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This problem is a little variation from this problem I made 2 months ago, which is inspired from this post.

When is the next date in the future such that all 8 digits of MM/DD/YYYY are all different and the product of MM, DD and YY is equal to YYYY, where YY is the last two digits of the year?

Please note that "product of MM, DD, YY" means MM × DD × YY, where YY is the last two digits of the year. For example, if YYYY=2024, then YY=24.

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  • $\begingroup$ Out of curiosity: When was the previous time? $\endgroup$ Commented Jun 29 at 3:39
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    $\begingroup$ @PierrePaquette I think the same as the accepted answer, but BC. I can't find any dates that work between 01/01/0001 and today $\endgroup$
    – Tim
    Commented Jun 29 at 16:47

1 Answer 1

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I believe the first such future date (represented as MM/DD/YYYY) is

04/19/2736

From the answer to the previous problem

Both 0 and either 1 or 2 must appear in the MM/DD part of the date so the earliest dates we are looking at are in the 2300s.

Since the last two digits of YYYY must divide into it, they also divide into YY00, that is, if there is a solution in the 2300s then the last two digits of the year must divide into 2300 and not contain any of the digits 0,1,2 or 3. The only two-digit factor of 2300 which satisfies this constraint is 46 and 2346/46 = 51. This very quickly leads us to MM=03, DD=17 but we have repeated the digit 3.

Next we look at the 2400s.
Here, the last two digits of the year must divide into 2400 and not contain the digits 0,1,2 or 4. The only two possibilites are 75 and 96.
2475/75 = 33 which forces our DD or MM to be divisible by 11 and repeat a digit.
2496/96 = 26. We cannot repeat the 2 and 6 so we must have DD=13 but then MM=02, repeating the 2 anyway.

Next, we look at the 2500s.
Here, the last two digits of the year must divide into 2500 and not contain 0,1,2 or 5 but there are no possibilities here.

In the 2600s, the last two digits must divide into 2600 and not contain 0,1,2 or 6 and again there are no possibilities.

In the 2700s, the last two digits must divide into 2700 and not contain 0,1,2 or 7. The possibilities here are 36, 45 and 54.
2736/36 = 76 and there is a solution here with MM=04 and DD=19. This is the only valid solution in this year (and indeed in this century as it turns out).

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    $\begingroup$ Nice explanation. $\endgroup$
    – z100
    Commented Jun 27 at 16:14
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    $\begingroup$ Running this forward, 03/17 is very common in future solutions, it just happened to not work here $\endgroup$
    – Tim
    Commented Jun 29 at 16:55

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