Difference between revisions of "2022 AMC 10A Problems/Problem 19"
Scarletsyc (talk | contribs) m (→Solution) |
Mathboy282 (talk | contribs) (→Solution) |
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Evaluating, we get: | Evaluating, we get: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
| − | -1 \cdot 9 \cdot 35 \cdot 11 \cdot 13 &\equiv 9 \cdot 11 \cdot | + | (-1) \cdot 9 \cdot 35 \cdot 11 \cdot 13 &\equiv (-1) \cdot 9 \cdot 1 \cdot 11 \cdot 13 \pmod{17} \\ |
| − | &\equiv 9 \cdot 11 \cdot | + | &\equiv 9 \cdot 11 \cdot (-13) \pmod{17} \\ |
| − | &\equiv | + | &\equiv 9 \cdot 11 \cdot 4\pmod{17} \\ |
&\equiv 2 \cdot 11 \pmod{17} \\ | &\equiv 2 \cdot 11 \pmod{17} \\ | ||
| − | |||
&\equiv 5\pmod{17} | &\equiv 5\pmod{17} | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| Line 32: | Line 31: | ||
~KingRavi | ~KingRavi | ||
| + | |||
| + | ~mathboy282 | ||
~Scarletsyc | ~Scarletsyc | ||
| − | |||
| − | |||
== Video Solution By ThePuzzlr == | == Video Solution By ThePuzzlr == | ||
Revision as of 21:15, 13 November 2022
Problem
Define 
as the least common multiple of all the integers from 
to 
inclusive. There is a unique integer 
such that
What is the remainder when is divided by 
?
Solution
Notice that 
contains the highest power of every prime below . Thus, 
.
When writing the sum under a common fraction, we multiply the denominators by divided by each denominator. However, since is a multiple of , all terms will be a multiple of until we divide out , and the only term that will do this is 
. Thus, the remainder of all other terms when divided by will be 
, so the problem is essentially asking us what the remainder of 
divided by is. This is equivalent to finding the remainder of 
divided by .
We use modular arithmetic to simplify our answer:
This is congruent to 
Evaluating, we get:
Therefore the remainder is 
and the answer is 
.
~KingRavi
~mathboy282
~Scarletsyc
Video Solution By ThePuzzlr
~ MathIsChess
See Also
| 2022 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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. 
