Find the greatest number which divides 2011 and 2623 leaving remainders 9 and 5 respectively.

The required number when divides 2011 and 2623
Leaves remainders 9 and the means
2011 – 9 = 2002 and 2623 – 5 = 2618 are completely divisible by the number
∴ The required number = HCF of 2002 and 2618
By applying Euclid’s division lemma
2618 = 2002 × 1 + 616
2002 = 616 × 3 + 154
616 = 754 × 4 + 0
∴ HCF of 2002 and 2618 = 154
Hence required number is 154

For the question it’s understood that, 

2011 – 9 = 2002 and 2623 – 5 = 2618 has to be exactly divisible by the number. 

Thus, the required number should be the H.C.F. of 2002 and 2618 

Applying Euclid’s division lemma, we get 

2618 = 2002 x 1 + 616 

2002 = 616 x 3 + 154 

616 = 154 x 4 + 0. (Here the remainder becomes 0) 

And hence the H.C.F. (2002, 2618) = 154 

∴ The required number is 154.