Difference between revisions of "2022 AMC 10B Problems/Problem 17"
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==Solution== | ==Solution== | ||
− | For A | + | For <math>\textbf{(A)}</math> modulo <math>3,</math> |
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
Line 18: | Line 18: | ||
Thus, <math>2^{606} - 1</math> is divisible by 3. | Thus, <math>2^{606} - 1</math> is divisible by 3. | ||
− | For B | + | For <math>\textbf{(B)}</math> modulo <math>5,</math> |
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
Line 30: | Line 30: | ||
Thus, <math>2^{606} + 1</math> is divisible by 5. | Thus, <math>2^{606} + 1</math> is divisible by 5. | ||
− | For D | + | For <math>\textbf{(D)}</math> modulo <math>3,</math> |
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
Line 41: | Line 41: | ||
Thus, <math>2^{607} + 1</math> is divisible by 3. | Thus, <math>2^{607} + 1</math> is divisible by 3. | ||
− | For E | + | For <math>\textbf{(E)}</math> modulo <math>5,</math> |
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} |
Revision as of 22:37, 24 November 2022
Contents
Problem
One of the following numbers is not divisible by any prime number less than Which is it?
Solution
For modulo
Thus, is divisible by 3.
For modulo
Thus, is divisible by 5.
For modulo
Thus, is divisible by 3.
For modulo
Thus, is divisible by 5.
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~MrThinker (LaTeX Error)
Solution 2 (Factoring)
. A is divisible by 3.
. B is divisible by 5.
. D is divisible by 3.
. E is divisible by 5.
Since all of the other choices have been eliminated, we are left with .
~not_slay
Solution 3 (Elimination)
Mersenne Primes are primes of the form , where
is prime. Using the process of elimination, we can eliminate every option except for
and
. Clearly,
isn't prime, so the answer must be
.
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by OmegaLearn Using Digit Cycles
~ pi_is_3.14
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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.