CPhill

Agree with EP......there might be an algebraic way to solve this [ using a Taylor expansion of cos (2x) ] or using  Newton's method......this might be tedious, but i'm not sure.... a graph seems the easiest way to go....

See here

https://www.desmos.com/calculator/jewm1ikbfc

This one shouldn't be too tough to figure out

16 * 4 = 64

And 4*4*4 = 64

So....3 fours

9

To see why....

Since each is divisible by 3

The first number  can be 3p and the second number can be 3q

So....their product  is  (3p) (3q) = 9pq

So....the product is divisible by 9  ....!!!

Note that we have the following sequence 

1, 8, 15, 22, 29

When the first day is Thursday....the last Saturday is on the 31st

When the first day is Friday......the last Saturday is on 30th

When the first day is Saturday.....the last Saurday is on the 29th

In each case....there will be 5 Saturdays

But when the first day is on Sunday......the next Saturday will fall on the 7th

And we have the following sequence for Saturdays

7 14  21  28

So...only 4 Saturdays when the first day is Sunday...

asdf is correct....

Since the integers do not have to be distinct (different).....we have....

2*2 + 2 + 2  =  8         4*4 + 4 + 4 =  24        6*6 + 6 + 6   =  48           8*8 + 8 + 8  =    80

2*4 + 2 + 4 =  14        4*6 + 4 + 6  = 34        6*8  +6 + 8  =   62           8*10 + 8 + 10 = 98 

........20                                    ........ 44                           ......76                             ......116

........26                                    .........54                           ......90                             ......134

........32                                    .........64                         ......104

........38                                    .........74

........44

10*10 + 10 + 10  =   120          12 * 12 + 12 + 12   = 168            14 * 14 + 14 + 14  = 224

10*12 + 10 +  12 =   142           12*14 +  12 + 14   = 194

                           ..... 164

We have     (7)*(8) / 2  - 1  =   27 different values.....

(x^2 + 3) (x^2 + 1)   = 0

There are no real number solutons to this......

Setting  both factors to 0   we have

x^2 + 3  = 0                    x^2 + 1 =  0

x^2  =  -3                       x^2 = -1

Note that any real number squared is either 0 or positive

So....no real numbers exist that make either equation true

Thanks, killersteve....!!!!

Your problem specifies "point"......so my answer is correct based on that....

However.....it should specify "lattice points" [ points with integer coordinates ]

Taking this into cosideration.....then 1/3 is correct......  

Mmmmm.....the area of the feasible region is  8 * 4  =  32 units^2

x + y ≤ 4  will form a triangle in this region with a base and height of 4.........so its area is 8

So....the probability of a point falling into this area   of the feasible region  =  8 / 32   =  1 / 4

Agree with Melody.....I never noticed that the LCM of the first n positive integers is just the product of all the primes from 2 to n....!!!!