Last Digits and High Exponents
Date: 09/26/98 at 16:44:57 From: Carl Subject: Exponent problem Hello, In my math book, the task is: Presume we have calculated the following expressions. What is the last digit of each? a) 3^1000 b) 7^134 I am not supposed to use a calculator. This is my way of solving task a: exponent ending digit 3^1 3 3^2 9 |__ 3^3 7 | 3^4 1 / 3^5 3 We can see the period (or series or whatever you call it) clearly: ____ 3971 which is 4 digits long When I divide 1000 by 4, I get a remainder of 0, and that's why 3^1000 ends with 1 (the last digit in the period. If the remainder of the quotient had been 1, then 3 would have been the last digit and so on). Task b is possible to solve to in this way too, but not without a calculator (or very complex sum outwritings). With a, this is the correct answer according to the key. But task b will be more difficult without a calculator (we will for example have to calculate 7^7 by head). My way of solving the problem is not the easiest. I said the answer was right according to the key. Not completely, however. Sometimes the key helps us a little with the solving. This time, you could read in the key: a) 3^1000 = (3^4)^250 ends with 1 b) 7^134 = (7^4)^33 * 7^2 ends with 9 They have solved the problem in another, probably better and by head possible, way. How? 3^4 is 81, ending with a one. But how do we know the ending digit will be the same when we raise this number to the 250th power? And, concerning b, 7^2 = 49, ending with a nine. Again, how do we know the ending digit will be the same? So, my question is: How should the problem be solved in the best way, without using a calculator? Thank you very much for your help, Carl, Sweden
Date: 09/27/98 at 04:03:51 From: Doctor Floor Subject: Re: Exponent problem Dear Carl, Thank you for sending your question with so much explanation. I think the most important step you need to make is the following: when you multiply two numbers, for the last digit you only have to multiply the last digits. So for example ...2 * ...7 = ...4. The earlier digits of ...2 and ...7 just bring multiples of 10, so they don't change the last digit. So this explains the keys: a) 3^1000 = (3^4)^250 ends with 1 3^4 = 81 = ...1, so: (3^4)^250 = (...1)^250 = ...1 b) 7^134 = (7^4)^33 * 7^2 ends with 9 7^2 = 49 = ...9 7^4 = 7^2*7^2 = ...9*...9 = ...1 7^132 = (7^4)^33 = (...1)^33 = ...1 7^134 = 7^132 * 7^2 = ...1 * ...9 = ...9 I hope this helps! Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
Date: 03/04/99 at 04:10:19 From: Anandh Pachaiappan Subject: Exponent Problem Hi Dr. Math, I was going through the archives. I came across the Last Digits and High Exponent problem. I want to share one interesting thing I noticed. This leads to a more mathematically complex, but easier to find solution. For any number n = 10*a + b where 0<=b<=9 and any value a: n^5 = 10*c + b for some value of c In other words, every number n, when raised to the power of 5, will give the last digit of n as its own last digit. To show why, we need only to check 1 through 9: 0^5 = 0 1^5 = 1 2^5 = 32 3^5 = 243 4^5 = 1024 5^5 = 3125 6^5 = 7776 7^5 = 16807 8^5 = 32768 9^5 = 59049 Then, using = for congruence, if n = a_k 10^k + ... + a_0: n^5 (mod 10) = (a_k 10^k + ... + a_0)^5 (mod 10) = (a_0)^5 (mod 10) The above says that to find the last digit of n^5, for any n we just need to look at the last digit of (a_0)^5, where a_0 is the last digit of n. In the list above, we showed that all of the possible last digits, when raised to the fifth power, end in the same digit. Therefore, we've shown that for any number n, n^5 ends in with the same last digit as n. In general, for any positive integer k, n^(4k+1) will have the same last digit as the last digit of n. We can show this by mathematical induction. Here is the base case, k = 2: n^9 (mod 10) = n^(5+4) (mod 10) = ( n^5 * n^4 ) (mod 10) = ( n*n^4) (mod 10) = n^5 (mod 10) = n (mod 10) In the actual problem, the question is to find the last digit of 7^134. We can do this by the following method: 7^134 (mod 10) = 7^(132+1+1) (mod 10) = 7^(4*33 + 1)*7 (mod 10) = 7*7 (mod 10) = 9 (mod 10) Thus, the last digit is 9.
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