Question

Verify the \(3x + 1\)conjecture for these integers.

a) 16

b) 11

c) 35

d) 113

Short answer

3x + 1is conjecture for 16;11;35;113 is true.

Step-by-step solution

Introduction

Let T the transformation that sends an even integer x to \(\frac{x}{2}\) and an odd integer x to 3x+1.

Proving the conjecture for 16

(a)

\(\begin{aligned}{}T(16) &= \frac{{16}}{2} = 8\\T(8) = \frac{8}{2} = 4\\T(4) = \frac{4}{2} = 2\\T(2) = \frac{2}{2} &= 1\end{aligned}\)

i.e.,\(16 \to 8 \to 4 \to 2 \to 1\)

Proving conjecture for 11

(b)

\(\begin{aligned}{}T(11) &= 3.11 + 1 = 34\\T(34) = \frac{{34}}{2} = 17\\T(17) = 3.17 + 1 = 52\\T(52) = \frac{{52}}{2} = 26\\T(26) = \frac{{26}}{2} = 13\\T(13) = 3.13 + 1 = 40\\T(40) = \frac{{40}}{2} = 20\\T(20) = \frac{{20}}{2} = 10\\T(10) = \frac{{10}}{2} = 5\\T(5) = 3.5 + 1 = 16\\T(16) = \frac{{16}}{2} = 8\\T(8) = \frac{8}{2} = 4\\T(4) = \frac{4}{2} = 2\\T(2) = \frac{2}{2} &= 1\end{aligned}\)

i.e.,\(34 \to 17 \to 52 \to 26 \to 13 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1\)

Proving conjecture for 35

(c)

\(\begin{aligned}{}T(35) &= 3.35 + 1 = 106\\T(106) = \frac{{106}}{2} = 53\\T(53) = 3.53 + 1 = 160\\T(160) = \frac{{160}}{2} = 80\\T(80) = \frac{{80}}{2} = 40\\T(40) = \frac{{40}}{2} = 20\\T(20) = \frac{{20}}{2} = 10\\T(10) = \frac{{10}}{2} = 5\\T(5) = 3.5 + 1 = 16\\T(16) = \frac{{16}}{2} = 8\\T(8) = \frac{8}{2} = 4\\T(4) = \frac{4}{2} = 2\\T(2) = \frac{2}{2} &= 1\end{aligned}\)

i.e.,\(35 \to 106 \to 53 \to 160 \to 80 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1\)

Proving conjecture for 113

(d)

\(\begin{aligned}{}T(113) &= 3.113 + 1 = 340\\T(340) = \frac{{340}}{2} = 170\\T(170) = \frac{{170}}{2} = 85\\T(85) = 3.85 + 1 = 256\\T(256) = \frac{{256}}{2} = 128\\T(128) = \frac{{128}}{2} = 64\\T(64) = \frac{{64}}{2} = 32\\T(32) = \frac{{32}}{2} = 16\\T(16) = \frac{{16}}{2} = 8\\T(8) = \frac{8}{2} = 4\\T(4) = \frac{4}{2} = 2\\T(2) = \frac{2}{2} &= 1\end{aligned}\)

i.e.,\(113 \to 340 \to 170 \to 85 \to 256 \to 128 \to 64 \to 32 \to 16 \to 8 \to 4 \to 2 \to 1\)