If Qs are numbers less then 200 when divided by 5 or 7 leaves remainder 2.Find the sum of all Qs.

• First find the l.c.m of 5,7=35. The no Q will be in the form 35p+2. Putting p=1,2 and so on we get the various values of Q which satisfy the above condition.putting p=1,2,3,.....5 then p will be 37,72,107,142,177 respectively...hence the sum is 535.
  
• 9 years agoHelpfull: Yes(41) No(6)
• 1st find for 5 u will get no like 2,7,12,......197
  Sn=3980
  2nd find for 7 u will get no like 2,9,17,.....198
  sn=2900
  find the no like 2,37,72...........=Sn =537
  
  total = 3980+2900 - 537
  = 6343
  
  
• 8 years agoHelpfull: Yes(8) No(0)
• @Ravi u should u consider 2 also , because when we divide 2/35 reminder will be 2
  hence some of Qs numbers= 535+2=537
• 9 years agoHelpfull: Yes(7) No(4)
• 1st find for 5 u will get no like 7,12,......197
  Sn=2277
  2nd find for 7 u will get no like 9,17,.....198
  sn=3978
  find the no like 37,72...........=Sn =535 and sub tract the sum with both the Sn
  and then add all Sn1=2277-535=1742
  Sn2=3978-535=3443
  Sn(final)=1742+3443+535=5720ans
• 9 years agoHelpfull: Yes(6) No(8)
• general form for forming N = 35 * a +2
  (lets a= 1,2,3,4,5)
  so , N can be ----> 37,72,107,142,177
  sum of all no is = 37 +72 +107 +142 +177 = 535
• 8 years agoHelpfull: Yes(2) No(1)
• 1>
  on divide by 5
  the series would be 7,12,17.........197
  last digit (l)=a+(n-1)d
  on solving n=39
  sum of all term (sn1)=n/2[2a+(n-1)d]
  =141 ;a=7, n=39,d=5
  similarly
  for 7
  series 9,........198
  (l)=a+(n-1)d
  n=28
  sn2=2898
  now the final ans sn=sn1+sn2
  =141+2989
  =3039
• 9 years agoHelpfull: Yes(1) No(5)
• we know if reminder is same then,
  General form=Lcm(divisor)*Y+reminder
  =Lcm(5,7)*Y+2,
  in questn we have to find sum of all Qs,Which is less than 200,
  so there are 6 nos are there which are less than 200,
  they are--->2,37,72,107,142,177
• 8 years agoHelpfull: Yes(1) No(1)
• First find the l.c.m of 5,7=35. The no Q will be in the form 35p+2. Putting p=1,2 and so on we get the various values of Q which satisfy the above condition. When p=1 Q=37. P=2 Q=72 p=3 Q=167. Sum of them = 276
• 9 years agoHelpfull: Yes(0) No(9)
• @RAYAGOUDA, its not the case of mod, its the case of division which is clearly mentioned in the que. so we cant take extra 2
  
• 9 years agoHelpfull: Yes(0) No(2)
• first of all we have numbers starting from 7 to 197 which gives the reminder 2 when devided 5 now as we can see the arithmetic progression so by using the formulae tn=a+(n-1)d wr the a is the starting no and d is the diffrence between the numbers now first we need to find tn means t197=7+(197-1)5=987 then to find the total sum of no mpresent in the Qs we need to apply this formulae sum of no=n/2(a+tn) i.e sum=197/2(7+987)=97909
• 9 years agoHelpfull: Yes(0) No(3)
• First find the l.c.m of 5,7=35. The no Q will be in the form 35p+2. After this put the value of p from 0,1,2
  so the total sum going to be 537
  
• 8 years agoHelpfull: Yes(0) No(0)
• First find the l.c.m of 5,7=35. The no Q will be in the form n*35-2. put p=1,2,3,4,5. find the values and add them ..
  33+68+103+138+173= 515
• 8 years agoHelpfull: Yes(0) No(0)
• 537
  take LCM of 5 and 7 that is 35
  and remainder is 2 so first no is 35+2 = 37
  and general case is (35x+2) put the values of that satisfy the the question condition
• 6 years agoHelpfull: Yes(0) No(0)