Difference between revisions of "2022 AMC 10B Problems/Problem 17"
MRENTHUSIASM (talk | contribs) m |
MRENTHUSIASM (talk | contribs) (→See Also) |
||
Line 87: | Line 87: | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2022|ab=B|num-b=16|num-a=18}} | {{AMC10 box|year=2022|ab=B|num-b=16|num-a=18}} | ||
+ | {{AMC12 box|year=2022|ab=B|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:15, 5 January 2023
- The following problem is from both the 2022 AMC 10B #17 and 2022 AMC 12B #15, so both problems redirect to this page.
Contents
Problem
One of the following numbers is not divisible by any prime number less than Which is it?
Solution 1 (Modular Arithmetic)
For modulo
Thus,
is divisible by
For modulo
Thus,
is divisible by
For modulo
Thus,
is divisible by
For modulo
Thus,
is divisible by
Therefore, the answer is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~MrThinker (LaTeX Error)
Solution 2 (Factoring)
We have
We conclude that
is divisible by
,
is divisible by
,
is divisible by
, and
is divisible by
.
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 |
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.