Difference between revisions of "2019 AMC 10B Problems/Problem 14"
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==Solution 1== | ==Solution 1== | ||
− | We can figure out <math>H = 0</math> by noticing that <math>19!</math> will end with <math>3</math> zeroes, as there are three <math>5</math> | + | We can figure out <math>H = 0</math> by noticing that <math>19!</math> will end with <math>3</math> zeroes, as there are three factors of <math>5</math> in its prime factorization. Next, we use the fact that <math>19!</math> is a multiple of both <math>11</math> and <math>9</math>. Their divisibility rules (see Solution 2) tell us that <math>T + M \equiv 3 \;(\bmod\; 9)</math> and that <math>T - M \equiv 7 \;(\bmod\; 11)</math>. By guess and checking, we see that <math>T = 4, M = 8</math> is a valid solution. Therefore the answer is <math>4 + 8 + 0 = \boxed{\textbf{(C) }12}</math>. |
==Solution 2 (similar to Solution 1)== | ==Solution 2 (similar to Solution 1)== |
Revision as of 13:15, 5 September 2021
Problem
The base-ten representation for is
, where
,
, and
denote digits that are not given. What is
?
Solution 1
We can figure out by noticing that
will end with
zeroes, as there are three factors of
in its prime factorization. Next, we use the fact that
is a multiple of both
and
. Their divisibility rules (see Solution 2) tell us that
and that
. By guess and checking, we see that
is a valid solution. Therefore the answer is
.
Solution 2 (similar to Solution 1)
We know that , because
ends in three zeroes (see Solution 1). Furthermore, we know that
and
are both factors of
. We can simply use the divisibility rules for
and
for this problem to find
and
. For
to be divisible by
, the sum of digits must simply be divisible by
. Summing the digits, we get that
must be divisible by
. This leaves either
or
as our answer choice. Now we test for divisibility by
. For a number to be divisible by
, the alternating sum must be divisible by
(for example, with the number
,
, so
is divisible by
). Applying the alternating sum test to this problem, we see that
must be divisible by 11. By inspection, we can see that this holds if
and
. The sum is
.
Video Solution
~IceMatrix
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.