Difference between revisions of "2021 Fall AMC 12B Problems/Problem 25"
Kevinmathz (talk | contribs) (Created page with "==Problem== For <math>n</math> a positive integer, let <math>R(n)</math> be the sum of the remainders when <math>n</math> is divided by <math>2</math>, <math>3</math>, <math>...") |
Kevinmathz (talk | contribs) (→Solution) |
||
| Line 9: | Line 9: | ||
Note that we can add <math>9</math> to <math>R(n)</math> to get <math>R(n+1)</math>, but must subtract <math>k</math> for all <math>k|n+1</math>. Hence, we see that there are four ways to do that because <math>9=7+2=6+3=5+4=4+3+2</math>. Note that only <math>7+2</math> is a plausible option, since <math>4+3+2</math> indicates <math>n+1</math> is divisible by <math>6</math>, <math>5+4</math> indicates that <math>n+1</math> is divisible by <math>2</math>, <math>6+3</math> indicates <math>n+1</math> is divisibly by <math>2</math>, and <math>9</math> itself indicates divisibility by <math>3</math>, too. So, <math>14|n+1</math> and <math>n+1</math> is not divisibly by any positive integers from <math>2</math> to <math>10</math>, inclusive, except <math>2</math> and <math>7</math>. We check and get that only <math>n+1=14 \cdot 1</math> and <math>n+1=14 \cdot 7</math> give possible solutions so our answer is <math>\boxed{\textbf{(C) }2}</math>. | Note that we can add <math>9</math> to <math>R(n)</math> to get <math>R(n+1)</math>, but must subtract <math>k</math> for all <math>k|n+1</math>. Hence, we see that there are four ways to do that because <math>9=7+2=6+3=5+4=4+3+2</math>. Note that only <math>7+2</math> is a plausible option, since <math>4+3+2</math> indicates <math>n+1</math> is divisible by <math>6</math>, <math>5+4</math> indicates that <math>n+1</math> is divisible by <math>2</math>, <math>6+3</math> indicates <math>n+1</math> is divisibly by <math>2</math>, and <math>9</math> itself indicates divisibility by <math>3</math>, too. So, <math>14|n+1</math> and <math>n+1</math> is not divisibly by any positive integers from <math>2</math> to <math>10</math>, inclusive, except <math>2</math> and <math>7</math>. We check and get that only <math>n+1=14 \cdot 1</math> and <math>n+1=14 \cdot 7</math> give possible solutions so our answer is <math>\boxed{\textbf{(C) }2}</math>. | ||
| + | -kevinmathz | ||
{{AMC12 box|year=2021 Fall|ab=B|num-b=24|after=Last problem}} | {{AMC12 box|year=2021 Fall|ab=B|num-b=24|after=Last problem}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:07, 25 November 2021
Problem
For 
a positive integer, let 
be the sum of the remainders when is divided by 
, 
, 
, 
, 
, 
, 
, 
, and 
. For example, 
. How many two-digit positive integers satisfy 
Solution
Note that we can add to to get 
, but must subtract 
for all 
. Hence, we see that there are four ways to do that because 
. Note that only 
is a plausible option, since 
indicates 
is divisible by , 
indicates that is divisible by , 
indicates is divisibly by , and itself indicates divisibility by , too. So, 
and is not divisibly by any positive integers from to , inclusive, except and . We check and get that only 
and 
give possible solutions so our answer is 
.
-kevinmathz
| 2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 24 |
Followed by Last problem |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
