Difference between revisions of "2016 AMC 8 Problems/Problem 5"
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To prove generalization vigorously, we can let <math>a</math> be the remainder when <math>z</math> is divided by <math>11.</math> Setting up a modular equation, we have <cmath>90c + 73 \equiv a \pmod {11}.</cmath> Simplifying, <cmath>90c+7 \equiv a \pmod {11}</cmath> If <math>c = 1,</math> then we don't have a 2 digit number! Thus, <math>c=0</math> and <math>a=\boxed { \textbf{(E) }7}</math> | To prove generalization vigorously, we can let <math>a</math> be the remainder when <math>z</math> is divided by <math>11.</math> Setting up a modular equation, we have <cmath>90c + 73 \equiv a \pmod {11}.</cmath> Simplifying, <cmath>90c+7 \equiv a \pmod {11}</cmath> If <math>c = 1,</math> then we don't have a 2 digit number! Thus, <math>c=0</math> and <math>a=\boxed { \textbf{(E) }7}</math> | ||
+ | |||
+ | ~CHECKMATE2021 | ||
==Solution 3== | ==Solution 3== |
Revision as of 21:10, 21 January 2024
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 , for a number divisible by
ends in zero. Since there is a remainder of
when
is divided by
, the multiple of
must end in a
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,
.
~CHECKMATE2021
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
~CHECKMATE2021
Solution 3
We know that the number has to be one more than a multiple of , because of the remainder of one, and the number has to be
more than a multiple of
, which means that it has to end in a
. Now, if we just list the first few multiples of
adding one to the number we get:
. As we can see from these numbers, the only one that has a three in the denominator is
, thus we divide
by
, getting
, hence,
.
-fn106068
We could also remember that, for a two-digit number to be divisible by , the sum of its digits has to be equal to
. Since the number is one more than a multiple of
, the multiple we are looking for has a ones digit of
, and therefore a tens digit of
, and then we could proceed as above. -vaisri
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=574
Video Solution
~savannahsolver
See Also
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.