Find Remainders: 3^2002/26, 5^2002/26Date: 09/21/2002 at 22:10:23 From: Mr. Hugehy Subject: Powers (a) Find the remainder obtained when 3^2002 is divided by 26 (b) Find the remainder obtained when 5^2002 is divided by 26 (c) Show that 3^2002 + 5^2002 is divisible by 26 Kids have asked me these questions and I can't answer them. Date: 09/22/2002 at 00:25:17 From: Doctor Greenie Subject: Re: Powers Hi, Mr. Hugehy - Here is a link to a page in the Dr. Math archives where there is some information on how you can go about answering these questions: 2^99 Divided by 99 http://mathforum.org/library/drmath/view/55599.html If you use the technique described on that page, you will find that the remainders when 3^n is divided by 26 for n=1 to n=10 are the following: 3, 9, 1, 3, 9, 1, 3, 9, 1, 3 As you can see, the remainders when 3^n is divided by 26 repeat for every 3rd power. Since the remainders of 3^n divided by 26 repeat every 3rd power, and since 2002/3 = 667 remainder 1, the remainder when 3^2002 is divided by 26 is the same as the remainder when 3^1 is divided by 26. So the answer to your first question is: the remainder when 3^2002 is divided by 26 is 3 And the remainders when 5^n is divided by 26 for n=1 to n=10 are the following: 5, 25, 21, 1, 5, 25, 21, 1, 5, 25 Here, you can see that the remainders when 5^n is divided by 26 repeat for every 4th power. Since the remainders of 5^n divided by 26 repeat every 4th power, and since 2002/4 = 500 remainder 2, the remainder when 5^2002 is divided by 26 is the same as the remainder when 5^2 is divided by 26. So the answer to your second question is: the remainder when 5^2002 is divided by 26 is 25 Once we have the answers to your first two questions, showing the third should be easy. If the sum 3^2002 + 5^2002 is to be divisible by 26, then the remainder when that sum is divisible by 26 should be 0; this means the sum of the remainders when 3^2002 and 5^2002 are divisible by 26 should be 0, or a multiple of 26. However, we have found that the remainders when 3^2002 and 5^2002 are divided by 26 are 3 and 25; and 3+25=28 is not a multiple of 26. The problem here is that I think you have misquoted the problem. I have seen this same problem here at Dr. Math several times previously in the last few months and every other time I have seen this problem, the last part was to show that 3^2002 + 29*5^2002 is divisible by 26. Since the remainders when 3^2002 and 5^2002 are divided by 26 are 3 and 25, respectively, then the remainder when 3^2002 + 29*5^2002 is divided by 26 is 3 + 29(25) = 3 + 725 = 728 = 28(26) This remainder is a multiple of 26, and so the number 3^2002 + 29*5^2002 is divisible by 26. I hope all this helps. Write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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