3415 base 7 ×(-432base 7)

Ignore the negative sign, since that only affects the sign of the final product

3415₇ × 432₇

You can compute the product by considering the digit expansion, by which I mean first expanding each number as sums of powers of 7:

3415₇ = 3000₇ + 400₇ + 10₇ + 5₇

(It's the same thing as writing, for example, the decimal 3415 as 3000 + 400 + 10 + 5, just in a different base.)

Similarly,

432₇ = 400₇ + 30₇ + 2₇

Then apply the distributive property:

= (3000₇ + 400₇ + 10₇ + 5₇) × (400₇ + 30₇ + 2₇)

= 3000₇ × (400₇ + 30₇ + 2₇)

… + 400₇ × (400₇ + 30₇ + 2₇)

… + 10₇ × (400₇ + 30₇ + 2₇)

… + 5₇ × (400₇ + 30₇ + 2₇)

= {12}00000₇ + {9}0000₇ + 6000₇

… + {16}0000₇ + {12}000₇ + {8}00₇

… + 4000₇ + 300₇ + 20₇

… + {20}00₇ + {15}0₇ + {10}₇

where I use the curly braces to indicate products (written in base 10) of digits that exceed 6. Anything in these braces needs to be properly converted to base 7. For example,

{12}₇ = 7 + 5 = 15₇

So you would have

= 1500000₇ + 120000₇ + 6000₇

… + 220000₇ + 15000₇ + 1100₇

… + 4000₇ + 300₇ + 20₇

… + 2600₇ + 210₇ + 13₇

Add everything up:

• In the 7⁰ place,

3₇

• In the 7¹ place,

1₇ + 1₇ + 2₇ = 4₇

• In the 7² place,

2₇ + 6₇ + 3₇ + 1₇ = {12}₇ = 15₇

• Carry the 1 (underlined). In the 7³ place,

2₇ + 4₇ + 1₇ + 5₇ + 6₇ + 1₇ = {19}₇ = 25₇

• Carry the 2. In the 7⁴ place,

1₇ + 2₇ + 2₇ + 2₇ = {7}₇ = 10₇

• Carry the 1. In the 7⁵ place,

2₇ + 1₇ + 5₇ + 1₇ = {9}₇ = 12₇

• Carry the 1. In the 7⁶ place,

1₇ + 1₇ = 2₇

So, we end up with

3415₇ × 432₇ = 22045543₇

and thus

3415₇ × (-432₇) = -22045543₇

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!