Professor absentminded 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 100 by remembering the three numbers obtained as remainders when the number is divided by 3, 5 and 7 respectively. For example (2,2,3) is 17. Professor remembers that he had (1,1,6) rupees in the purse, and he paid (2,0,6) rupees to the servant. How much money is left in the purse?
Choose the most appropriate option:
A) 56 B) 57 C) 58 D) 59 E) 60
Let the money with the professor = N   Then N = 3a +1 = 5b + 1 = 7c + 6.   Solving the above we get N = 181   (Explanation: See LCM formula 1 and 2: Click here)   When a number is divided by several numbers and we got same remainder in each case, then the general format of the number is LCM (divisors).x + remainder.   In this case 3, 5 are divisors. So N = 15x + 1. Now we will find the number which satisfies 15x + 1 and 7c + 6.   ⇒ 15x + 1 = 7c + 6 ⇒ c = 15x−5715x−57 ⇒ c = 2x+x−572x+x−57   Here x = 5 satisfies. So least number satisfies the condition is 5(15)+1 = 76.   (x = 12 also satisfies condition. So substituting in 15x + 1 we get, 181 which satisfies all the three equations but this is greater than 100)   Similarly Money given to servant = M = 3x + 2 = 5y 3  + 2= 
<gadget id="calculator">3  + 2</gadget>
<output>5</output>
5y= 7z + 6   Solving we get M = 25.   (125 also satisfies but this is next number)   Now N - M = 56   56nswer:	56
<result>A</result>
Professor absentminded 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 100 by remembering the three numbers obtained as remainders when the number is divided by 3, 5 and 7 respectively. For example (2,2,3) is 17. Professor remembers that he had (1,1,6) rupees in the purse, and he paid (2,0,6) rupees to the servant. How much money is left in the purse?