'Aroused' by mathlover's working, memories of Smiling 哥's hated primary school maths lessons came flooding back, and he decided to 'repackage' the a,b,c,d in order to construct the algebraic equation/formula using just one variable (coz he was one of those straight-minded students who refused to believe that a single equation can solve for 4 unknowns)

Let ~~Miss Loi's birthday~~ the special number be `x`.

"five-digit number formed from adding a ‘3′ after it"

⇒ 10`x` + 3 (×10 to 'shift all digits one space to the left')

"five-digit number formed from adding a ‘3′ before it"

⇒ 30000 + `x`

∴ (30000 + `x`) − (10`x` + 3) = 3843

Thus, he solved the above and got `x` = 2906.

And with all the clues and promptings from here and over at the Maths Tutor's Facebook Wall, a sudden realization came to him and he took a deep breath and shouted

Happy Birthday! Loi Loi!

^^ Q.E.D. ^^

]]>Let the desired 4 digits be 'a' 'b' 'c' and 'd' in that order. Hence the number we want to find will be 1000a + 100b + 10c + d.

30000 + 1000a + 100b + 10c + d - (10000a + 1000b + 100c + 10d + 3) = 3843

9000a + 900b + 90c + 9d = 26154

1000a + 100b + 10c + d = 2906

Q.E.D. ^^

And why do I have a feeling you'll have something to say about students complaining this is 1 equation with 4 unknowns and hence 'unsolvable'?

P.S. @ clarion-x : Having to admit I was wrong about Miss Loi's age being 77 is worse than death...

Happy 38th birthday Miss Loi! See, I halved 77 and so kindly rounded it down. 😀

]]>Disembodied voice (which most definitely sounds <77 yrs old) speaks again

]]>Actually this question is adapted from another question that required the formation and evaluation of an algebraic equation. But it was difficult to fit the hallowed number of 2906 into the original question.

So for the purpose of practice, as well as to atone for your 77-year-old heresy, can you obtain the answer through a suitable algebraic equation?

you looking for death ah you -.- ]]>

If you let the number be abcd, then use 3abcd - abcd3, and then use simple subtraction, you will get abcd = 2906.

And considering you used the number 33 in this question, and you are certainly unlikely to be 33 years of age, the only logical explanation is that you were born in 1933.

*runs*

]]>EEEEEEEEKS D:

Miss Loi I'm having exams now... Next time can?

Disembodied voice speaks again

]]>Oh I haven't dealt with you yet for storming The Temple the last time!