Difference between revisions of "2022 AMC 10A Problems/Problem 7"
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<math>\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12</math> | <math>\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12</math> | ||
| − | == Solution == | + | == Solution 1 == |
Note that | Note that | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
| Line 20: | Line 20: | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
| + | |||
| + | == Solution 2 == | ||
| + | ==Problem== | ||
| + | |||
| + | The least common multiple of a positive integer <math>n</math> and <math>18</math> is <math>180</math>, and the greatest common divisor of <math>n</math> and <math>45</math> is <math>15</math>. What is the sum of the digits of <math>n</math> ? | ||
| + | |||
| + | |||
| + | <math>\textbf{(A)} 3 \textbf{ (B)} 6 \textbf{ (C)} 8 \textbf{ (D)} 9 \textbf{ (E)} 12</math> | ||
| + | |||
| + | == Solution 2 == | ||
| + | Since the <math>lcm</math> contains only factors of <math>2</math>, <math>3</math>, and <math>5</math>, <math>n</math> cannot be divisible by any other prime. | ||
| + | Let n = <math>2^a</math> <math>3^b</math> <math>5^c</math>, where <math> a</math> ,<math>b</math>, and <math>c</math> are nonnegative integers. | ||
| + | We know that <math>lcm(n, 18)</math> = <math>lcm(n, 3^2 * 2)</math> = <math>lcm(2^a * 3^b * 5^c, 3^2 * 2)</math> = | ||
| + | <math>lcm(2^ {max(a,1)} \cdot 3^ {max(a,2)} \cdot 5^b)</math> = <math>180</math> = <math>2^2 \cdot 3^2 \cdot 5.</math> | ||
| + | Thus | ||
| + | |||
| + | <math> (1) max(a,1) = 2 -> a = 2 </math> | ||
| + | |||
| + | <math> (2) max(a,2) = 2 -> 0 <= a <= 2</math> | ||
| + | |||
| + | <math>(3) c = 1.</math> | ||
| + | From the gcf information, <math>gcf(n,45)</math> = gcf(n, <math>3^2 \cdot 5</math>) = <math>gcf(2^a \cdot 3^b \cdot 5^c, 3^2 \cdot 5) </math> | ||
| + | = <math>3^min(b,2) \cdot 5^min(c,1) = 15</math> | ||
| + | This means, that since c = 1, 3^min(b,2) \cmath 5 = 15, so min(b,2) = 1 and b = 1. | ||
| + | Hence, multiplying using a = 2, b = 1, c = 1 gives n = 60 and the sum of digits is hence 6. | ||
== See Also == | == See Also == | ||
Revision as of 00:38, 12 November 2022
Problem
The least common multiple of a positive divisor 
and 
is 
, and the greatest common divisor of and 
is 
. What is the sum of the digits of ?
Solution 1
Note that
From the least common multiple condition, we conclude that 
where 
From the greatest common divisor condition, we conclude that that 
Therefore, we have 
The sum of its digits is 
~MRENTHUSIASM
Solution 2
Problem
The least common multiple of a positive integer and is , and the greatest common divisor of and is . What is the sum of the digits of ?
Solution 2
Since the 
contains only factors of 
, 
, and 
, cannot be divisible by any other prime.
Let n = 
, where 
,
, and 
are nonnegative integers.
We know that 
= 
= 
=
= = 
Thus
From the gcf information, 
= gcf(n, 
) = 
= 
This means, that since c = 1, 3^min(b,2) \cmath 5 = 15, so min(b,2) = 1 and b = 1.
Hence, multiplying using a = 2, b = 1, c = 1 gives n = 60 and the sum of digits is hence 6.
See Also
| 2022 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 | ||
| 2022 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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. 
