This is the last time in a century that we will be able to do the dates like this, with one digit for each day, month and year.
Thanks to LP for sharing the concept.
The next time will be January 01, 2101.
This is the 252 day of the year. The sum of those digits equals 9 (2+5+2)
Multiply 9 by any other single-digit number, and then look at the product. The sum of the two digits will equal 9.
REALLY?
I don't think so. But let's try it.
9 x 1 = 9 ( 9 = 9)
9 x 2 = 18 (1 + 8 = 9)
9 x 3 = 27 ( 2 + 7 = 9)
9 x 4 = 36 ( 3 + 6 = 9)
9 x 5 = 45 ( 4 + 5 = 9)
9 x 6 = 54 ( 5 + 4 = 9)
9 x 7 = 63 ( 6 + 3 = 9)
9 x 8 = 72 ( 7 + 2 = 9)
9 x 9 = 81 ( 8 + 1 = 9)
Multiply 9 by any two or three-digit number, and then look at the product.
The sum of the digits will equal 9 or multiples of 9.
9 x 15 = 135 ( 1 + 3 + 5 = 9)
9 x 46 = 414 ( 4 + 1 + 4 = 9)
9 x 77 = 693 ( 6 + 9 + 3 = 18 and we find that 1+ 8 = 9)
9 x 864 = 7776 ( 7 + 7 + 7 + 6 = 27 and 2 + 7 = 9)
Wow!
Let's try a 4-digit number
9 x 2009 = 18081 you can already see this is going to work! (1 + 8 + 8 + 1 = 18 and 1 + 8 = 9)
OK, now a 5-digit number
9 x 12525 = 112725 = ( 1 + 1 + 2 + 7 + 2 + 5 = 18 and 1 + 8 = 9)
How about a 6-digit number?
9 x 654987 = 5,894,883 = ( 5 + 8 + 9 + 4 + 8 + 8 + 3)
Can we simplify the process? Look at digits in the product.
We have one 9 standing on its own.
If we add the 5 and the 4 we get 9.
We have a 3 and three 8s left.
Let's split the 3 into ones and add them to each of the 8s. We have 3 more 9s now.
Let's use this on a 7-digit number
9 x 1234567 = 11111103 = 9
Will it work on an 8-digit number?
9 x 22446688 = 202020192 (4 2s and a 1 + 9)
Finally, will this same scheme still work on an 9-digit number?
9 x 555888222 = 5002993998 (5 + 2 + 3 + 8 = 18 and 4 nines)
YES!!
No, I can't quite explain the theory on this ... I'm a publisher, not a mathematician!
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