A number divides 206 , 368 , 449 leaving the same remainder in each case. The laegest such number is?

Question

anonymous · Answer

@Arnab09 @experimentX @apoorvk @yash2651995 @lgbasallote

experimentx · Answer

i guess it is best to start with factorizing it

apoorvk · Answer

Similar to yesterday's question.

Let the number be 'n'. And the common remainder left be 'x'.
Now, (206-x), (368-x), (449-x) are perfectly divisible by 'n'.

Also, the difference of any of these two numbers should be divisible by 'n'.
So, 449-368=81, 449-206=243, and 368-206=162, i.e. 81, 243 and 162 should be completely divisible by that 'n'.
Now since, we need to find the largest possible value of 'n', a no. which divides the above there perfectly, if we find the HCF of the three nos. just above, we can find 'n'.

So, HCF of 81, 162 and 243 = 9.

And my friend, if you check, 9 completely satisfies every criteria in the question. So, 9 it is.