find the sum of all 2-digit numbers which leaves remainder 6 when divided by 8.

• 594
  
  Such a two digit number will form a series 14,22,30,........,94 with difference of 8
  If total numbers are 'n', then 94=14+(n-1)*8, n=9
  So sum=(n/2)[2*14+(n-1)*8]=(11/2)[(2*14 +(11-1)*8]=594
• 9 years agoHelpfull: Yes(0) No(0)
• Pl. read number of terms n=11 in my earlier solution.
• 9 years agoHelpfull: Yes(0) No(0)