Difference between revisions of "2003 AMC 12A Problems/Problem 18"
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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{(B)}</math>. |
==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 23:13, 15 February 2016
Contents
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
.
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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