Question

the maximum value of n such that 2^n divides 21! is ______
a) 20 b)15 c)18 d)7

Answer 1

21 ! = 2 *3 *4 *5 *6 *7 *8 *9 *10 *11 *12 *13 *14 *15*16 *17 *18 *19* 20*21


there are 10 numbers with a factor of 2
there are 5 numbers with a factor of 4
                2  numbers with a factor of 8
                1                                         16

So total = 10+5+2+1 = 18

Answer = 2^18  divides 21!     n = 18

Answer 2

Answer:

$\huge\star\underline{\mathtt\color{aqua}{⫷❥A} \mathfrak\color{yellow}{N~ }\mathfrak\color{blue}{S} \mathbb\color{purple}{W} \mathtt\color{orange}{E} \mathbb\color{red}{R⫸}}\star\:⋆$Diagonal in a rhombus splits an angle in two, which means that this angle would be 140'. The sum of all angles equals 360' and two of those are 140'.

The sum of both acute angles= 360'-280' = 80'. Then, the last thing you need to do is split it into two (as there are two acute angles and they are the same) which gives us 40'.