Difference between revisions of "2021 Fall AMC 10B Problems/Problem 19"
Ihatemath123 (talk | contribs) |
|||
Line 41: | Line 41: | ||
~Arcticturn | ~Arcticturn | ||
+ | |||
+ | == Solution 3 == | ||
+ | First, we compute <math>f \left( 2 \right)</math>. | ||
+ | |||
+ | Because <math>N > 4 \cdot 10^{312}</math>, <math>\sqrt{N} > 2 \cdot 10^{166}</math>. | ||
+ | Because <math>N < 9 \cdot 10^{312}</math>, <math>\sqrt{N} < 3 \cdot 10^{166}</math>. | ||
+ | |||
+ | Therefore, <math>f \left( 2 \right) = 2</math>. | ||
+ | |||
+ | Second, we compute <math>f \left( 3 \right)</math>. | ||
+ | |||
+ | Because <math>N > 1 \cdot 10^{312}</math>, <math>\sqrt[3]{N} > 1 \cdot 10^{104}</math>. | ||
+ | Because <math>N < 8 \cdot 10^{312}</math>, <math>\sqrt[3]{N} < 2 \cdot 10^{104}</math>. | ||
+ | |||
+ | Therefore, <math>f \left( 3 \right) = 1</math>. | ||
+ | |||
+ | Third, we compute <math>f \left( 4 \right)</math>. | ||
+ | |||
+ | Because <math>N > 1 \cdot 10^{312}</math>, <math>\sqrt[4]{N} > 1 \cdot 10^{78}</math>. | ||
+ | Because <math>N < 16 \cdot 10^{312}</math>, <math>\sqrt[4]{N} < 2 \cdot 10^{78}</math>. | ||
+ | |||
+ | Therefore, <math>f \left( 4 \right) = 1</math>. | ||
+ | |||
+ | |||
+ | Fourth, we compute <math>f \left( 5 \right)</math>. | ||
+ | |||
+ | Because <math>N > 3^5 \cdot 10^{310}</math>, <math>\sqrt[5]{N} > 3 \cdot 10^{62}</math>. | ||
+ | Because <math>N < 4^5 \cdot 10^{310}</math>, <math>\sqrt[5]{N} < 4 \cdot 10^{62}</math>. | ||
+ | |||
+ | Therefore, <math>f \left( 5 \right) = 3</math>. | ||
+ | |||
+ | Fifth, we compute <math>f \left( 6 \right)</math>. | ||
+ | |||
+ | Because <math>N > 1 \cdot 10^{312}</math>, <math>\sqrt[6]{N} > 1 \cdot 10^{52}</math>. | ||
+ | Because <math>N < 2^6 \cdot 10^{312}</math>, <math>\sqrt[6]{N} < 2 \cdot 10^{52}</math>. | ||
+ | |||
+ | Therefore, <math>f \left( 6 \right) = 1</math>. | ||
+ | |||
+ | Therefore, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | f \left( 2 \right) + f \left( 3 \right) + f \left( 4 \right) + f \left( 5 \right) + f \left( 6 \right) | ||
+ | & = 2 + 1 + 1 + 3 + 1 \\ | ||
+ | & = 8 . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{\textbf{(A) }8}</math>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-a=20|num-b=18}} | {{AMC10 box|year=2021 Fall|ab=B|num-a=20|num-b=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:57, 25 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.