17^23 is divided

i have simply done 17 divide by 16 gives remainder 1 for 23 numbers of 17 will give reaminder 1.

is this true for all this type of questions

pradeepkaushal9518 wrote:i have simply done 17 divide by 16 gives remainder 1 for 23 numbers of 17 will give reaminder 1.

is this true for all this type of questions

yup thats the correct approach and thats what i wanted to explin

nathanalgren wrote:Guys, if you know what modular arithmetic is, this is the easiest and fastest way to solve these kind of questions.


17^1 = 1 (mod 16)

So any power of 17 = 1 (mod 16)

This could be implied to any kind of remainder questions. The other approaches written here are okay, but this is far best.

Another example: Suppose it asked the remainder of 18^34 divided by 13.

18^1=5 (mod 13)
18^2=12 (mod 13)
18^3=8 (mod 13)
18^4=1 (mod 13)

The remainder of 34/4 is 2. So we look at 18^2=12 (mod 13).

Hence the remainder of 18^34 divided by 13 is equal to 12.

could u explin in more detail how it works? how are u calculating these mods?

thanks

amising6 wrote: You will get 2*2*2*2*..23 times(if I have correctly understood your point) ...then how will you proceed from here...??
now make a group of four 2's i.e 16 *16*16*16*16*8 divide by 15
so every 16 will give 1so u will be left with 1*8 mod 15=8

Yup !! It works and thanks for making me understand this concept.

I would personally prefer little binomial concept over using cyclicity, mod arithmetic, or any dicey imagination whatsoever to save time, on such questions in particular. Most of the GMATians are unfamiliar with mod arithmetic which is not as complex as the cyclicity rules. [spoiler]Never mind Govind if a smile is not paid, one day people would burst to laughter on your posts; leave alone a smile[/spoiler]

https://people.richland.edu/james/lectur ... omial.html

We can realize that in the expansion (a + b) ^n, all terms EXCEPT b^n have a as factor. Hence when a divides the expansion (a + b) ^n, the remainder is decided by b^n only.