Difference between revisions of "2003 AMC 12A Problems/Problem 18"
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Therefore, <math>q</math> can be any integer from <math>100</math> to <math>999</math> inclusive, and <math>r</math> can be any integer from <math>0</math> to <math>99</math> inclusive. | Therefore, <math>q</math> can be any integer from <math>100</math> to <math>999</math> inclusive, and <math>r</math> can be any integer from <math>0</math> to <math>99</math> inclusive. | ||
− | For each of the <math>9\cdot10\cdot10=900</math> possible values of <math>q</math>, there are at least <math>\lfloor \frac{100}{11} \rfloor = 9</math> possible values of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>. | + | For each of the <math>9\cdot10\cdot10=900</math> possible values of <math>q</math>, there are at least <math>\left\lfloor \frac{100}{11} \right\rfloor = 9</math> possible values of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>. |
− | Since there is <math>1</math> "extra" possible value of <math>r</math> that is congruent to <math>0\pmod{11}</math>, each of the <math>\lfloor \frac{900}{11} \rfloor = 81</math> values of <math>q</math> that are congruent to <math>0\pmod{11}</math> have <math>1</math> more possible value of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>. | + | Since there is <math>1</math> "extra" possible value of <math>r</math> that is congruent to <math>0\pmod{11}</math>, each of the <math>\left\lfloor \frac{900}{11} \right\rfloor = 81</math> values of <math>q</math> that are congruent to <math>0\pmod{11}</math> have <math>1</math> more possible value of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>. |
Therefore, the number of possible values of <math>n</math> such that <math>q+r \equiv 0\pmod{11}</math> is <math>900\cdot9+81\cdot1=8181 \Rightarrow\boxed{(B)} </math>. | Therefore, the number of possible values of <math>n</math> such that <math>q+r \equiv 0\pmod{11}</math> is <math>900\cdot9+81\cdot1=8181 \Rightarrow\boxed{(B)} </math>. | ||
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== Solution 2 == | == Solution 2 == | ||
− | Notice that <math>q+r | + | Notice that <math>q+r\equiv0\pmod{11}\Rightarrow100q+r\equiv0\pmod{11}</math>. This means that any number whose quotient and remainder sum is divisible by 11 must also be divisible by 11. Therefore, there are <math>\frac{99990-10010}{11}+1=8181</math> possible values. The answer is <math>\boxed{\textbf{(B) }8181}</math>. |
+ | |||
+ | == Solution 3 == | ||
+ | Let <math>abcde</math> be the five digits of <math>n</math>. Then <math>q = abc</math> and <math>r = de</math>. By the divisibility rules of <math>11</math>, <math>q = a - b + c \pmod{11}</math> and <math>r = -d + e \pmod{11}</math>, so <math>q + r = a - b + c - d + e = abcde = n \pmod{11}</math>. Thus, <math>n</math> must be divisble by <math>11</math>. There are <math>\frac{99990 - 10010}{11} + 1 = 8181</math> five-digit multiples of <math>11</math>, so the answer is <math>\boxed{\textbf{(B) }8181}</math>. | ||
+ | ==Video Solution 1== | ||
+ | https://youtu.be/OpGHj-B0_hg?t=672 | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
+ | {{AMC10 box|year=2003|ab=A|num-b=24|after=Last Question}} | ||
{{AMC12 box|year=2003|ab=A|num-b=17|num-a=19}} | {{AMC12 box|year=2003|ab=A|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:20, 18 July 2022
Problem
Let be a
-digit number, and let
and
be the quotient and the remainder, respectively, when
is divided by
. For how many values of
is
divisible by
?
Solution 1
When a -digit number is divided by
, the first
digits become the quotient,
, and the last
digits become the remainder,
.
Therefore, can be any integer from
to
inclusive, and
can be any integer from
to
inclusive.
For each of the possible values of
, there are at least
possible values of
such that
.
Since there is "extra" possible value of
that is congruent to
, each of the
values of
that are congruent to
have
more possible value of
such that
.
Therefore, the number of possible values of such that
is
.
Solution 2
Notice that . This means that any number whose quotient and remainder sum is divisible by 11 must also be divisible by 11. Therefore, there are
possible values. The answer is
.
Solution 3
Let be the five digits of
. Then
and
. By the divisibility rules of
,
and
, so
. Thus,
must be divisble by
. There are
five-digit multiples of
, so the answer is
.
Video Solution 1
https://youtu.be/OpGHj-B0_hg?t=672
~IceMatrix
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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 |
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.