Happy birthday to me (part 1)

Let see the properties:

• If $01/01^{st}$ is a thursday, then $M/D$ is a friday. The fridays of each month are:

$January: 2, 9, 16, 23, 30$

$February: 6, 13, 20, 27$

$March: 6, 13, 20, 27$

$April: 3, 10, 17, 24$

$May: 1, 8, 15, 22, 29$

$June: 5, 12, 19, 26$

$July: 3, 10, 17, 24, 31$

$August: 7, 14, 21, 28$

$September: 4, 11, 18, 25$

$October: 2, 9, 16, 23, 30$

$November: 6, 13, 20, 27$

$December: 4, 11, 18, 25$

• $D$ is divisible by $M$ , then the days that satisfy this property are:

$January: 2, 9, 16, 23, 30$

$February: 6, 20$

$March: 6, 27$

$April: 24$

$May: 15$

$June: 12$

$July: N/A$

$August: N/A$

$September: 18$

$October: 30$

$November: N/A$

$December: N/A$

• $M$ represent the number of prime factors of $D$ . The days that satisfy this property are:

$January: 2, 23$

$February: 6$

$March: 27$

$April: 24$

$May: N/A$

$June: N/A$

$July: N/A$

$August: N/A$

$September: N/A$

$October: N/A$

$November: N/A$

$December: N/A$

• And finally, The smallest possible sum of days in a possible month are: $6$ in February $(2)$

The birthday is: $02/06$

$2+6= \boxed{8}$

that's 02/06th = 2 + 6 = 8

My birthday is at $02/06^{th}$ ­.

First of all, we can restrict the possible months of my birthday. Since $2­$ ­ is the smallest prime number, $2^{4}=16­$ ­ is the maximum power of two that can represent a day, because $2^{5}=32 \rightarrow 32>31­$ ­. Replacing any of the prime factors of $32$ , it still would be greater than $31$ . So, the greater number of prime factors only can be $4$ , and the maximum month is april.

Looking for fridays since $01/01^{st}$ until $04/30^{th}$ (I used modular arithmetic to determine rather a day is a friday or not, but here, I'm going to list them all, already factorised), we would have:

• january : $2(2), \ 9(3\cdot{3}), \ 16(2\cdot{2}\cdot{2}\cdot{2}), \ 23(23), \ 30(2\cdot{3}\cdot{5})$ ;
• february : $6(2\cdot{3}), \ 13(13), \ 20(2\cdot{2}\cdot{5}), \ 27(3\cdot{3}\cdot{3})$ ;
• march : $6(2\cdot{3}), \ 13(13), \ 20(2\cdot{2}\cdot{5}), \ 27(3\cdot{3}\cdot{3})$ ;
• april : $3(3), \ 10(2\cdot{5}), \ 17(17), \ 24(2\cdot{2}\cdot{2}\cdot{3})$ ;

Eliminating the days which have a number of prime factors different than the number of the month, and checking for divisibility of these days by the number of the month (only $03/20^{th}$ falls down after applying the "divisibility criterion" if we have used the "number of prime factors" before it), for every possible month we'll get:

• january : $S({M_1})=\{2, \ 23\}$ ;
• february : $S({M_2})=\{6\}$ ;
• march : $S({M_3})=\{27\}$ ;
• april : $S({M_4})=\{24\}$ ;

After we add up the elements on each set (last criterion), the smallest sum is $6$ , which pertains to february. Hence, the answer is $M/D=02/06^{th}\rightarrow \boxed{M+D=8}$ ­.