Difference between revisions of "2022 AMC 10A Problems/Problem 19"
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This is congruent to <math>-1 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \pmod{17}</math> | This is congruent to <math>-1 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \pmod{17}</math> | ||
− | Evaluating, we get: <cmath>-45 \cdot 7 \cdot 11 \cdot 13 \pmod{17} \ | + | Evaluating, we get: <cmath>-45 \cdot 7 \cdot 11 \cdot 13 \pmod{17} \equiv 6 \cdot 7 \cdot 11 \cdot 13 \pmod{17}</cmath> |
− | <cmath> \ | + | <cmath> \equiv 42 \cdot 11 \cdot 13 \pmod{17} \equiv 8 \cdot 11 \cdot 13 \pmod{17}</cmath> |
− | <cmath> \ | + | <cmath> \equiv 88 \cdot 13 \pmod{17} \equiv 3 \cdot 13 \pmod{17}</cmath> |
− | <cmath>\ | + | <cmath>\equiv 39 \pmod{17} \equiv 5\pmod{17}</cmath> |
Therefore the remainder is <math>5</math> and the answer is <math>\boxed{C}</math> | Therefore the remainder is <math>5</math> and the answer is <math>\boxed{C}</math> |
Revision as of 14:32, 12 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
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.