Difference between revisions of "2017 AMC 12B Problems/Problem 19"
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==Solution 2== | ==Solution 2== | ||
− | We know that this number is divisible by <math>9</math> because the sum of the digits is <math>270</math>, which is divisible by <math>9</math>. If we subtracted <math>9</math> from the integer we would get <math>1234 \cdots 4335</math>, which is also divisible by <math>5</math> and by <math>45</math>. Thus the remainder is <math>9</math>, or <math>\boxed{\textbf{ | + | We know that this number is divisible by <math>9</math> because the sum of the digits is <math>270</math>, which is divisible by <math>9</math>. If we subtracted <math>9</math> from the integer we would get <math>1234 \cdots 4335</math>, which is also divisible by <math>5</math> and by <math>45</math>. Thus the remainder is <math>9</math>, or <math>\boxed{\textbf{C}}</math>. |
==Solution 3 (Beginner's Method)== | ==Solution 3 (Beginner's Method)== |
Revision as of 16:16, 11 February 2019
Problem
Let be the
-digit number that is formed by writing the integers from
to
in order, one after the other. What is the remainder when
is divided by
?
Solution
We will consider this number and
. By looking at the last digit, it is obvious that the number is
. To calculate the number
, note that
so it is equivalent to
Let be the remainder when this number is divided by
. We know that
and
, so by the Chinese remainder theorem, since
,
, or
. So the answer is
.
Solution 2
We know that this number is divisible by because the sum of the digits is
, which is divisible by
. If we subtracted
from the integer we would get
, which is also divisible by
and by
. Thus the remainder is
, or
.
Solution 3 (Beginner's Method)
To find the sum of digits of our number, we break it up into cases, starting with
,
,
,
, or
.
Case 1:
Case 2:
(We add 10 to the previous cases, as we are in the next ten's place)
Case 3:
Case 4:
Case 5:
Thus the sum of the digits is , so the number is divisible by
. We notice that the number ends in "
", which is
more than a multiple of
. Thus if we subtracted
from our number it would be divisible by
, and
. (Multiple of n - n = Multiple of n)
So our remainder is , the value we need to add to the multiple of
to get to our number.
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 |
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.