1950 AHSME Problems/Problem 20: Difference between revisions

Latest revision as of 15:47, 9 November 2025

Problem

When $x^{13}+1$ is divided by $x-1$ , the remainder is:

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$

Solution

Solution 1

Using synthetic division, we get that the remainder is $\boxed{\textbf{(D)}\ 2}$ .

Solution 2

By the remainder theorem, the remainder is equal to the expression $x^{13}+1$ when $x=1.$ This gives the answer of $\boxed{(\mathrm{D})\ 2.}$

Solution 3

Note that $x^{13} - 1 = (x - 1)(x^{12} + x^{11} \cdots + 1)$ , so $x^{13} - 1$ is divisible by $x-1$ , meaning $(x^{13} - 1) + 2$ leaves a remainder of $\boxed{\mathrm{(D)}\ 2.}$

Solution 4

Say the quotient of the polynomial is $ q $ and the remainder is $ r $. Then we have

$x^{13} + 1 = (x - 1)q + r$

To eliminate $ q $, we put $ x = 1 $ in which we get

$1^{13} + 1 = r$

$1 + 1 = r$

$2 = r$

Thus the answer is $\boxed{(\mathrm{D})\ 2.}$

Video Solution

https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=2485 - AMBRIGGS It's fake.

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

@@ Line 8: / Line 8: @@
 ===Solution 1===
-Use synthetic division, and get that the remainder is <math>\boxed{\mathrm{(D)}\ 2.}</math>
+Using synthetic division, we get that the remainder is <imath>\boxed{\textbf{(D)}\ 2}</imath>.
 ===Solution 2===
-By the remainder theorem, the remainder is equal to the expression <math>x^{13}+1</math> when <math>x=1.</math> This gives the answer of <math> \boxed{(\mathrm{D})\ 2.} </math>
+By the remainder theorem, the remainder is equal to the expression <imath>x^{13}+1</imath> when <imath>x=1.</imath> This gives the answer of <imath> \boxed{(\mathrm{D})\ 2.} </imath>
+===Solution 3===
+Note that <imath>x^{13} - 1 = (x - 1)(x^{12} + x^{11} \cdots + 1)</imath>, so <imath>x^{13} - 1</imath> is divisible by <imath>x-1</imath>, meaning <imath>(x^{13} - 1) + 2</imath> leaves a remainder of <imath>\boxed{\mathrm{(D)}\ 2.}</imath>
+===Solution 4 ===
+Say the quotient of the polynomial is \( q \) and the remainder is \( r \). Then we have
+<cmath>
+x^{13} + 1 = (x - 1)q + r
+</cmath>
+To eliminate \( q \), we put \( x = 1 \) in which we get
+<cmath>
+^{13} + 1 = r
+</cmath>
+<cmath>
++ 1 = r
+</cmath>
+<cmath>
+= r
+</cmath>
+Thus the answer is <imath> \boxed{(\mathrm{D})\ 2.} </imath>
+===Video Solution===
+https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=2485 - AMBRIGGS
+It's fake.
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