Difference between revisions of "2021 Fall AMC 10B Problems/Problem 19"
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==Solution 2 (Condensed Solution 1)== | ==Solution 2 (Condensed Solution 1)== | ||
− | Since <math>7777..7</math> is a <math>313</math> digit number and <math>\sqrt {7}</math> is around <math>2.5</math>, we have <math>f(2)</math> is <math>2</math>. <math>f(3)</math> is the same story, so <math>f(3)</math> is <math>1</math>. It is the same as <math>f(4)</math> as well, so <math>f(4)</math> is also <math>1</math>. However, <math>313</math> is <math>3</math> mod <math>5</math>, so we need to take the 5th root of <math>777</math>, which is between <math>3</math> and <math>4</math>, and therefore, <math>f(5)</math> is <math>3</math>. <math>f(6)</math> is the same as <math>f(4)</math>, since it is <math>1</math> more than a multiple of <math>6</math>. Therefore, we have <math>2+1+1+3+1</math> which is <math>\boxed {(A) 8}</math>. | + | Since <math>7777..7</math> is a <math>313</math> digit number and <math>\sqrt {7}</math> is around <math>2.5</math>, we have <math>f(2)</math> is <math>2</math>. <math>f(3)</math> is the same story, so <math>f(3)</math> is <math>1</math>. It is the same as <math>f(4)</math> as well, so <math>f(4)</math> is also <math>1</math>. However, <math>313</math> is <math>3</math> mod <math>5</math>, so we need to take the 5th root of <math>777</math>, which is between <math>3</math> and <math>4</math>, and therefore, <math>f(5)</math> is <math>3</math>. <math>f(6)</math> is the same as <math>f(4)</math>, since it is <math>1</math> more than a multiple of <math>6</math>. Therefore, we have <math>2+1+1+3+1</math> which is <math>\boxed {\textbf{(A)8}}</math>. |
~Arcticturn | ~Arcticturn |
Revision as of 15:28, 26 November 2021
Problem
Let be the positive integer
, a
-digit number where each digit is a
. Let
be the leading digit of the
th root of
. What is
Solution 1
For notation purposes, let be the number
with
digits, and let
be the leading digit of
. As an example,
, because
, and the first digit of that is
.
Notice that for all numbers
; this is because
, and dividing by
does not affect the leading digit of a number. Similarly,
In general, for positive integers
and real numbers
, it is true that
Behind all this complex notation, all that we're really saying is that the first digit of something like
has the same first digit as
and
.
The problem asks for
From our previous observation, we know that
Therefore,
. We can evaluate
, the leading digit of
, to be
. Therefore,
.
Similarly, we have
Therefore,
. We know
, so
.
Next,
and
, so
.
We also have
and
, so
.
Finally,
and
, so
.
We have that .
~ihatemath123
Solution 2 (Condensed Solution 1)
Since is a
digit number and
is around
, we have
is
.
is the same story, so
is
. It is the same as
as well, so
is also
. However,
is
mod
, so we need to take the 5th root of
, which is between
and
, and therefore,
is
.
is the same as
, since it is
more than a multiple of
. Therefore, we have
which is
.
~Arcticturn
Solution 3
First, we compute .
Because ,
.
Because
,
.
Therefore, .
Second, we compute .
Because ,
.
Because
,
.
Therefore, .
Third, we compute .
Because ,
.
Because
,
.
Therefore, .
Fourth, we compute .
Because ,
.
Because
,
.
Therefore, .
Fifth, we compute .
Because ,
.
Because
,
.
Therefore, .
Therefore,
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
See Also
2021 Fall AMC 10B (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.