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Finding last digit
- Thread starter lPing7
- Start date
Again, I have no clue how to start the problem
Find the last digit in 2003^2002 + 2001^2002
if you have a number which ends in three, what does the square of that number end in, i.e. a number ending in 3 can be written as
n = 10 a1 + 3
so
n2 = 100 a12 + 60 a1 + 9 = 10 a2 + 9
So a number ending in 3 squared ends in 9. How about a number ending in 9 times a number ending in 3? Pretty soon you will start to repeat and you will have it ends in a certain value mod some number.
Now do the same thing for a number ending in 1.
For example, take a number ending in 2, we would have ending in 2, 4, 8, 6, 2, 4, 8, 6, 2, .... Notice the cycle is 4 long so if you raised the number ending in 2 to the power n>0, it would end in
2 if n=1 mod(4)
4 if n=2 mod(4)
8 if n=3 mod(4)
6 if n=0 mod(4)
Now a number ending in 4:
4 if n=1 mod(2)
6 if n=0 mod(2)
Suppose we have 27 + 46. 7=3 mod(4), so 27 ends in 8. 6=0 mod(2) so 46 ends in 6. If we add those numbers together we get a number ending in 4.
Hello, lPing7!
\(\displaystyle \begin{array}{cccccc}\text{We see that:} &2003^1 \text{ ends in }3 \\ & 2003^2 \text{ ends in }9 \\ & 2003^3 \text{ ends in }7 \\ &2003^4\text{ ends in }1 \end{array}\)
Hence: \(\displaystyle \:2003^{2002} \:=\: (2003^{2000})(2003^2) \:=\: (2003^4)^{500}(2003^2) \)
\(\displaystyle \quad \text{which ends in: }\: (1^{500})(9) \:=\:9\)
We know that \(\displaystyle 2001^{2002}\) ends in 1.
Therefore: \(\displaystyle \;2003^{2002} + 2001^{2002}\,\) ends in \(\displaystyle \, 9 + 1 \;\Rightarrow\; 0\)
Find the last digit of \(\displaystyle \: 2003^{2002} + 2001^{2002}\)
\(\displaystyle \begin{array}{cccccc}\text{We see that:} &2003^1 \text{ ends in }3 \\ & 2003^2 \text{ ends in }9 \\ & 2003^3 \text{ ends in }7 \\ &2003^4\text{ ends in }1 \end{array}\)
Hence: \(\displaystyle \:2003^{2002} \:=\: (2003^{2000})(2003^2) \:=\: (2003^4)^{500}(2003^2) \)
\(\displaystyle \quad \text{which ends in: }\: (1^{500})(9) \:=\:9\)
We know that \(\displaystyle 2001^{2002}\) ends in 1.
Therefore: \(\displaystyle \;2003^{2002} + 2001^{2002}\,\) ends in \(\displaystyle \, 9 + 1 \;\Rightarrow\; 0\)