Difference between revisions of "2016 AMC 10B Problems/Problem 8"
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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. | 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. | 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 \\ | |
− | 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- | + | 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=0 \Longrightarrow \boxed{\textbf{(A)}\ 0}</math>. | ||
~JH. L | ~JH. L | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/jDHIJ8o4VSA | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
==Video Solution== | ==Video Solution== | ||
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== Video Solution == | == Video Solution == | ||
https://youtu.be/zfChnbMGLVQ?t=2683 | https://youtu.be/zfChnbMGLVQ?t=2683 | ||
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==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 12:46, 2 July 2023
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.
From here we only need to multiply
and we can ignore the hundreds digits.
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 |
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