If the product of 3 consecutive integers is 210 then the sum of the two smaller integers is:

(a) 17
(b) 13
(c) 11
(d) 10

• let the numbers be x-1 , x , x+1
  (x-1)(x)(x+1)=210
  210=2*105
  =2*3*35
  =2*3*5*7
  this can be written as 6*5*7
  =>x-1=5
  x=6
  x+1=7
  sum of two smallest numbers=6+5=11
• 8 years agoHelpfull: Yes(42) No(3)
• factors of 210 are 2,3,5,7 = 6,5,7 therefore sum of two smallest integers is 6+5=11
• 8 years agoHelpfull: Yes(19) No(2)
• Logically...
  1's digit is "0" so the number is multiple of 5 or 10.
  if we divide 210 by 10 then we get quotient as 21 so factors are 3 and 7 which are not consecutive
  So 5 is one factor and we get 42 as quotient when we divide 210 by 5 so factors of 42 are 6 and 7.
  so the factors are 5,6,7
  and sum of two smaller integers is 11(5+6).
• 8 years agoHelpfull: Yes(7) No(0)
• the product is 210.then we have to take least numbers which can get the product of 210.if we go for 5,6,7 wee get the product 210.in these three numbers least numbers are 5,6 as compare to 7.so if we add 5,,6 then wee get as 11 as sum.
  ans :C
• 8 years agoHelpfull: Yes(3) No(1)
• 5*6*7=210
  so
  5+6=11
  option(c)
• 8 years agoHelpfull: Yes(1) No(0)
• 10
  the three consecutive integers be 3,7,10 then sum of two smaller integers is 10
• 8 years agoHelpfull: Yes(0) No(9)
• 11
  From the factors we get no's as 5,6,7
• 8 years agoHelpfull: Yes(0) No(0)
• 11 is the ans
• 8 years agoHelpfull: Yes(0) No(0)
• ans is 11
  5*6*7=210
• 8 years agoHelpfull: Yes(0) No(0)
• 5*6*7=210 then 5+6=11
• 8 years agoHelpfull: Yes(0) No(0)
• 5*6*7=21
  sum of two smaller=11
• 8 years agoHelpfull: Yes(0) No(0)
• answer is option C.
  the 3 consecutive integers are 5,6,7.
  therefore the sum of 2 samller integers= 5+6= 11
• 8 years agoHelpfull: Yes(0) No(0)