE><B>Question 181858</B>
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*[tex \LARGE \text{          }\math 3 \times 5 \times 8 = 120]


Multiply 120 by any integer in the range *[tex \Large 9 \leq x \leq 83]


That&#39;s because *[tex \Large {1000 \over 120} = 8.\overline{3}], so you need the next higher integer to make sure you have at least 4 digits and, *[tex \Large {10000 \over 120} = 83.\overline{3}], so you need the next lower integer so that you make sure you have no more than 4 digits.  Lastly, you don&#39;t care about the divisibility of this last factor because you have already ensured divisibility of the 120 factor by creating it from the product of the three given divisors.


In fact you could probably get extra credit for this question if you answered it by saying:


*[tex \LARGE \text{          }\math \{x | x, y\ \in \ I, \ x = 120y,\  9 \leq y \leq 83\}]


Meaning, the set of all &lt;i&gt;x&lt;/i&gt; such that &lt;i&gt;x&lt;/i&gt; and &lt;i&gt;y&lt;/i&gt; are integers, &lt;i&gt;x&lt;/i&gt; is equal to 120 times &lt;i&gt;y&lt;/i&gt; and &lt;i&gt;y&lt;/i&gt; is in the inclusive interval 9 to 83, which is a description of the set of  &lt;i&gt;&lt;b&gt;all&lt;/b&gt;&lt;/i&gt; four-digit numbers evenly divisible by 3, 5, and 8.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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