Difference between revisions of "2016 AMC 8 Problems/Problem 5"
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− | ==Solution 2== | + | ==Solution 2 ~ More efficient for proofs== |
− | + | This two digit number must take the form of <math>10x+y,</math> where <math>x</math> and <math>y</math> are integers <math>0</math> to <math>9.</math> However, if x is an integer, we must have <math>y=3.</math> So, the number's new form is <math>10x+3.</math> This needs to have a remainder of <math>1</math> when divided by <math>9.</math> Because of the <math>9</math> divisibility rule, we have <cmath>10x+3 \equiv 1 \pmod 9.</cmath> | |
+ | We subtract the three, getting <cmath>10x \equiv -2 \pmod 9.</cmath> | ||
+ | which simplifies to <cmath>10x \equiv 7 \pmod 9.</cmath> | ||
+ | However, <math>9x \equiv 0 \pmod 9,</math> so <cmath>10x - 9x \equiv 7 - 0 \pmod 9</cmath> and <cmath>x \equiv 7 \pmod 9.</cmath> | ||
+ | |||
+ | Let the quotient of <math>9</math> in our modular equation be <math>c,</math> and let our desired number be <math>z,</math> so <math>x=9c+7</math> and <math>z = 10x+3.</math> We substitute these values into <math>z = 10x+3,</math> and get <cmath> z = 10(9c+7) + 3</cmath> so <cmath> z = 90c+73.</cmath> As a result, <math>z \equiv 73 \pmod {90}.</math> | ||
+ | |||
+ | *Alternatively, we could have also used a system of modular equations to immediately receive <math>z \equiv 73 \pmod {90}.</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> | ||
==Video Solution== | ==Video Solution== |
Revision as of 23:40, 1 November 2020
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
?
Video Solution
https://youtu.be/7an5wU9Q5hk?t=574
Solution
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
Video Solution
https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)
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 |
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