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I don't understand this math question for my discrete math 2 class.

FOrmulate a conjecture about the decimal digits that appear as the final decimal digit of the fourth power of an integer. Prove your conjecture using a proof by cases.

So that is the question. I'm having trouble figuring out the answer. ANy insight appreciated.

Mandy Marie
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    The question asks you to come up with a conjecture. I'd try taking the 4th powers of the integers 1 through 10 and see if you notice any pattern in the final digit. – Josh B. Sep 03 '14 at 21:27
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    Try calculating $n^4$ mod $10$, for $0 \leq n \leq 9$. –  Sep 03 '14 at 21:28
  • Well, that is the boring part, so here are a few 1,16,81,256,625,1296,2401,4096,6561,10000,14641,20736,28561. Now think what happens to the last digit of $n^4$ if you increase $n$ by some suitable number. – almagest Sep 03 '14 at 21:31
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    It's not the boring part, it's the whole point. If we write $[n]$ to denote the residue of $n$ mod $10$, then $[n^4] = [n]^4$, so it suffices to consider $0 \leq n \leq 9$. You won't get any new final digits by considering larger values of $n$. –  Sep 03 '14 at 21:37
  • Everyone's saying the same thing in different ways. (That's a good thing, by the way.) – David K Sep 03 '14 at 22:18
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    A better question title: "Formulate a conjecture about the final decimal digit of the fourth power of an integer." – David K Sep 03 '14 at 22:21
  • Yet another way of stating the hint: let $n$ be any integer. How many of the digits of $n$ (and which ones) would you have to know in order to know what the last digit of $n^4$ is? – David K Sep 03 '14 at 22:29

2 Answers2

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try this :-

first see what is $n^4$ for first integers $n$

$ 0^4 ={\color{Red} 0}\\ 1^4 ={\color{Red} 1}\\ 2^4=1{\color{Red} 6}\\ 3^4=8{\color{Red} 1}\\ 4^4=25{\color{Red} 6}\\ 5^4=62{\color{Red} 5}\\ 6^4=129{\color{Red} 6}\\ 7^4=240{\color{Red} 1}\\ 8^4=409{\color{Red} 6}\\ 9^4=656{\color{Red} 1}\\ 10^4=1000{\color{Red} 0}\\ 11^4=1464{\color{Red} 1}\\ 12^4=2073{\color{Red} 6}\\ ...$

conjecture : final decimal digit of the fourth power of an integer would be (0,1,6,5)

Proof :- let $n_i$ be any integer number s.t $i$ is number of digits then $n_i=a_{i-1} 10^{i-1}+a_{i-2}10^{i-2}+...+a_110^1+a_010^0$ ,s.t $a_j$ is unit digit

then $n_i \bmod 10=a_{i-1} 10^{i-1}+a_{i-2}10^{i-2}+...+a_110^1+a_010^0 \bmod 10= {\color{Red} {a_0}} \bmod 10$ thus :- $n_i^4\bmod 10 ={\color{Red} {a_0^4}} \bmod 10 $

but $ 0\le a_0 \le 9 $ so we would have 10 cases for $n_i^4\bmod 10$

by cases :-

$ n_i^4\bmod 10 = 0^4 \mod10 ={\color{Red} 0}\mod 10\\n_i^4\bmod 10 =1^4 \mod 10 ={\color{Red} 1}\mod 10\\ n_i^4\bmod 10 =2^4 \bmod 10= {\color{Red} 6}\bmod 10\\ n_i^4\bmod 10 =3^4\bmod 10= {\color{Red} 1}\bmod 10\\ n_i^4\bmod 10 =4^4\bmod 10= {\color{Red} 6}\bmod 10\\ n_i^4\bmod 10 =5^4\bmod 10= {\color{Red} 5}\bmod 10\\ n_i^4\bmod 10 =6^4\bmod 10= {\color{Red} 6}\bmod 10\\ n_i^4\bmod 10 =7^4\bmod 10= {\color{Red} 1}\bmod 10\\ n_i^4\bmod 10 =8^4\bmod 10= {\color{Red} 6}\bmod 10\\ n_i^4\bmod 10 =9^4\bmod 10= {\color{Red} 1}\bmod 10\\$

other method to prove it , $n_i^4=(n_i^2)^2$ and last digit of a square $\in \left \{ 0,1,4,9,5,6 \right \} $ so you would have 6 cases to check

$n_i^2\bmod 10 =0^2 \bmod 10= {\color{Red} 0}\bmod 10\\ n_i^2\bmod 10 =1^2\bmod 10= {\color{Red} 1}\bmod 10\\ n_i^2\bmod 10 =4^2\bmod 10= {\color{Red} 6}\bmod 10\\ n_i^2\bmod 10 =5^2\bmod 10= {\color{Red} 5}\bmod 10\\ n_i^2\bmod 10 =6^2\bmod 10= {\color{Red} 6}\bmod 10\\ n_i^2\bmod 10 =9^2\bmod 10= {\color{Red} 1}\bmod 10\\$

Bswan
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  • So using the proof by cases I would list the values of n? This may sound stupid but what does mod mean? I am getting a masters in CS and I havent taken discrete math 1 in like 5 years! – Mandy Marie Sep 04 '14 at 02:36
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    proof by cases means list of possible values of the remainder when n divided 10 not for n itself . , mod means "modulo" which is the remainder from division operation [http://en.wikipedia.org/wiki/Modulo_operation], now if you have integer n then n mod 10 gives the remainder when we divide by 10 , which give the first digit right ? – Bswan Sep 04 '14 at 04:17
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Here is another way to prove it.

Let's formulate a conjecture.

$ (-3)^4 = 8 \\ (-2)^4 = 16 \\ (-1)^4 = 1\\ 0^4 = 0\\ 1^4 = 1\\ 2^4 = 16\\ 3^4 = 81\\ 4^4 = 256\\ 5^4 = 625\\ 6^4 = 1296\\ 7^4 = 2401\\ $

Conjecture: The final decimal digit of the fourth power of an integer is either 0,1,5 or 6.

Proof: We use proof by cases. Assume n is an integer. Then n=10a+b, where a and b are positive integers and b, is 0,1,2,3,4,5,6,7,8,9. Taking fourth power on both sides gives us: $ \hspace{1cm} n^4 = (10a + b)^4 \\ \hspace{1cm} n^4 = 10000a^4 + 4000a^3b + 600a^2b^2 +40ab^3 + b^4 \\ \hspace{1cm} n^4 = 10(1000a^4 + 400a^3b + 60a^2b^2 +4ab^3) + b^4\\ \text{thus, } n^4 = 10k + b^4 \text{ where } k = 1000a^4 + 400a^3b + 60a^2b^2 +4ab^3.$

Notice that the last digit of $n^4$ is the same as the last digit of $b^4$. At this point, we can check the last digit of all possible $n^4$ by checking only the last digit of all $b^4$, and we can do that by exhaustion since our b only has 10 integers as declared earlier. We have reduced it to just 10 cases, but we can do better.

since $(10-b)^4 = 10(1000 + 400b + 60b^2 + 4b^3) + b^4$, the last digit of $b^4$ is the same as the last digit of $(10-b)^4$. This tells us that the final digit of 1 and 9, 2 and 8, 3 and 7, 4 and 6 to the fourth power are the same. This reduces our cases to just 6:

case1: When the last digit of n is 1 or 9, then the final digit of $n^4$ is the same as the final digit of $1^4 = 1$ or $9^4 = 6561$ which is 1.

case2: When the last digit of n is 2 or 8, then the final digit of $n^4$ is the same as the final digit of $2^4 = 16$ or $8^4 = 4096$ which is 6.

case3: When the last digit of n is 3 or 7, then the final digit of $n^4$ is the same as the final digit of $3^4 = 81$ or $7^4 = 2401$ which is 1.

case4: When the last digit of n is 4 or 6, then the final digit of $n^4$ is the same as the final digit of $4^4 = 256$ or $6^4 = 1296$ which is 6.

case5: When the last digit of n is 5, then the final digit of $n^4$ is the same as the final digit of $5^4 = 625$ which is 5.

case6: When the last digit of n is 0, then the final digit of $n^4$ is the same as the final digit of $0^4 = 0$ which is 0.

∴In all cases, the final digit of the fourth power of an integer is either 0,1,5 or 6 as desired. □

Chhangsreng P
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