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Sari Dates

Posted by Denny in Mathematics and Life's Details on 12 27th, 2009 | no responses

My daughter Sandy’s friend, Sari, noted on December 16, 2009, that on that particular day, when the date was expressed as 12/16/09, the square of the month equaled the product of the day of the month and the year. That is,

December squared = (the 16th) times (09), or better yet,

$12^2 = 16 \cdot 9$

I think this is a fun relationship and will give such dates the name Sari Dates. So, of course, we need to know how many Sari Dates there are.

I counted 40 of them. I listed each month and just counted. Here is how.

In general we want to count the number of occurrences of the event

$\hbox{month}^2 = \hbox{day} \cdot \hbox{year}$

Let’s narrow it down to this century.

I am just going to make a straight count.

Let $a$ = the month number
$b$ = day number
$c$ = year number

So, we are looking for how many times between 2000 and 2099 that

$a^2 = b\cdot c$

Not all numbers work. There are some restrictions on the values $a$ , $b$ , and $c$ .

Because $a$ represents a month, it must be between and including 1 and 12. That is,

Because $b$ represents a day, it must be between and including 1 and 28, or 1 and 29, or 1 and 30, or 1 and 31.

Because $c$ represents a year in this century, it must be between and including 0 (2000) and 99 (2099).

To make the count, start with the square of the month then find all the factors of that number that meet the

Let’s count.

January:
$a^2 = b\cdot c$

$01^2 = b\cdot c$

$1 = b\cdot c$

The only factors of 1 are 1 and 1. So,

$b=1, c=1$

Then for January, we get only one date, 01/01/01

February:
$a^2 = b\cdot c$

$02^2 = b\cdot c$

$4 = b\cdot c$

The factors of 4 are 4,1; 2,2. So

$b=4, c=1 \rightarrow 2^2 = 4 \cdot 1$

$b=1, c=4 \rightarrow 2^2 = 1 \cdot 4$

$b = 2, c=2 \rightarrow 2^2 = 2 \cdot 2$

Then for February, we get 3 dates: 02/02/01, 02/02/02, and 01/01/04.

March:
$a^2 = b\cdot c$

$03^2 = b\cdot c$

$9 = b\cdot c$

The factors of 9 are 9,1; 3,3. So

$b=9, c=1 \rightarrow 3^2 = 9 \cdot 1$

$b=1, c=9 \rightarrow 3^2 = 1 \cdot 9$

$b = 3, c=3 \rightarrow 3^2 = 3 \cdot 3$

Then for March, we get 3 dates: 03/09/01, 03/03/03, and 01/01/09.

April:
$a^2 = b\cdot c$

$04^2 = b\cdot c$

$16 = b\cdot c$

The factors of 16 are 16,1; 8,2; 4,4. So

$b=16, c=1 \rightarrow 4^2 = 16 \cdot 1$

$b=1, c=16 \rightarrow 4^2 = 1 \cdot 16$

$b=8, c=2 \rightarrow 4^2 = 8 \cdot 2$

$b=2, c=8 \rightarrow 4^2 = 2 \cdot 8$

$b=4, c=4 \rightarrow 4^2 = 4 \cdot 4$

Then for April, we get 5 dates: 05/16/01, 04/08/02, 04/04/04, 04/02/08, and 04/01/16.

May:
$a^2 = b\cdot c$

$05^2 = b\cdot c$

$25 = b\cdot c$

The factors of 25 are 25,1; 5,5. So

$b=25, c=1$ But this cannot happen because we can’t have a day of 25!

$b=1, c=25 \rightarrow 5^2 = 1 \cdot 25$

$b=5, c=5 \rightarrow 5^2 = 5 \cdot 5$

Then for May, we get 2 dates: 05/05/05, and 05/01/25.

June:
$a^2 = b\cdot c$

$06^2 = b\cdot c$

$36 = b\cdot c$

The factors of 36 are 36,1; 18,2; 9,4; 3,12. So

$b=36, c=1$ But this cannot happen because we can’t have a day of 36!

$b=1, c=36 \rightarrow 6^2 = 1 \cdot 36$

$b=18, c=2 \rightarrow 6^2 = 18 \cdot 2$

$b=2, c=18 \rightarrow 6^2 = 2 \cdot 18$

$b=9, c=4 \rightarrow 6^2 = 9 \cdot 4$

$b=4, c=9 \rightarrow 6^2 = 4 \cdot 9$

$b=3, c=12 \rightarrow 6^2 = 3 \cdot 12$

$b=12, c=3 \rightarrow 6^2 = 12 \cdot 2$

Then for June, we get 6 dates: 06/18/02, 06/09/04, 06/04/09, 06/03/12, 06/02/18, and 06/01/36.

July:
$a^2 = b\cdot c$

$07^2 = b\cdot c$

$49 = b\cdot c$

The factors of 49 are 49,1; 7,7. So

$b=49, c=1$ But this cannot happen because we can’t have a day of 49!

$b=1, c=49 \rightarrow 7^2 = 1 \cdot 49$ B

$b=7, c=7 \rightarrow 7^2 = 7 \cdot 7$

Then for July, we get 6 dates: 06/18/02, 06/09/04, 06/04/09, 06/03/12, 06/02/18, and 06/01/36.

August:
$a^2 = b\cdot c$

$08^2 = b\cdot c$

$64 = b\cdot c$

The factors of 64 are 64,1; 32,2; 16,4, 8,8. So

$b=64, c=1$ But this cannot happen since we can’t have a day of 64!

$b=1, c=64 \rightarrow 8^2 = 1 \cdot 64$

$b=32, c=2$ But this cannot happen since we can’t have a day of 32!

$b=2, c=32 \rightarrow 8^2 = 2 \cdot 32$

$b=16, c=4 \rightarrow 8^2 = 16 \cdot 4$

$b=4, c=16 \rightarrow 8^2 = 4 \cdot 16$

$b=8, c=8 \rightarrow 8^2 = 8 \cdot 8$

Then for August, we get 5 dates: 08/16/04, 08/08/08, 08/04/16, 08/02/32, 08/01/64.

September:
$a^2 = b\cdot c$

$09^2 = b\cdot c$

$81 = b\cdot c$

The factors of 81 are 81,1; 9,9. So

$b=81, c=1$ But this cannot happen since we can’t have a day of 81!

$b=1, c=81 \rightarrow 9^2 = 1 \cdot 81$

$b=9, c=9 \rightarrow 9^2 = 9 \cdot 9$

Then for September, we get 2 dates: 09/09/09 and 09/01/64.

October:
$a^2 = b\cdot c$

$10^2 = b\cdot c$

$100 = b\cdot c$

The factors of 100 are 100,1; 50,2; 25,4; 20,5; 10,10. So

$b=100, c=1$ But this cannot happen since we can’t have a day of 100!

$b=1, c=100$ But this cannot happen since we can’t have a year of 100! That you would be the year 3000.

$b=50, c=2$ But this cannot happen since we can’t have a day of 50!

$b=2, c=50 \rightarrow 10^2 = 2 \cdot 50$

$b=25, c=4 \rightarrow 10^2 = 25 \cdot 4$

$b=4, c=25 \rightarrow 10^2 = 4 \cdot 25$

$b=20, c=5 \rightarrow 10^2 = 20 \cdot 5$

$b=5, c=20 \rightarrow 10^2 = 5 \cdot 20$

$b=10, c=10 \rightarrow 10^2 = 10 \cdot 10$

Then for October, we get 6 dates: 10/25/04, 10/20/05, 10/10/10, 10/05/20, 10/04/25, and 10/02/50.

November:
$a^2 = b\cdot c$

$11^2 = b\cdot c$

$121 = b\cdot c$

The factors of 121 are 121,1; 11,11. So

$b=121, c=1$ But this cannot happen since we can’t have a day of 121!

$b=1, c=121$ But this cannot happen since we can’t have a year of 121!

$b=11, c=11 \rightarrow 11^2 = 11 \cdot 11$

Then for November, we get 1 date: 111/11/11.

December:
$a^2 = b\cdot c$

$12^2 = b\cdot c$

$144 = b\cdot c$

The factors of 144 are 144,1; 72,2; 48,3; 36,4; 24,6; 18,8; 16,9; and 12,12. So

$b=144 c=1$ But this cannot happen since we can’t have a day of 144!

$b=1, c=144$ But this cannot happen since we can’t have a year of 144!

$b=72, c=2$ But this cannot happen since we can’t have a day of 72!

$b=2, c=72\rightarrow 12^2 = 2 \cdot 72$

$b=48, c=3$ But this cannot happen since we can’t have a day of 48!

$b=3, c=48 \rightarrow 12^2 = 2 \cdot 48$

$b=36 c=4$ But this cannot happen since we can’t have a day of 36!

$b=4, c=36 \rightarrow 12^2 = 4 \cdot 36$

$b=24, c=6 \rightarrow 12^2 = 24 \cdot 6$

$b=6, c=24 \rightarrow 12^2 = 6 \cdot 24$

$b=18, c=8 \rightarrow 12^2 = 18 \cdot 8$

$b=8 c=18 \rightarrow 12^2 = 8 \cdot18$

$b=16, c=9 \rightarrow 12^2 = 16 \cdot 9$

$b=9, c=16 \rightarrow 12^2 = 9 \cdot 16$

$b=12, c=12 \rightarrow 12^2 = 12 \cdot 12$

Then for December, we get 7 dates: 12/24/06, 12/16/09, 12/18/08, 12/12/12, 12/09/16, 12/06/24, and 12/03/48.

To summarize, letting SD stand for Sari Date, I counted:

January: 1 SD
February: 3 SDs
March: 3 SDs
April: 5 SDs
May: 2 SDs
June: 6 SDs
July: 2 SDs
August: 5 SDs
September: 2 SDs
October: 3 SDs
November: 1 SD
December: 7 SDs

Then,

$1\hbox{SD} + 3\hbox{SD} + 3\hbox{SD} + 5\hbox{SD} + 2\hbox{SD} + 6\hbox{SD} + 2\hbox{SD} + 5\hbox{SD} + 2\hbox{SD} + 3\hbox{SD} + 1\hbox{SD} + 7\hbox{SD} = 40\hbox{SD}$