. sum of the consequent 7 integers some 1613. Find the no. of prime numbers in that sequence

• WRONG QUESTION!!
  CORRECT QUESTION: Sum of the consequent 7 integers equals 1617. Find the no. of prime numbers in that sequence.
  
  SOLUTION:
  Let the numbers be x, x+1, x+2, x+3, x+4, x+5, x+6
  Since their sum equals 1617, we have-
  x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)=1617
  7x+21=1617
  7x=1596
  x=228
  
  Hence the numbers are: 228,229,230,231,234,235,236
  Prime Numbers: 229, 231
• 12 years agoHelpfull: Yes(56) No(19)
• wrong question..please submit correct1..thank you bruce!!!
• 12 years agoHelpfull: Yes(8) No(4)
• Let the numbers be x, x+1, x+2, x+3, x+4, x+5, x+6
  Since their sum equals 1617, we have-
  x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)=1617
  7x+21=1617
  7x=1596
  x=228
  
  Hence the numbers are: 228,229,230,231,232,233,234
  Prime Numbers: 229, 231,233
• 11 years agoHelpfull: Yes(7) No(5)
• How 231 become a prime no.?? 231 is divisible by 3 231/3=77
• 11 years agoHelpfull: Yes(6) No(0)
• Let the numbers be x, x+1, x+2, x+3, x+4, x+5, x+6
  Since their sum equals 1617, we have-
  x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)=1617
  7x+21=1617
  7x=1596
  x=228
  
  Hence the numbers are: 228,229,230,231,234,235,236
  Prime Numbers: 229, 231
• 11 years agoHelpfull: Yes(5) No(11)
• Let the numbers be x, x+1, x+2, x+3, x+4, x+5, x+6
  Since their sum equals 1617, we have-
  x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)=1617
  7x+21=1617
  7x=1596
  x=228
  
  Hence the numbers are: 228,229,230,231,234,235,236
  
  Prime numbers are:- 229 and 233
• 10 years agoHelpfull: Yes(3) No(0)
• Let the numbers be x, x+1, x+2, x+3, x+4, x+5, x+6
  Since their sum equals 1617,
  we have-
  x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)=1617
  7x+21=1617
  7x=1596
  x=228
  
  Hence the numbers are: 228,229,230,231,232,233,234
  Prime Numbers: 229,231,233
• 11 years agoHelpfull: Yes(1) No(2)