The 29th of February

Leap years occur approximately* every 4 years, so there are $365 \times 4 + 1 = 1461$ days between leap years. 1461 has a remainder of 5 mod 7. If you're familiar with number theory + modular arithmetic, that check that the remainder isn't 0 is enough to clue you off that there will be a full 7 4-year periods, or 28 years between Fridays , as "for all primes, multiplication groups modulo any integer, n, are cyclic" :)

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But, if you're not familiar with that theorem, here's a method that solves this problem without a computer/calculator:

Let the day of birth be day 0. Then Fridays are the days thereafter that are multiples of 7: day 7, 14, 28. Aka, days which have remainder 0 mod 7.

Leap years occur every $365 \times 4 + 1 = 1461$ days, so we need to check day $1461$ , day $2 \times 1461 = 2922$ , day $3 \times 1461 = 4383$ , etc. Until, one of these multiples of 1461 is divisible by 7. But, heck, it's not even pleasant to find the multiples of 1461, let alone checking for divisibility by 7.

So there's a powerful shortcut we can use here: knowing that 1461 has a remainder of 5 mod 7 (via one nasty long-division check if you please) is actually a huge asset in checking all of the other multiples of 1461 for their remainders mod 7.

Think about it like this: if you have 1461 stones, that's like having a bunch of perfect sets of 7 stones, plus 5 left over. So two sets of 1461 stones will mean twice that many perfect sets of 7, plus 10 left over. In other words, we get the remainder of $2 \times 1461$ , just by multiplying its remainder mod 7 by 2: $10 = 5 \times 2$ .

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What do do with the resulting remainders

If the remainder is 0, then it's a Friday. If it's not zero, then the number of days past Friday. So, if the remainder is 1, then it's Saturday; 2=Sunday; 3=Monday; etc.

Now we can easily check all of the multiples of 5 by hand.

$5 \times 0 = 0 \bmod{7}$ (Friday)
$5\times1 = 5 \bmod{7}$ (Wednesday)
$5\times2 = 3 \bmod{7}$ (Monday)
$5\times3 = 1 \bmod{7}$ (Saturday)
$5\times4 = 6 \bmod{7}$ (Thursday)
$5\times5 = 4 \bmod{7}$ (Tuesday)
$5\times6 = 2 \bmod{7}$ (Sunday)
$5\times7 = 0 \bmod{7}$ (Friday)

... no multiple $= 0 \bmod{7}$ until $5\times7$ , so, a Friday will only occur every 7 leap years, or every $\fbox{28}$ years.

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approximately*

This is true for leap years around now, but there are some exceptions to the "every 4 year rule." If you're curious, here's the algorithm for determining whether or not a year is a leap year:

if (year is not exactly divisible by 4) then (it is a common year)
else
if (year is not exactly divisible by 100) then (it is a leap year)
else
if (year is not exactly divisible by 400) then (it is a common year)
else (it is a leap year)

The calendar repeats itself every 28 yrs

It will take 7 leap years to have the same weekday so 7 x 4 = 28.

To check the answer, the weekday shift between 4 years is 5 days or (365,25x4)MOD7 = 5 (except for year ending in 00 but not ending in 400)

check for which min value of X, ( (4 365+1) * X ) is a multiple of 7. and then 4X is the age. 4 365+1 = 1461 which is not a multiple of 7. ie, smallest value of X has to be 7. ie, Age = 7*4 = 28 :)

each date falls on same day every 7 years but since 29th February comes after every 4 years so the the right age will be 7*4=28