Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively.

The require number when divides 285 and 1249, leaves remainder 9 and 7, 

this means 285 – 9 = 276 and 1249 – 7 = 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 remainder is = 0
Hence required number is 138

From the question it can be understood that the required number when divides 285 and 1249, leaves remainder 9 and 7 respectively should be the number which 

285 – 9 = 276 and 1249 - 7 = 1242 can divide them exactly. 

So, if the H.C.F. of 276 and 1242 is found then that will be the required number. 

Now, by applying Euclid’s division lemma, we get 

1242 = 276 x 4 + 138 

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

So, the H.C.F. = 138 

∴ The required number is 138.

The new numbers after subtracting remainders are:

285 - 9 = 276

1249 - 7 = 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 285 and 1249 leaving remainder 9 and 7 respectively is 138

Let number be N
276(285-9) is multiple of N
1242(1249-7) is multiple of N
N is HCF of (276,1242)
LCM(276)=2*2*3*23
LCM(1242)=2*3*3*3*23
N=HCF=23*2*3=138.