My Birthday Problem

$90! = 1 \times 2 \times 3 \times 4 \times 5 \times ... \times 89 \times 90$

We can rearrange the factors of 90! into:

$(5 \times 10 \times 15 \times ... \times 85 \times 90) \times (1 \times 2 \times 3 \times 4 ) \times (6 \times 7 \times 8 \times 9) \times (11 \times 12 \times 13 \times 14) \times (16 \times 17 \times 18 \times 19 ) \times ... \times (81 \times 82 \times 83 \times 84) \times (86 \times 87 \times 88 \times 89)$

Factoring out the 5's,

$5^{18} \times 18! \times (1 \times 2 \times 3 \times 4 ) \times (6 \times 7 \times 8 \times 9) \times (11 \times 12 \times 13 \times 14) \times (16 \times 17 \times 18 \times 19 ) \times ... \times (81 \times 82 \times 83 \times 84) \times (86 \times 87 \times 88 \times 89)$

Let us consider (10 a + 1)(10 a + 2)(10 a + 3)(10 a +4) and (10 a + 6)(10 a + 7)(10 a + 8)(10 a + 9)

This expression is a generalized format of the terms (1 \times 2 \times 3 \times 4 ) and (6 \times 7 \times 8 \times 9), respectively.

Simplifying these two expressions, we'll get:

$10000a^{4} + 10000a^{3}+ 3500a^{2}+500a+24$ and $10000a^{4} + 30000a^{3}+ 33500a^{2}+16500a+3024$

We can disregard the first four terms because they contain at least 2 trailing zeroes

We can generalize that the last two digits of the terms of the form (10 a + 1)(10 a + 2)(10 a + 3)(10 a +4) and (10 a + 6)(10 a + 7)(10 a + 8)(10 a + 9) is 24 .

Therefore, the last two digits of 90! is equal to the last two digits of $5^{18} \times 18! \times 24^{18} = 5^{18} \times 18! \times 2^{18} \times 2^{36} \times 3^{18}$

By observation,

1. the last digit of $(2^{10})^{even}$ is 76

2. the last digit of $(2^{10})^{odd}$ is 24

3. the last digit of $(3^{4})^{1 mod 5}$ is 81

4. the last digit of $(3^{4})^{2 mod 5}$ is 61

5. the last digit of $(3^{4})^{3 mod 5}$ is 41

6. the last digit of $(3^{4})^{4 mod 5}$ is 21

7. the last digit of $(3^{4})^{5 mod 5}$ is 01

Finding the last two digits of $18! \times 2^{36} \times 3^{18} => (2^{10})^{3} \times 2^{6} \times (3^{4})^{4} \times 3^{2} \times 18!$

$=> (24) \times 2^{6} \times (21) \times 3^{2} \times 18! => (24)(64)(21)(9) \times 18! => 04 \times 18!$

The last two digits of 18! can be solved like how we solved earlier:

$(5 \times 10 \times 15) \times (1 \times 2 \times 3 \times 4 ) \times (6 \times 7 \times 8 \times 9) \times (11 \times 12 \times 13 \times 14) \times (16 \times 17 \times 18)$

= $5^{3} \times 3! \times (24^{3} \times (16 \times 17 \times 18) => 6 \times 5^{3} \times 2^{3} \times 12^{3} \times 96 => 6 \times 12^{3} \times 96$

Therefore, the last two digits of 90! is equivalent to the last two digits of $04 \times 6 \times 12^3 \times 96$ = $\boxed{12}$

Using The calculator