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What do you have in mind to start off?

Squares: there are 44 of them (44*44=1936 but 45*45=2025)

Cubes: there are 12 of them (up to 12*12*12=1728) but exclude 1^3, 4^3, and 9^3 as those are also squares, so just 9 unique ones.

Fourth powers (and all non-prime powers)—are included under lower-ranked exponents

Fifth powers: 4 of them (up to 4*4*4*4*4=1024, but exclude 1^5 and 4^5 as they are also squares, so just 2 unique ones

Seventh powers: Just 1 unique one: 2^7=128.

Once we get up to 11th powers or higher, the lowest unique example is 2^11=2048 which is too big.

So in total, there are 44+9+2+1=56 such numbers up to 2000.

f(2x)=x+1

f(2)=1+1

f(2)=2

G(x) = |x|-2 in the frst one
g(x) = -|x| in the second one
 

This may not be correct, but I think its for the left side.

Because the only time you plug in for the right side is when it is like this f(1).

Since its just x = 1,

then its on the left side

so f(2*1) = x + 1

f(2) = x + 1

2  + 1 = 3

a=2; b=2;c= a^b; if(c<=2000, goto4, goto5);printc,", ",; a++;if(a<45, goto2, 0);a=2;b++;if(b<45, goto2,0)

4  8  9  16  25  27  32  36  49  64  81  100  121  125  128  144  169  196  216  225  243  256  289  324  343  361  400  441  484  512  529  576  625  676  729  784  841  900  961  1000  1024  1089  1156  1225  1296  1331  1369  1444  1521  1600  1681  1728  1764  1849  1936  >>Total =  58

Thank you so much for the Help!

See the answer here:  https://web2.0calc.com/questions/can-u-please-help_1

Least Common Multiple is 3150       3150/7 = 450 pkgs

By prime factorization:

7 = 7¹

9=  3²

10 = 2 x 5

50 = 2 x 5²  

^{Take the red numbers (this is where that number occurs most often in the factorizations) and multiply them together:}

^{7 x 3 x 3 x 2 x 5 x 5 = 3150}

sumfor(n, 1, 132, n^2) =775,390 mod 11 = 0 

sumfor(n, 1, 100, 2^n) = 2 5353012004 5645880299 3406410750 mod 7 = 2

n mod 6 =5, n mod 10 = 9, solve for n

n = 30 m + 29 and m =0, 1, 2, 3.......etc.

So, the smallest n = 29

29 mod 30 = 29

CORRECTION OF ABOVE. SORRY ABOUT THAT:

simplify {n/2*(2*(3/4) + (n -1)*(3/8)) - [3/4*((1/2)^n) / (1/2 - 1)]} >=200

3 (n^2 + 3 n + 2^(3 - n))>=3200

n<=root of 2^x (3 x^2 + 9 x - 3200) + 24 near x = -7.02021

n>=root of 2^x (3 x^2 + 9 x - 3200) + 24 near x = 31.1943

Thanks for the help Rom!

....

simplify {n/2*(2*(3/4) + (n -1)*(3/8)) - 3/4*((1/2)^n) / (1/ - 1)} >=200

3 (n^2 + 3 n + 2^(2 - n))>=3200

n<=root of 2^x (3 x^2 + 9 x - 3200) + 12 near x = -8.00368

n>=root of 2^x (3 x^2 + 9 x - 3200) + 12 near x = 31.1943

I just multiplied through by  8  on both sides.........this eliminates those pesky fractions !!!

8 * 8   =   64 ways

How many ordered pairs of positive integers $(m,n)$ satisfy $\text{GCD}(m,n) = 3$ and $\text{LCM}(m,n) = 108?$

lcm      gcd       m           n

108        3        108        3

108        3         3        108

108        3         27        12

108        3         12        27

....

p=0;print 1/1005;print 1/1003;print 1/502;print 2/503;a=1;b=168;cycle1:c=a/b;printc;p=p+1;a++;b++;if(a<167, goto cycle1, 0);a=1;b=2;cycle:c=a/b;printc;p=p+1;a++;b++;if(a<2005, goto cycle,0);printp+4

1 / 1005
1 / 1003
1 / 502
2 / 503
1 / 168
2 / 169
3 / 170
This continues until you get: 166 / 333 + 1
165 / 332
166 / 333
1 / 2
2 / 3
3 / 4 - This continues until you get: 2004 / 2005
2001 / 2002
2002 / 2003
2003 / 2004
2004 / 2005
TOTAL TERMS = 2,174

All the odd numbers between 1 and 1000 which is 500 numbers. From that you must subtract 100 numbers that end in 5 such as: 5, 25, 35, 45........etc.

So, 500 - 100 = 400 numbers - which is what you want.

....

My apologies, but what happens to the (1/8) during, (1/8)n^2 + (1/8)n - 8n  =  0, before becoming, n^2 + n - 64n  = 0.

Let the x axis be the number of bags of popcorn

Let the y axis be the number of drinks

We have this equality

5x + 3y = 30

The intercepts are  (0,10)  and (6, 0)

Here is the graph : https://www.desmos.com/calculator/efez6wdpej

This graph  shows the combintions of bags of popcorn / drinks  that  can be bought for $30

y  =  5 ^x+1      take the log of both sides

log y  =  log 5 ^x+1      and by a log property we can write

log y  = (x + 1) log 5

log y  =  x log 5  + log 5

log y - log 5  = x log 5      divide both sides by  log 5

logy / log5  - 1 =  x        and by another log property we can write

log₅ y - 1  = x       "swap"  x  and y

log ₅ x  - 1  =  y

g(x)  = log ₅ x - 1        this is the inverse.......this answer goes in the first box

left         this answer goes in the second box

down      this answer goes in the last box

....