Question

2³³+2³²+2³¹/2³⁴+2³³-2³²​

Answer 1

Answer:

If we use the above approach I'd work with prime as a base.

now if we expand this, all terms but the last one will have 28 as a multiple and thus will be divisible by 7. The last term will be . So we should find the remainder when is divided by 7.

2^1 divided by 7 yields remainder of 2;

2^2 divided by 7 yields remainder of 4;

2^3 divided by 7 yields remainder of 1;

2^4 divided by 7 yields remainder of 2;

2^5 divided by 7 yields remainder of 4;

2^6 divided by 7 yields remainder of 1;

...

The remainder repeats the pattern of 3: 2-4-1.

So we should find (the power of 2) is 1st, 2nd or 3rd number in the above pattern of 3. is 2 in odd power, 2 in odd power gives remainder of 2 when divided by cyclicity number 3, so it's the second number in pattern. Which means that remainder of divided by 7 would be the same as divided by 7. divided by 7 yields remainder of 4.