Question

# Find the greatest number which on dividing 1657 and 2037 leaves a remainder of 6 and 5 respectively.A) 123B) 127C) 235D) 305

Hint: Since, we need the specific remainders, we will subtract the remainder from their respective numbers. Now, we need to find the greatest number which divides both the numbers together. So, we are basically looking for HCF of both the numbers.

We have with us, the two numbers which are: 1657 and 2037.
When we divide these numbers by the required number, we have the remainders of 6 and 5 respectively.
So, let us first just remove these remainders to get a clean and perfect division.
Now, we will have: 1657 – 6 = 1651
And 2037 – 5 = 2032.
Now, we have created two new numbers 1651 and 2032 which are perfectly divisible by the number we are required to find.
If two numbers are divisible by some number that means that number is a factor of both the numbers.
The number is given to be greatest.
Hence, we just need to find HCF.
So, let us first do the prime factorization of $1651 = 13 \times 127$ and $2032 = 2 \times 2 \times 2 \times 2 \times 127$.
We see that they only have 127 common in their prime factorization.
Hence, HCF(1651, 2032) = 127.
Hence, the greatest number which divides 1657 and 2037 leaves a remainder of 6 and 5 respectively is 127.

Note: Trial and error can be used to solve this problem. With the help of Trial and error method we can get the answer easily for simple calculations. But for complex calculations it is better to solve the problem using the concept.
Remember to remove the remainders from the actual numbers. If we do not remove the remainder the factors and the HCF changes and we will get the wrong answer.
We do not necessarily need to use the prime factorization method, we can also use less multiples but that might create some problems as if we have a factor 6 in one and 9 in other, we will ignore it but they have 3 in common, which we require.