Difference between revisions of "2020 AMC 10B Problems/Problem 22"
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==MAA Official Solution== | ==MAA Official Solution== | ||
| − | + | <cmath>2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\ | |
| − | 2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\ | ||
&= (2^{101} + 1)^2 - 2^{102} + 201\\ | &= (2^{101} + 1)^2 - 2^{102} + 201\\ | ||
| − | &= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201. | + | &= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.</cmath> |
| − | |||
==See Also== | ==See Also== | ||
Revision as of 23:41, 7 February 2020
Problem
What is the remainder when 
is divided by 
?
Solution
Let 
. We are now looking for the remainder of 
.
We could proceed with polynomial division, but the denominator looks awfully similar to the Sophie Germain Identity, which states that ![\[a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)\]](http://latex.artofproblemsolving.com/b/1/0/b100b5bcacb4bcafb2d261cb92d9f8d145d89ebc.png)
Let's use the identity, with 
and 
, so we have
![\[1+4x^4=(1+2x^2+2x)(1+2x^2-2x)\]](http://latex.artofproblemsolving.com/6/d/8/6d81826c91cc16610cc5b8b43a5d25dba7ac0ffe.png)
Rearranging, we can see that this is exactly what we need:
![\[\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1\]](http://latex.artofproblemsolving.com/c/0/e/c0e010372051a195c274880c199e616600c4a8ad.png)
So ![\[\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1}\]](http://latex.artofproblemsolving.com/b/e/0/be06c275b6cbeb278bb73d756532e63391bdc2af.png)
Since the first half divides cleanly as shown earlier, the remainder must be 
~quacker88
Solution 2
Similar to Solution 1, let . It suffices to find remainder of . Dividing polynomials results in a remainder of 
.
MAA Official Solution
\[2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\
&= (2^{101} + 1)^2 - 2^{102} + 201\\
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.\] (Error compiling LaTeX. Unknown error_msg)
See Also
| 2020 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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. 
