The Time Machine by H.G. Wells - a classic science fiction story exploring time travel and human evolution.

The image you've uploaded contains a logic puzzle titled "The Time Machine". Let's analyze and solve it step by step.

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Problem Summary:

We are told:

- The time machine is built in the 21st century.
- It can travel to any year between 1000 AD and 2000 AD, but not before 1000 AD.
- The machine has a digital display showing the current year.
- The machine’s control panel has two buttons:
- A red button that moves the machine backwards in time by 10 years.
- A blue button that moves the machine forwards in time by 14 years.
- The machine starts at 2000 AD.
- The goal is to reach 1000 AD using only these two buttons.

We are asked:
> Can you reach 1000 AD? If so, how many steps does it take?

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Step-by-step Analysis

We start at 2000 AD and want to reach 1000 AD.

We have two operations:
- Red button: Subtract 10 → $ x \to x - 10 $
- Blue button: Add 14 → $ x \to x + 14 $

But we cannot go below 1000 AD (the machine won't work before then).

So, we must only use the red button to go back in time, because the blue button moves forward — which would take us away from 1000 AD.

Wait — this seems problematic.

Let’s think carefully.

We start at 2000 AD.

We want to go to 1000 AD, which is 1000 years earlier.

We can only move backward by 10 years (red button) or forward by 14 years (blue button).

But if we press the blue button, we go further into the future (e.g., 2000 → 2014), which is away from 1000 AD.

So using the blue button would make it harder to reach 1000 AD.

Is there any way to use the blue button to help?

Let’s suppose we try to go forward first, then come back.

But that seems counterproductive unless we can loop around somehow.

But the machine only allows travel between 1000 and 2000 AD — no mention of going beyond 2000.

So if we press the blue button from 2000, we go to 2014, but that’s outside the allowed range.

The problem says: *"It can travel to any year between 1000 AD and 2000 AD"*, so likely it cannot go beyond 2000 AD.

Thus, pressing the blue button from 2000 might be invalid.

Similarly, pressing the red button from 1000 would go to 990, which is before 1000 AD, and that’s not allowed.

So:
- We cannot go below 1000 AD
- We cannot go above 2000 AD

Therefore:
- From 2000 AD, we can only press the red button (back 10 years) → 1990
- From 1000 AD, we cannot press the red button again (would go to 990, not allowed)

So, the only way to get to 1000 AD is by repeatedly pressing the red button.

But can we do that?

Let’s see:

Start: 2000

Press red: 1990
Press red: 1980
...
Each time we subtract 10.

We want to reach 1000.

So, difference = 2000 - 1000 = 1000 years.

Each red button move: -10 years.

So number of steps: $ \frac{1000}{10} = 100 $ steps.

So: Yes, we can reach 1000 AD, by pressing the red button 100 times.

But wait — is there any restriction on the path?

For example, could we press the blue button at some point to help?

Suppose we press the blue button from some year.

But if we press blue from any year $ y $, we go to $ y + 14 $.

But if $ y > 1986 $, then $ y + 14 > 2000 $, which is outside the allowed range.

So we cannot press the blue button unless we are at a year ≤ 1986.

But even if we press it, say from 1986 → 2000, then we’re back at 2000.

That doesn’t help.

What if we go from 1986 → 2000, then go back?

But that just adds extra steps.

For example:
- Start at 2000
- Press red 10 times → 1900
- Press blue → 1914
- Then press red → 1904, etc.

This seems to complicate things and may not help us reach 1000.

But is there a way to reach 1000 AD using a combination of red and blue?

Let’s suppose we want to find integers $ r $ (number of red presses) and $ b $ (number of blue presses) such that:

$$
2000 - 10r + 14b = 1000
$$

And all intermediate years must stay within [1000, 2000].

Simplify:

$$
-10r + 14b = -1000 \\
\Rightarrow 10r - 14b = 1000 \\
\Rightarrow 5r - 7b = 500 \quad \text{(divide by 2)}
$$

Now solve for integers $ r, b \geq 0 $

We want non-negative integer solutions.

Try solving:

$ 5r = 500 + 7b $

So $ r = \frac{500 + 7b}{5} = 100 + \frac{7b}{5} $

So $ \frac{7b}{5} $ must be integer → $ b $ must be divisible by 5.

Let $ b = 5k $, then:

$ r = 100 + \frac{7 \cdot 5k}{5} = 100 + 7k $

So general solution:
- $ b = 5k $
- $ r = 100 + 7k $

Total steps: $ r + b = 100 + 7k + 5k = 100 + 12k $

Minimum steps when $ k = 0 $: $ r = 100, b = 0 $ → 100 steps

If $ k = 1 $: $ r = 107, b = 5 $ → 112 steps

But now check: Can we actually perform these operations without going out of bounds?

Let’s test $ k = 1 $: $ r = 107 $, $ b = 5 $

We need to press blue 5 times and red 107 times.

But we start at 2000.

Each blue press adds 14, each red subtracts 10.

But if we press blue too early, we might go over 2000.

For example, if we press blue from 2000 → 2014 → invalid.

So we cannot press blue at 2000.

We can only press blue from years ≤ 1986.

So we need to first go down to a year ≤ 1986, then press blue.

But let’s suppose we do:

- Press red several times to get to 1986
- Then press blue → 2000
- Then continue with red

But that’s just a detour.

For instance:

- From 2000 → press red 14 times → 1986
- Press blue → 2000
- Now we're back at 2000

Net effect: wasted 15 steps (14 red + 1 blue), no progress.

So it’s inefficient.

Can we use blue to jump ahead and then come back?

But we’re trying to go back to 1000.

No benefit.

Alternatively, suppose we go to a lower year, then use blue to jump forward, then continue back.

But since we’re going backward overall, it seems unnecessary.

Moreover, the simplest path is just to press red 100 times:

- 2000 → 1990 → 1980 → ... → 1000

All intermediate years: 2000, 1990, ..., 1000 — all valid.

And we never go below 1000 or above 2000.

So it works.

Now, is there a shorter path?

From earlier, the total steps are $ 100 + 12k $. For $ k = 0 $, it's 100 steps. For $ k > 0 $, more steps.

So minimum number of steps is 100, achieved by only pressing the red button.

But wait — what if we use blue in reverse?

No — we can’t go forward and then come back in fewer steps.

Let’s suppose we press blue at some point.

Say we go to 1986, press blue → 2000, then go back.

But that’s longer.

Alternatively, can we use blue to “jump” over some years?

But we’re moving backward, so blue moves us forward — opposite direction.

So it can’t help reduce the number of steps.

Hence, the minimal number of steps is 100, using only the red button.

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Final Answer:

✔ Yes, you can reach 1000 AD.

➡️ It takes 100 steps, by pressing the red button (back 10 years) exactly 100 times.

You do not need to use the blue button.

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Why not use the blue button?

- It moves forward in time.
- To go from 2000 to 1000, you need to go backward.
- Using blue would require going forward first, then coming back — increasing the total number of steps.
- Also, pressing blue from 2000 leads to 2014, which is invalid (outside allowed range).
- So blue button is useless for reaching 1000 AD.

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✔ Conclusion:

> Yes, it is possible to reach 1000 AD. It takes exactly 100 steps by repeatedly pressing the red button.

Let me know if you'd like a step-by-step list!