5. Sum of the digits of a three digit number is 17. Sum of squares of digits of the given number is 109. If we subtract 495 from that number we will get a number written in reverse order. Find the number.

• let a,b,c
  a+b+c=17....eq1
  a^2+b^2+c^2=109......eq2
  100a+10b+c-495=100c+10b+a.......eq3
  eq3- 99a-99c=495
  a-c=5
  a=5+c
  for c=1 a=6 and b=10 but eq2 not true.
  for c=2 a=7 and b=8 but eq2 not true
  for c=3 a=8 and b=6 so eq2 is true
  digit is 863 ans
• 11 years agoHelpfull: Yes(32) No(2)
• 863,
  
  SINCE, 8^2+6^2+3^2=109 ,
  AND 863-495=368,
  SO, NUMBER IS 863 ANS.
• 11 years agoHelpfull: Yes(7) No(0)
• dummy question.....just do option verification...u don't need any tricks.
• 11 years agoHelpfull: Yes(3) No(0)
• 863
  take 3 digit as a+b+c=17
  sum of square is a^2+b^2+c^2=109
  100a+10b+c-495=100c+10b+a
  99(a-c)=495
  a-c=5
  
• 11 years agoHelpfull: Yes(1) No(5)
• no. is 863 8*8+6*6+3*3=109
  863-495=368
• 11 years agoHelpfull: Yes(1) No(3)
• 8+6+3=17
  now by squaring we get=64+36+9=109
  now 863-495=368
  so by subs. we are getting the reverse of that no.
  
• 11 years agoHelpfull: Yes(0) No(1)
• 863 863-495=368 proved
• 11 years agoHelpfull: Yes(0) No(1)
• u can go with the given answer...
  this is the best way to find the no...
• 11 years agoHelpfull: Yes(0) No(0)
• 782-495=287
• 10 years agoHelpfull: Yes(0) No(0)