Number Theory
Power Cycle
If we observe pattern of last digit of powers of 2, we find that,
\({2^1}\) ends with 2
\({2^2}\) ends with 4
\({2^3}\) ends with 8
\({2^4}\) ends with 6
\({2^5}\) ends with 2
\({2^6}\) ends with 4
\({2^7}\) ends with 8
\({2^8}\) ends with 6
Clearly there is a cycle of 4. We can conclude that every power of 2 ends with either 2, 4, 6 or 8 depending on the power.
If the power is multiple of 4, then the last digit is 6,
If the power is one more than multiple of 4, then the last digit is 2,
If the power is two more than multiple of 4, then the last digit is 4,
If the power is three more than multiple of 4, then the last digit is 8
Power cycle of 3
\({3^1}\) ends with 3
\({3^2}\) ends with 9
\({3^3}\) ends with 7
\({3^4}\) ends with 1
\({3^5}\) ends with 3
\({3^6}\) ends with 9
\({3^7}\) ends with 7
\({3^8}\) ends with 1
Clearly there is a cycle of 4. We can conclude that every power of 3 ends with either 3, 9, 7 or 1 depending on the power.
If the power is multiple of 4, then the last digit is 1,
If the power is one more than multiple of 4, then the last digit is 3,
If the power is two more than multiple of 4, then the last digit is 9,
If the power is three more than multiple of 4, then the last digit is 7
Power cycle of 7
\({7^1}\) ends with 07
\({7^2}\) ends with 49
\({7^3}\) ends with 43
\({7^4}\) ends with 01
If the power is one more than multiple of 4, then the last two digits are 07,
If the power is two more than multiple of 4, then the last two digits are 49,
If the power is three more than multiple of 4, then the last digit are 43
If the power is multiple of 4, then the last digit is 01
Example 32: Find the last two digits of the number \({7^2008}\)
Solution: Since the power, 2008 is multiple of 4, hence the last two digits will be 01
Example 33: Find the last digit of the number. \({123^{321}} \times {217^{312}} \times {122^{221}}\)
Solution: Last digit of the product depends on only the last digits of the input numbers; hence the question is equivalent to:
\({3^{321}} \times {7^{312}} \times {2^{221}}\)
Last digit of \({3^321} =3\),
Last digit of \({7^312} =1\),
Last digit of \({2^221} =2\),
Hence the last digit of the product is 6
Following numbers show a unique pattern
\({5^n}\) always ends with 25
\({5^{{\rm{odd\,\ number}}}}\) ends with 125
\({5^{{\rm{even\,\ number}}}}\) ends with 625
\({76^n}\) always ends with 76
\({125^{{\rm{odd\,\ number}}}}\) ends with 125
\({125^{{\rm{even\,\ number}}}}\) ends with 625
\({625^{{\rm{any\,\ number}}}}\) ends with 625
\({376^{{\rm{any\,\ number}}}}\) ends with 376
\({9376^{{\rm{any\,\ number}}}}\) ends with 9376
To find number of digits in a product:
Example 34: Find the number of digits in the number \(212354 \times 6524\).
Solution: The number 212354 can be written as \(2.12354 \times {10^5}\) and the number 6524 can be written as \(6.524 \times {10^3}\). Now the product is \(\left[ {\left( {2.12354 \times {{10}^5}} \right)\left( {{\rm{ }}6.524 \times {{10}^3}} \right)} \right]\)
That is approximately equal to \(13 \times {10^8} = {\rm{ }}1.3 \times {10^9}\)
Hence number of digits = 10.
Example 35: Find the number of digits before decimal in the product \(\left( {\frac{{1235 \times 98134}}{{1125}}} \right)\)
Solution: The number can be written as
\(\frac{{{\rm{(1}}{\rm{.235}} \times {\rm{1}}{{\rm{0}}^{\rm{3}}}{\rm{)(9}}{\rm{.8134}} \times {\rm{1}}{{\rm{0}}^{\rm{4}}}{\rm{)}}}}{{{\rm{1}}{\rm{.115}} \times {\rm{1}}{{\rm{0}}^{\rm{3}}}}}\)
\( \approx 10 \times {10^4} \approx {10^5}\). Hence the number has 6 digits.