2016 AMC 8 Problems/Problem 5
Contents
Problem
The number is a two-digit number.
• When is divided by
, the remainder is
.
• When is divided by
, the remainder is
.
What is the remainder when is divided by
?
Solution 1
From the second bullet point, we know that the second digit must be . Because there is a remainder of
when it is divided by
, the multiple of
must end in a
in order for it to have the desired remainder
We now look for this one:
The number satisfies both conditions. We subtract the biggest multiple of
less than
to get the remainder. Thus,
.
Solution 2 ~ More efficient for proofs
This two digit number must take the form of where
and
are integers
to
However, if x is an integer, we must have
So, the number's new form is
This needs to have a remainder of
when divided by
Because of the
divisibility rule, we have
We subtract the three, getting
which simplifies to
However,
so
and
Let the quotient of in our modular equation be
and let our desired number be
so
and
We substitute these values into
and get
so
As a result,
- Alternatively, we could have also used a system of modular equations to immediately receive
To prove generalization vigorously, we can let be the remainder when
is divided by
Setting up a modular equation, we have
Simplifying,
If
then we don't have a 2 digit number! Thus,
and
Solution 3
Video Solution
https://youtu.be/7an5wU9Q5hk?t=574
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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