|
|
Line 1: |
Line 1: |
− | ==Problem 8==
| + | #REDIRECT [[2021_Fall_AMC_12B_Problems/Problem_1]] |
− | The largest prime factor of <math>16384</math> is <math>2</math>, because <math>16384 = 2^{14}</math>. What is the sum of the digits of the largest prime factor of <math>16383</math>?
| |
− | | |
− | <math>\textbf{(A) }3\qquad\textbf{(B) }7\qquad\textbf{(C) }10\qquad\textbf{(D) }16\qquad\textbf{(E) }22</math>
| |
− | | |
− | ==Solution 1==
| |
− | We have
| |
− | | |
− | <cmath>16383=16384-1=2^{14}-1=(2^7+1)(2^7-1)=129\cdot127=3\cdot43\cdot127.</cmath>
| |
− | Since <math>127</math> is prime, our answer is <math>\boxed{\textbf{(C) }10}</math>.
| |
− | | |
− | ~kingofpineapplz
| |
− | | |
− | ==Solution 2==
| |
− | Since <math>16384</math> is <math>2^14</math>, we can consider it as <math>(2^7)^2</math>. <math>16383</math> is <math>1</math> less than <math>16384</math>, so it can be considered as <math>1</math> less than a square. Therefore, it can be expressed as <math>(x-1)(x+1)</math>. Since <math>2^7</math> is <math>128, 16383</math> is <math>127 \cdot 129</math>. <math>129</math> is <math>3 \cdot 43</math>, and since <math>127</math> is larger, our answer is <math>\boxed {(C) 10}</math>.
| |
− | | |
− | ~Arcticturn
| |
− | | |
− | ==See Also==
| |
− | {{AMC10 box|year=2021 Fall|ab=B|num-a=9|num-b=7}}
| |
− | {{MAA Notice}}
| |