Difference between revisions of "2021 JMPSC Sprint Problems/Problem 15"
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<math>0-25=-25</math>. <math>-25\equiv\boxed{75}\pmod{100}</math> | <math>0-25=-25</math>. <math>-25\equiv\boxed{75}\pmod{100}</math> | ||
| + | |||
| + | == 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...\boxed{75}</math>. | ||
| + | |||
| + | ~Mathdreams | ||
| + | |||
| + | == Solution 3 == | ||
| + | <cmath>100^{10} \equiv 0 \mod 100</cmath><cmath>5^{10} \equiv 25 \mod 100</cmath>Therefore, the answer is <math>75</math> | ||
| + | |||
| + | - kante314 - | ||
| + | |||
| + | ==See also== | ||
| + | #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]] | ||
| + | #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]] | ||
| + | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
| + | {{JMPSC Notice}} | ||
Latest revision as of 19:54, 7 September 2021
Problem
Find the last two digits of 
Solution
Note that 
and 
.
. 
Solution 2
By multiplying out several powers of 
, we can observe that the last 
digits are always 
(with the exception of 
where 
). Also, 
ends with several zeros, so the answer is 
.
~Mathdreams
Solution 3
![\[100^{10} \equiv 0 \mod 100\]](http://latex.artofproblemsolving.com/b/a/0/ba0ec0b3d29299e35f0ef779a6a76a64cc7f6889.png)
![\[5^{10} \equiv 25 \mod 100\]](http://latex.artofproblemsolving.com/a/5/3/a531d4ef151426b71ae88922bda2fa695d562dc6.png)
Therefore, the answer is 
- kante314 -
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. 
