Difference between revisions of "2021 JMPSC Sprint Problems/Problem 15"

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== Solution 2 ==
 
== Solution 2 ==
  
By multiplying out several powers of <math>5</math>, we can observe that the last <math>2</math> digits are always <math>25</math> (with the exception of <math>5^n</math> where <math>n \le 1</math>). Also, <math>10^10</math> ends with several zeros, so the answer is <math>100...00 - 25 = 99...99 - 24 = 999...75</math>.
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By multiplying out several powers of <math>5</math>, we can observe that the last <math>2</math> digits are always <math>25</math> (with the exception of <math>5^n</math> where <math>n \le 1</math>). Also, <math>10^{10}</math> ends with several zeros, so the answer is <math>100...00 - 25 = 99...99 - 24 = 999...\boxed{75}</math>.
  
 
~Mathdreams
 
~Mathdreams
  
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== Solution 3 ==
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<cmath>100^{10} \equiv 0 \mod 100</cmath><cmath>5^{10} \equiv 25 \mod 100</cmath>Therefore, the answer is <math>75</math>
  
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- kante314 -
  
 
==See also==
 
==See also==

Latest revision as of 20:54, 7 September 2021

Problem

Find the last two digits of $10^{10}-5^{10}.$

Solution

Note that $10^{10}\equiv0\pmod{100}$ and $5^{10}\equiv25\pmod{100}$.

$0-25=-25$. $-25\equiv\boxed{75}\pmod{100}$

Solution 2

By multiplying out several powers of $5$, we can observe that the last $2$ digits are always $25$ (with the exception of $5^n$ where $n \le 1$). Also, $10^{10}$ ends with several zeros, so the answer is $100...00 - 25 = 99...99 - 24 = 999...\boxed{75}$.

~Mathdreams

Solution 3

\[100^{10} \equiv 0 \mod 100\]\[5^{10} \equiv 25 \mod 100\]Therefore, the answer is $75$

- kante314 -

See also

  1. Other 2021 JMPSC Sprint Problems
  2. 2021 JMPSC Sprint Answer Key
  3. All JMPSC Problems and Solutions

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