What is the least 3 digit number which when divided by 7 and 11 leaves remainders of 6 and 8 respectively? Please tell me how to do it...
Answers
Answer:
If you look at the respective Divisors and the Remainders, you will notice that the difference between the Divisors and the Remainders is 6 in both the cases. What this means is that if you take the number as 7 (in the first case) and subtract 6 from it, you get 1, which is the required number as the Remainder required is 1. Similarly, if you take the number as 9 (in the second case) and subtract 6 from it, you get 3, which is the required number as the Remainder required here is 3. Since we are looking for a common number, satisfying both the conditions, we need to arrive at the LCM and then subtract 6 from it. This makes it easy to solve.
The LCM of 7 and 9 is 63 and if you subtract 6 from it, you get 57, which is the number that partly meets the given criteria.
We know that Number = Divisor*Quotient + Remainder
57 = 7*8 + 1
57 = 9*6 + 3
But we are looking for the smallest 3 digit number, therefore, 57 is not the answer. But now we know that we have to look for certain multiple of 63 and then subtract 6 from it to arrive at an answer.
63*2 = 126. This is the smallest multiple of 63 after 63 (63*1) itself. This also happens to be a 3 digit number and this is smallest 3 digit number that could be divided by both 7 and 9. All we have to do now is to subtract 6 from it to arrive at the requiried answer.
126 - 6 = 120.
This 120 is the required answer.
120 = 7*17 + 1
120 = 9*13 + 3
Step-by-step explanation:
please follow me