how many 3 digit no which on dividing by 5,6,7 leaves remainder 2,3,5

• for 5 the 3 digit no last digit must 2,7 then only remainder will be 2
  for 6 the 3 digit no last digit must 1,3,5,7,9 then only remainder will be 3
  for 7 the 3 digit no last digit must 1,2,3,4,5,6,7,8,9 then only remainder will be 2
  compare all last digits we have only 7 as common so last digit must be 7
  for 7 if we want 5 as remainder so digits must 117,259,327,397,467,537,617,677,747,,827,887,957
  from above digits digits which give 2&3 as remainders when number divided by 5&6 are 117,327,537,747,957 so total of 5 digits
  answer 5
• 11 years agoHelpfull: Yes(37) No(6)
• LCM of 5,6,7 is 210, so the number will be exactly divisible with difference of 210. similary with remainder 2,3,5 will be on difference of 210
  numbers divisible by 5 leaving remainder 2 are: 102,107,112,117.....
  number divisible by 6 leaving remainder 3 are 117......
  number divisible by 7 leaving remainder 5 are 117.......
  so the first three digit no by 5 leaving remainder 2 and 6 leaving remainder 3 and 7 leaving remainder 5 is : 117
  the difference is :210
  so the nos are : 117,327,537,747,957
  hence there are 5 such nos..
  ans is 5
• 11 years agoHelpfull: Yes(29) No(0)
• Numbers that leave remainder 2 on dividing by 5 are:
  102,107,112,......997
  Quotients on dividing by 5 are
  ->> 20,21,22,23.....199
  
  now out of (20,21,22....199) numbers which give remainder as 3 on division are
  21,27,33,39........195
  quotients obtained on dividing these numbers by 6 are:
  3,4,5,6,7,8,.....32
  
  now out of (3,4,5,6...32) numbers which give 5 as remainder on division by 7 are:
  12,19,26 only
  
  so answer should be 3
  
  please give your valuable feedback .
• 11 years agoHelpfull: Yes(25) No(18)
• numbers divisible by 5 leaving remainder 2 are: 102,107,112,117,.......997(differ by 5)
  common numbers divisible by 5 nd 6 leaving remainder 2 nd 3 respectively are:117,147,177,207,237,267,297,327.......987.(differ by 30)
  now out of these the common numbers divisible by 5,6,7 leaving remainder 2,3,5 respectively are:117,327,537,747,957(differ by 210).
  so the answer is 5.
• 11 years agoHelpfull: Yes(18) No(1)
• general no is of d form:210x-2 put x=1,2,3,4
  ans=4
• 11 years agoHelpfull: Yes(6) No(6)
• YES THERE ARE ONLY 5 OPTION...117,324,537,747,957
  LOGIC IS IF U DIVIDE A NUMBER BY 5 AND IT LEAVES A REMAINDER 2 MEANS NUMBER HAVING LAST DIGIT EITHER 7,2...AND IF SAME NUMBER DIVIDED BY 6 LEAVES REMAINDER 3 MEANS MULTIPLE OF 6 HAVING LAST DIGIT EITHER 4,9...9 NEVER COMES IN LAST DIGIT IN MULTIPLE OF 6 SO ALL THE OPTION HAVING ONLY LAST DIGIT 7 GUYS IT'S JUST A HINT TRYING YOURSELF.
• 11 years agoHelpfull: Yes(5) No(0)
• n*LCM(5,6,7)-3
  put n=1,2,3,4
  ans=4
• 11 years agoHelpfull: Yes(3) No(8)
• 117,327,537,747,957 thus 5 in all
• 11 years agoHelpfull: Yes(3) No(5)
• When a three digit number is divided by
  5,6,7 successively, the final quotient will be
  a single digit (as 5*6*7 = 210 and the max
  three digit number 999 is not more than
  210*5).
  Let the number be x and the quotient after
  division by 5 be a and the quotient after
  division by 6 be b and the quotient after
  division by 7 be c.
  Then, x = 5a+2,
  a = 6b +3
  b = 7c + 5.
  Now start with c=1, b=12, a=75, x=377
  when c=2, b=19, a=117, x=587,
  c=3, b=22, a=113, x=454 and so on with
  an increment of 210. Thus the 3 digit
  numbers possible are 377,587,797
  The answer is 3 such numbers are
  possible.
• 9 years agoHelpfull: Yes(0) No(0)