Originally published October 24, 2010 at 10:35 PM | Page modified November 2, 2010 at 5:17 PM

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# Abstract gift for Bill Gates

The Microsoft founder and philanthropist is turning 55 this week and since his age coincides with the rightmost two digits of his birth year, it's particularly special — mathematically, that is.

Special to The Seattle Times

**Aziz S. Inan**

**Age:** 55.

**Position:** Professor of electrical engineering, University of Portland (Ore.)

**Born:** May 15, 1955 (or 5-15-55) in Istanbul

**Education:** B.S. in electrical engineering, San Jose State University; M.S. and Ph.D. in electrical engineering, Stanford University

*Source: University of Portland website*

Bill Gates was born in 1955 and, this week, on Oct. 28, the Microsoft chairman and philanthropist will turn 55, a distinct age because it coincides with the rightmost two digits of his birth year.

Because I also had my 55th birthday this year, I got so excited about this rare occurrence that I went ahead and explored this unique age further, especially in the context of Gates' upcoming birthday. My findings were so fascinating that I decided to report them here as a birthday gift for Bill in celebration of his 55th birthday.

First of all, the number 55 equals the sum of all integers from 1 to 10, and 10 equals the sum of the digits of 55.

Second, 55 equals the sum of the squares of all integers from 1 to 5, where 5^{2} is the product of the digits of 55.

Third, 55 = 15 + 40, where 15 equals the sum of all powers of 2 from 0 to 3 (that is, 1 + 2 + 4 + 8 = 15) and 40 is the sum of the powers of 3 from 0 to 3 (expressed by 1 + 3 + 9 + 27 = 40).

Fourth, 55 is a Fibonacci number. Fibonacci numbers consist of a simple series of numbers first introduced in a book titled "Liber Abaci," published in 1202 by Italian mathematician Leonardo Fibonacci (1170-1250).

The series begins with 1 and 1. After that, each number is obtained by adding the previous two numbers of the series:

1 + 1 = 2

1 + 2 = 3

2 + 3 = 5

3 + 5 = 8

5 + 8 = 13

In other words, the Fibonacci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, etc.

There are only a handful number of double-digit Fibonacci birthdays in one's lifetime. Gates' last one was his 34th, which occurred 21 years ago. His upcoming 55th will be his next Fibonacci birthday and coincides with a Fibonacci (21st) century, an extremely rare occurrence. After this one, Bill will hopefully celebrate one more Fibonacci birthday in this Fibonacci century, in 2044, when he turns 89.

I also discovered that Bill's 89th birthday written in full date format as 10-28-2044 (or simply, 10282044) possesses other unique features.

First, split this date into two numbers 1028 and 2044. What you'll see is that 2044 can be derived from 1028 by doubling the first and third leftmost digits (1 and 2) and halving the second and the fourth (0 and 8).

Second, 1028 and 2044 each differ from perfect powers of number two by four (2^{2}) — that is, 1028 is 2^{2} more than 2^{10}, whereas 2044 is 2^{2} less than 2^{11}.

Next, 55 is also a Kaprekar number, named after Indian mathematician Dattatreya Ramachandra Kaprekar (1905-1986). A Kaprekar number is a positive integer whose square split in the middle adds up to the original number.

For example, 9 is a Kaprekar number since 9^{2} = 81. Split 81 into 8 and 1 and one has 8 + 1 = 9. The number 45 is also a Kaprekar number since 45^{2} = 2025 and 20 + 25 = 45.

The first few Kaprekar numbers are 1, 9, 45, 55, 99, 297, 703, etc. Note that 55^{2} = 3025 and 30 + 25 yields 55.

Gates experienced his last Kaprekar birthday 10 years ago when he turned 45. This birthday was also unique in the following way: The number 45 equals the sum of 5, 17 and 23, which are the prime factors of his birth year, namely, 1955 = 5 x 17 x 23. After his 55th birthday, Gates could experience one more Kaprekar birthday in his lifetime, 44 years later, when he turns 99.

Interestingly enough, I discovered another connection between Bill's birth year, 1955, and his new age 55. If 1955 is split into 19 and 55, their product is 19 x 55 = 1045, and if 1045 is split as 10 and 45, 10 + 45 yields 55. The same is also true for the reverse of 1955; that is, split 5591 into 55 and 91, you'll find 55 x 91 = 5005 and — guess what? — 50 + 05 = 55!

Next, 55 is a palindrome number; it reads the same forward or backward. Gates' last palindrome birthday was 11 years ago, when he turned 44, and his next one will be his 66th. That Gates' palindrome 55th birthday occurs in 2010 is distinctive for two reasons.

First, the prime factors of 2010 (that is, 2010 = 2 x 3 x 5 x 67) add up to 77 and the prime factors of 102 (102 = 2 x 3 x 17) — the reverse of 2010 — add up to 22. The difference of these two palindrome numbers (77 — 22) is 55, the number of Gates' upcoming palindrome birthday!

Second, and more fascinating, the reverse of the Gates' palindrome 55th — 10-28-2010 or 10282010 — is 1028201, surprisingly another palindrome. (Instead of reversing this date, one could also insert an extra zero as a 9th digit on the left-hand-side of 10282010, yielding 010282010, a palindrome!) This is an extremely rare occurrence. (Note that the year Gates' birthday is to coincide with a full palindrome date is 6,191 years later, in 8201, on 10288201, Oct. 28, 8201.)

There are 10 dates this year that are like 10282010, from Oct. 20 to 29. I refer to them as "reverse palindrome dates" (that is a full non-palindrome date possessing a palindrome reverse) expressed as 102A2010, where digit A can take any value between 0 and 9. Last time such eight-digit reverse palindrome dates occurred was in 1900 (from Sept. 10 to 19) written as 091A1900.

After this month, the next time such reverse palindrome dates will occur is in 2100 (from Jan. 20 to 29) as 012A2100, followed by year 2110 (between Nov. 20 and 29) in the form 112A2110. So Gates' 55th is special since it coincides with a reverse palindrome date, 10282010!

As an aside, after his 55th, Bill Gates will have other special birthdays to come. For example, his 59th birthday on 10282014 is notable since 10 is half 20 and 28 is double 14. In 2016, Bill will turn 61 which is the reverse of the last two digits of 2016. Similarly, he will turn 72 in 2027, 83 in 2038, and 94 in 2049. Gates' 70th birthday will occur in perfect square year 2025 (2025 is 45^{2}) but will miss the fifth perfect square date in that year, Oct. 27, 2025 (since 10272025 = 3205^{2}), by only one (2^{0}) day.

Gates' 73rd birthday will be on 10282028. His 93rd, on 10282048, will miss the remarkably special date 10242048 (that is, 2^{10} 2^{11}) by four (2^{2}) days! His 97th birthday, in 2052 (10282052), is also interesting because 2052 equals three times the difference of the squares of numbers 10 and 28, together representing Oct. 28. Last, his 101st birthday, to occur on 10282056, will also be notable because 2056 is twice 1028!

My hope is that Gates lives long enough to experience all of these birthdays and more, and cherish all of them. Happy birthday, Bill, and welcome to the club of 55-year-olds!

This story was originally published Oct. 25, 2010, and corrected Nov. 2, 2010. The Fibonacci series of numbers starts with the number 1, not 0 as the original story incorrectly said.

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