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 the remainder 4 and 3, this means
280- 4 = 276 and 1245 – 3 = 1245 – 3 = 1242 are completely divisible by the number
∴ The required number = HCF of 276 and 1242
By applying Euclid’s division lemma
1242 = 276 × 4 + 138
276 = 138 × 2 + 0
∴ HCF = 138
Hence the required numbers is 138

From the question it’s understood that, 

280 – 4 = 276 and 1245 – 3 = 1242 has to be exactly divisible by the number. 

Thus, the required number should be the H.C.F. of 276 and 1242. 

Now, applying Euclid’s division lemma 

1242 = 276 x 4 + 138 

276 = 138 x 2 + 0 (the remainder becomes 0 here) 

And hence the H.C.F (280, 1245) = 138 

∴ The required number is 138.

The new numbers after subtracting remainders are:

280 - 4 = 276

1245 - 3 = 1242

Prime factors of 276 = 23 × 3 × 2

Prime factors of 1242 = 2 × 3 × 3 × 3 × 23

Therefore HCF of 276 and 1242 is:

2 × 3 × 23 = 138

Hence the greatest number which divides 280 and 1245 leaving remainder 4 and 3 respectively is 138