989901
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1, 3, 9, 11, 33, 99, 101, 303, 909, 1111, 3333 and 9999.
To be palindromic, the year must be of the form XYYX where X and Y are digits, so we need only consider the first half of the number, namely the XY and the question becomes how many numbers XY are there between 20 and 99 such that XYYX is between 2000 and 9999.The answer depends upon the definition of "between":If "between" means strictly between, that is the year cannot be 2000 or 9999XY = 20 ⇒ 2002 > 2000 so allowed XY = 99 ⇒ 9999 = 9999 so not allowed⇒ number of years is 99 - 20 = 79If "between" means between but also including the limits, that is the year could be 2000 or 9999XY = 20 ⇒ 2002 > 2000 so allowed XY = 99 ⇒ 9999 = 9999 so still allowed⇒ number of years is 99 - 20 + 1 = 80
9999 as scientific notation is 9.999 x 10^3.
95963541399
1 x 99 = 99 2 x 99 = 198 3 x 99 = 297 4 x 99 = 396 5 x 99 = 495 6 x 99 = 594 7 x 99 = 693 8 x 99 = 792 9 x 99 = 891 10 x 99 = 990 11 x 99 = 1089 12 x 99 = 1188