In a set of five consecutive integers, the smallest integer is more than 2/3 the largest. what is the smallest possible value of the sum of the five integers?

The answer is: "55".
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"55" is the smallest possible value of the sum of the five integers.
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We are given five (5) consecutive integers.
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Let us represent the first integer as: "x"; 
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The second integer as: "x + 1" ; 
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The third integer as: "x + 2" ; 
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The fourth integer as: "x + 3" ; 
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The fifth integer as: "x + 4" ; 
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We are given: The smallest integer is MORE THAN (⅔ of the largest integer.
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The smallest integer, "x" is greater than: "(⅔) (x + 4)"
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Note the distributive property of multiplication:
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→  a*(b + c) = ab + ac ;
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→ As such; (⅔) (x + 4) = [(⅔)*(x)] + [⅔)*(4); 
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→ (⅔) (x + 4) = [(⅔)*(x)] + [⅔)*()
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{Note: [⅔)*()==}; 
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→ Rewrite the equation: → ⅔ of the largest integer =
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→ (⅔) *(x + 4) = (⅔)x + ;
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The smallest, "x", is GREATER than: (⅔)x +  ; 
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→ Write as:  x > (⅔)x + 
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→ Multiply the ENTIRE "inequality" (BOTH SIDES) by "3", to get rid of the "fractions":
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3*{  x > (⅔)x +  } ; to get: 
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→ 3x > 2x + 8 ; → Now, subtract "2x" from EACH SIDE of the inequality;
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 → 3x − 2x > 2x + 8 − 2x ;  
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→ to get:  →  x > 8
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Now, we want to to know the "smallest possible value of the sum five          integers".
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→That is, the smallest possible value of the sum of :
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→x + (x + 1) + (x + 2) + (x + 3) + (x + 4) = 5x + 10.
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So, we want to know the smallest value of "(5x + 10)" ; x > 8, 
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→ Let us solve for "5x + 10" ; when x = 10; by substituting "8" for "x"
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→ 5x + 10 = (5*8) + 10 = 40 + 10 = 50
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So, if "x > 8"; then "x" must be greater than "8"; 
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When "x = 8", the sum of the 5 (five) integers in our problem = 50.
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Since "x" in the first of the consecutive integers in the problem; we know that "x > 8"; then the smallest possible value for "x" would be "9"; since "x" has to be an integer.
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So, we know that:
The smallest possible value of sum of the value integers =
 5x + 10; when x = 9; 
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→ So, we plug "9" for the value of "x" and solve:
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→ 5(9) + 10 = 45 + 10 = 55 ; → which is our answer.
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Let us check our
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→ x + (x + 1) + (x + 2) + (x + 3) + (x + 4) = ? 55?
                                                           (when "x = 9" ?)?? ;
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→  9 + (9 + 1) + (9 + 2) + (9 + 3) + (9 + 4) = ? 55? ;
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→  9  + 10 +  11 + 12 + 13 = ? 55 ?? ;  Yes!.
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Also, is the 55 , Reasonable? Yes, since it is an integer, and 5 consecutive integers added together would "add up" to an integer value.
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Hope this answer and lengthy explanation is helpful.
Best of luck!
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