Question:
Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.
Solution:
We need to find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.
The required number when divides 280 and 1245 , leaves remainder 4 and 3 , this means $280-4=276$ and $1245-3=1242$ are completely divisible by the number.
Therefore, the required number = H.C.F. of 276 and 1242.
By applying Euclid’s division lemma
$1242=276 \times 4+138$
$276=138 \times 2+0$
Therefore, H.C.F. = 138.
Hence, the required number is138