Today is my birthday!

Let $M$ be the product of the hours, minutes and seconds. We can see that $M=7\cdot 46\cdot 23 \Rightarrow \boxed{M=7406}$ .

Now, in order for us to find the number of trailing zeroes in $M!$ , let us note two things:

$1^{\circ}$ Trailing zeros are a sequence of $0$ 's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

$2^{\circ}$ The only way to get trailing zeros (or a trailing zero) in a number is to be able to factorize the given number $P$ as $P=M\cdot 2^{m}\cdot 5^{n}$ , where $M$ is the product of the remaining prime factors and $m,n \geq 1$ . This basically means that $2$ and $5$ have got to be prime factors of the number.

Using these two statements, we can see that the number of trailing zeros equals the number of pairs of $2$ 's and $5$ 's in the prime factorization of $7406!$ .

Now take into account that we can "extract" $5$ 's and $2$ 's from $7406!$ from their multiples that appear as multipliers in $7406!$ (for example: $6=2\cdot 3$ , giving us a $2$ ). It is also rather obvious that there will be a lot more $2$ 's than $5$ 's in the prime factorization, therefore we only need to count the number of $5$ 's that appear there since we are certain that there will be enough $2$ 's to cover them.

We must also not forget to count the number of multiples of every power of $5$ that is smaller than $7406$ , since in that case we get $2$ $5$ 's instead of $1$ in the extraction (for example: $75=3\cdot 5^{2}$ , giving us $2$ $5$ 's instead of just $1$ )!

In order to count the number of multiples $G$ of a number $P$ inside another number $T$ we use the following formula: $G=\lfloor \frac{T}{P} \rfloor$ .

Since the biggest power of $5$ that is smaller than $7406$ is $5^{5}=3125$ , the total number $L$ of extracted $5$ 's is:

$L=\lfloor \frac{7406}{5} \rfloor + \lfloor \frac{7406}{5^{2}} \rfloor + \lfloor \frac{7406}{5^{3}} \rfloor + \lfloor \frac{7406}{5^{4}} \rfloor + \lfloor \frac{7406}{5^{5}} \rfloor \Rightarrow$

$\Rightarrow L= \lfloor \frac{7406}{5} \rfloor + \lfloor \frac{7406}{25} \rfloor + \lfloor \frac{7406}{125} \rfloor + \lfloor \frac{7406}{625} \rfloor + \lfloor \frac{7406}{3125} \rfloor \Rightarrow$

$\Rightarrow L=1481+296+59+11+2 \Rightarrow \boxed{L=1849}$ .

Thus, the number of trailing zeros in $M!$ is $1849$ .

Note: We only count the squares, the cubes and the other powers once because we have already counted them in the number of multiples of $5$ and/or the smaller powers consequently!

Now, all we need to do is find $x$ , which is the remainder when $1849^{2}$ is divided by $15$ (if she is $18$ years old now, $3$ years ago she was $15$ ). We get:

$1849 \equiv 4 \pmod {15}$

$1849^{2} \equiv (4)^{2} \pmod {15}$

$1849^{2} \equiv 16 \pmod {15}$

$1849^{2} \equiv 1 \pmod {15}$

This means that finally $\boxed{x=1}$ !

At last, we can calculate the answer, which is

$x^{2}+3x+4=1+3+4=\boxed{8}$ .

I know question is tricky while reading but try to read carefully.

First of all, product of hours-minutes-seconds is = (7)(46)(23) = 7406

Now, number of trailing zeroes in 7406! (factorial) can be calculated as follows:

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Adding all results we get, 1481+296+59+11+2 = 1849

This is number of trailing zeroes.

Squaring it we get, 3418801, which gives out remainder 1 when divided by 15.

This is value of X .

Put in given equation: X ^2+3 X +4

Answer comes out to be 8.

the product of h,m,s = 7406 ,the trailing zeroes = 1849 ,the square = 3418801 ,the square devided by 15 (his age 3 years ago) = 227920.0667 (remainder = 1) so (x^2 +3x +4 = 8) so the answer is 8