Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3,

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