remainder prob

digitaltejas · Joined: 03 Feb 2006 Posts: 20

What is the remainder of 17^24 when divided by 300?
A) 121 b) 168 c) 241 d)none

plz ans wid steps

PrasadShah · Joined: 09 Aug 2006 Posts: 2

Hi Tejas,

Please find the solution below. Hope it helps you with the remainder problem.

Regards,
Prasad

Solution :
17^24 = (17^2)^12 = 289^12=(300-11)^12

So, (300-11)^12 has 13 terms with 12 of them having 300 in them. So they will leave no remainder. Last term (11^12) by 300 will posibly leave a remainder.

Lets solve 11^ 12 further to some simple term

11^12= (11^2)^6= (121)^6 = (120 +1)^6
This would have 7 terms in it, with the first 5 terms divisible by 300 if you expand using Binomial theorem.

This will leave us with last 2 terms which are 6C5 * 120^1 * 1^5 and 6C6 * 120^0 * 1^6 to be divided by 300, which can be solved to leave a remainder of 121.

So the answer for the remainder comes to be (A) 121.

digitaltejas · Joined: 03 Feb 2006 Posts: 20

Thx for the great answer..