Difference between revisions of "2016 AMC 10B Problems/Problem 8"
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\textbf{(E)}\ 8</math> | \textbf{(E)}\ 8</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Notice that, for <math>n\ge 2</math>, <math>2015^n\equiv 15^n</math> is congruent to <math>25\pmod{100}</math> when <math>n</math> is even and <math>75\pmod{100}</math> when <math>n</math> is odd. (Check for yourself). Since <math>2016</math> is even, <math>2015^{2016} \equiv 25\pmod{100}</math> and <math>2015^{2016}-2017 \equiv 25 - 17 \equiv \underline{0}8\pmod{100}</math>. | Notice that, for <math>n\ge 2</math>, <math>2015^n\equiv 15^n</math> is congruent to <math>25\pmod{100}</math> when <math>n</math> is even and <math>75\pmod{100}</math> when <math>n</math> is odd. (Check for yourself). Since <math>2016</math> is even, <math>2015^{2016} \equiv 25\pmod{100}</math> and <math>2015^{2016}-2017 \equiv 25 - 17 \equiv \underline{0}8\pmod{100}</math>. | ||
| − | So the answer is <math>\textbf{(A)}\ 0</math>. | + | So the answer is <math>\textbf{(A)}\ \boxed{0}</math>. |
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | In a very similar fashion, we find that <math>2015^{2016} \equiv 15^{2016} \pmod{100}</math>, which equals <math>225^{1008}</math>. Next, since every power (greater than <math>0</math>) of every number ending in <math>25</math> will end in <math>25</math> (which can easily be verified), we get <math>225^{1008} \equiv 25 \pmod{100}</math>. (In this way, we don't have to worry about the exponent very much.) Finally, <math>2017 \equiv 17 \pmod{100}</math>, and thus <math>2015^{2016}-2017 \equiv 25-17 \equiv 08 \pmod{100}</math>, as above. | ||
| + | |||
| + | ==Solution 3 (exponent pattern)== | ||
| + | Since we only need the tens digits, we only need to care about the multiplication of tens and ones. (If you want to use mathematical terms then we only need to look at the exponents in <math>mod 100</math>.) We will use the "<math>\equiv</math>" sign to denote congruence in modulus, basically taking the last two digits and ignoring everything else. | ||
| + | |||
| + | <cmath>\begin{split} | ||
| + | 15^{1}\equiv15 \\ | ||
| + | 15^{2}\equiv25 | ||
| + | \end{split}</cmath> | ||
| + | From here we only need to multiply <math>15\cdot25</math> and we can ignore the hundreds digits. | ||
| + | <cmath>\begin{split} | ||
| + | 15^{3}\equiv75 \\ | ||
| + | 15^{4}\equiv25 \\ | ||
| + | 15^{5}\equiv75 | ||
| + | \end{split}</cmath> | ||
| + | Notice that for every <math>x\neq1</math>, <math>15^{x}\equiv25</math> if <math>x</math> is even, and <math>15^{x}\equiv75</math> if <math>x</math> is odd. Since <math>2015^{2016}</math> has an even exponent, we conclude that the last two digits will be <math>25</math>, and subtracting <math>25-17=\underline{0}8 \Longrightarrow \boxed{\textbf{(A)}\ 0}</math>. | ||
| + | ~JH. L | ||
| + | |||
| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/jDHIJ8o4VSA | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/GVPF6CcCugU | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | == Video Solution == | ||
| + | https://youtu.be/zfChnbMGLVQ?t=2683 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=7|num-a=9}} | {{AMC10 box|year=2016|ab=B|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 21:41, 20 November 2024
Contents
Problem
What is the tens digit of 
Solution 1
Notice that, for 
, 
is congruent to 
when 
is even and 
when is odd. (Check for yourself). Since 
is even, 
and 
.
So the answer is 
.
Solution 2
In a very similar fashion, we find that 
, which equals 
. Next, since every power (greater than 
) of every number ending in 
will end in (which can easily be verified), we get 
. (In this way, we don't have to worry about the exponent very much.) Finally, 
, and thus 
, as above.
Solution 3 (exponent pattern)
Since we only need the tens digits, we only need to care about the multiplication of tens and ones. (If you want to use mathematical terms then we only need to look at the exponents in 
.) We will use the "
" sign to denote congruence in modulus, basically taking the last two digits and ignoring everything else.
![\[\begin{split} 15^{1}\equiv15 \\ 15^{2}\equiv25 \end{split}\]](http://latex.artofproblemsolving.com/7/6/f/76fd818661e17c795c8fceb8a2202b124962322b.png)
From here we only need to multiply 
and we can ignore the hundreds digits.
![\[\begin{split} 15^{3}\equiv75 \\ 15^{4}\equiv25 \\ 15^{5}\equiv75 \end{split}\]](http://latex.artofproblemsolving.com/d/7/e/d7e62bb97c3e25642c8506c77b50110c2b8e6da8.png)
Notice that for every 
, 
if 
is even, and 
if is odd. Since 
has an even exponent, we conclude that the last two digits will be , and subtracting 
.
~JH. L
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
Video Solution
https://youtu.be/zfChnbMGLVQ?t=2683
See Also
| 2016 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
