74 of 100: Number Flip

The possible digits are 0,1,6,8,9.
So, after the year 2017, the possible smallest first digit is 6.
For 6 we need 9 as the last digit of the year.so year is in the form of 6 _ _ 9
Now the required year will be smallest if we put 0 in the 2 blanks
Hence, the answer is 6009

Let the number of interest be $\overline{abcd}$ . This means digits $a$ and $d$ must be strobogrammatic themselves. Because we are looking for the next year $> 2017$ , this eliminates $0$ and $1$ . To minimize $\overline{abcd}$ with the remaining options, we let $a = 6$ , and it follows that $d = 9$ .

What remain are the middle digits, $b$ and $c$ . Letting $b = c = 0$ will allow us to minimize the year.

The answer is $\overline{abcd} = \boxed{6009}$ .

$\text{simple and easy to understand.}$

as the possible digits are- $0,1, 6,8,9$ and 1961 is the example[digit -1 has been used here]

$\text{so, after 2017 the possible smallest digit is 6}$

$\text{we could have used 6000,6001,6006,6008-----but if we rotate these numbers it will look different}$

$\text{so,the smallest possible year after 2017 must be 6009}$

2 can not be put in a strobogrammatic arrangement, and the next number that can be put in one is $6$ . So, $6$ will have to be put in the thousands place. After that, the smallest numbers that can be in the hundreds and tens place is $0$ . The ones digit will have to be $9$ , which in total is * $6009$ *

Sharing this fact as seen from a Vsauce video:

Reference:

Stevens, M. [Vsauce]. (2013). What will we miss? [Video]. YouTube. https://youtu.be/7uiv6tKtoKg

Well, after 2017 you know that the first digit can't be one, because then it would be before 2017, and it can't be 2,3,4, or 5 ether, because if you turn those upside then you don't get the a number. Therefore, the smallest first number that it could be is 6 (and you can see that this is the smallest number in the problem). Well, 6 turned upside down is 9, and when it is turned upside down it is the last number. That means that 9 has to be the last number. That makes your number so far: 6_ _9. The problem is asking for the smallest number, so that means that you should replace the two blanks with 0. That makes your answer 6009.

would 5002 work?

if you flip 6009, it turns into 6009.

I would argue that 2112 is a better answer, written in a certain font. I don't understand the exclusion if digits 2 and 5, really. In a block font they would satisfy the criteria. It's also an awesome album!

The fist digit has to be greater than 2, that leaves with 6, 8, 9. The smallest is 6. So to have the property it has to start with 6 and end with 9. To make it the smallest year after 2017, it has to be 6009.

When written using standard characters (ASCII), the numbers, $0, 1, 8$ are symmetrical around the horizontal axis, and $6$ and $9$ are the same as each other when rotated $180$ degrees. Thus, the answer is $\boxed{6009}$ .

The smallest valid digit after $2$ for the thousands place is $6$ . Therefore, $6$ must be the thousands digit.

Because $6$ is the thousands digit, then $9$ must be the ones digit, because it is what you get by flipping $6$ around 180 degrees.

In order to have the smallest number, we must fill in the tens and hundreds values with the smallest, valid, possible, which is $0$ .

Therefore, the number/year must be $\boxed{6009}$ .