Question

Find the two smallest integers which divided by 3,7,11 leave remainders 1, 6, 5 respectively.

Answer 1

Hint:By using the Euclid’s division lemma, we can write a number in the following form
\[a=b\times q+r\]
(Where a is a number, b is the divisor, q is the quotient and r is the remainder)
Now, we can write the unknown numbers using the above lemma and then get to the answer.

Complete step-by-step answer:
As mentioned in the question, we are asked to find the two smallest integers which divide by 3, 7, and 11 leave remainders 1, 6, 5 respectively.
 Let the positive integer is x.
Now, we can represented the number as follows
x=3p+1 (x could be ...1,4,7,10,13 ...181...)
x=7q+6 (x could be ...6, 13, 20, 27...181...)
x=11r+5 (x could be ...5, 16, 27...181…)
 (By using the information given in the hint)
 Therefore, the divisor will be LCM of 3,7,11 and that can be evaluated as follows
 \[LCM\ of\ 3,7,11=231\]
 Now, the remainder will be the first common factor of three series that we have mentioned above and it will be equal to 181.
So, the general form of such number can be written or represented as follows
\[x=231k+181\]
Now, for k=0, the number comes out to be
x=181
And for k=1, the number comes out to be
x=412
(Because 0 and 1 are the smallest integers for which the required number comes out to be a positive integer)
So , 0 and 1 are the smallest integers which divided by 3,7,11 leave remainders 1, 6, 5 respectively.

Note: -The students can make an error in solving and getting to the right answer if they don’t know about the Euclid’s division lemma as without the knowledge of it one could not get to the correct solution.We can check the solution by substituting 0 and 1 in general form.If we substitute k=0 ,we get 181 in which the remainder will be 1,6,5 when it is divided by 3,7 and 11.Similarly we can check for k=1.