A set of five positive integers has an arithmetic mean of 150. A particular number among the five exceeds another by 100. The rest of the three numbers lie between these two numbers and are equal. How many different values can the largest number among the five take? Choose one:	A) 18	B) 19	C) 21	D) 42	E) 59
Lets say numbers are in this order: x &lt; y = y = y &lt; x+100. Question is, how many different values can x take.   x+3y+x+100 = 750   2x+3y = 650   We have condition x &lt; y &lt; x + 100 so lets use it   2x + 3y &lt; 2x + 3*(x+100) = 5x + 300   2x + 3y &gt; 2x + 3x = 5x   2x + 3y  2  + 3= 
<gadget id="calculator"> 2  + 3</gadget>
<output>5</output>
5x   2x + 3y= 650   So we got double inequality:   5x &lt; 650 &lt; 5x + 300 which turns into   350 &lt; 5x &lt; 650   70 &lt; x &lt; 130   x can take   130-70-1= 
<gadget id="calculator">  130-70-1</gadget>
<output>59</output>
59 values   ANSW59R:59
<result>E</result>
A set of five positive integers has an arithmetic mean of 150. A particular number among the five exceeds another by 100. The rest of the three numbers lie between these two numbers and are equal. How many different values can the largest number among the five take?