54. An absentminded professor has a very peculiar problem, in that he cannot remember numbers larger than 15. However. he tells his wife. I can remember any number up to Ioo by remembering the three numbers obtained as remainders when the number is divided by 3, 5 and 7 respectively. For example. (2.23) is 17. Professor remembers that he had (l.l.6) rupees in the purse, and he paid (10.6) rupees to the servant. How much money is left in the purse?

• 21
  he had : 76 rupees(gives rem 1,1,6)
  he gave : 55 rupees (gives rem 1,0,6)
  now he has : 21 rupees
• 10 years agoHelpfull: Yes(8) No(7)
• 11.6-10.6 is money left=01.0
  so 21 is the LCM for 7 and 3 which gives 0 as remainder.
  
• 10 years agoHelpfull: Yes(4) No(2)
• please explain how we get 76 and 55 clearly with procedure
• 10 years agoHelpfull: Yes(3) No(0)
• He had 76 rupees,according to questions, he remember any no. larger than 15 as remainder when divided by 3,5 and 7 respectively.
  Divide 76 by 3,5 and 7, we will get (1,1,6)as remainder.
  He paid 20 rupees bcoz when 20 divided by 3,5 & 7 , we will get (2,0,6) as remainder.
  Now money left=76-20
  =56 (Ans) :) :) :)
• 9 years agoHelpfull: Yes(2) No(1)
• answer plz
  
• 9 years agoHelpfull: Yes(0) No(0)
• plz explain how 2 get 76 and 75 clearly
• 9 years agoHelpfull: Yes(0) No(0)
• He had (1,1,6) When a number divided by 7 the remainder is 6
  Possible values less than 100 will be 6,13,20,27,34,41,48,55,62,69,76,83,90,98 When divided by #&5 the remainder is 1 so only possible value=76
  Similarly 1,0,6- Divisible by 5 (20 or 55) when divided by 3 remainder is 1 so 55 he spent
  76-55=21(saving)
• 7 years agoHelpfull: Yes(0) No(0)