Once you attempt the question then PrepInsta explanation will be displayed.

Assuming that you know binomial theorem… The solution is as follows

or

“Remainder when 17 power 23 is divided by 16" is a bit challenging question for the people who are getting prepared for competitive exams.

Actually it is not a difficult question and it can be answered easily once we know the stuff.

Let us take exponents 0, 1, 2, 3, ....one by one to "17"

For example, if we take exponent "0" for 17, we get 17?

** = 1**
Here 1 is less than the divisor 16 and 1 can not be divided by 16.

If the dividend is less than the divisor, then that dividend itself to be considered as "Remainder"

So, if 17? is divided by 16, we get the remainder "1"

If the dividend is greater than the divisor, then we have to divide the dividend by the divisor and get remainder.

Let us deal our problem in this way.

When we look at the above table carefully, 17? is divided by 16, we get the remainder "1".

Again we get remainder "1" for the exponent "1".

Next we get remainder "1" for the exponent "2" and so on.

So, we get remainder "1", for all the exponents we take for 17.

**Since we get the remainder 1 for all the exponents we take for 17,**

**when we divide 17²³ by 16, the remainder will be "1 ".**

17 ^1 ends with 7 in unit's digit

17 ^2 ends with 9

17^3 ends with 3

17 ^4 ends with 1

17^5 ends with 7....

So from here on it follows a cyclicty of 4. after that it follows a pattern.

Now 23 can be written as 4*5 +

**3**
So if u find the remainder of

17 ^3 , it will be same as

17^23.

remainder of 17^ 3 when divided by 16 is 1 .

So for

17^23 also it will be

**1.**
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