Difference between revisions of "2003 AMC 12A Problems/Problem 18"
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== Solution 2 == | == Solution 2 == | ||
− | Notice that <math>q+r=0\pmod{11}\Rightarrow100q+r=0\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{99990-11010}{11}+1=8181 | + | Notice that <math>q+r=0\pmod{11}\Rightarrow100q+r=0\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{99990-11010}{11}+1=8181 (B)</math> possible values. |
==See Also== | ==See Also== | ||
{{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}} |
Revision as of 14:06, 23 December 2013
Contents
[hide]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.
See Also
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 |
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