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Nutka1998 [239]
3 years ago
7

3415 base 7 ×(-432base 7)

Mathematics
1 answer:
docker41 [41]3 years ago
5 0

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 = <u>1</u>5₇

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

2₇ + 4₇ + 1₇ + 5₇ + 6₇ + <u>1₇</u> = {19}₇ = <u>2</u>5₇

• Carry the 2. In the 7⁴ place,

1₇ + 2₇ + 2₇ + <u>2₇</u> = {7}₇ = <u>1</u>0₇

• Carry the 1. In the 7⁵ place,

2₇ + 1₇ + 5₇ + <u>1₇</u> = {9}₇ = <u>1</u>2₇

• Carry the 1. In the 7⁶ place,

1₇ + <u>1₇</u> = 2₇

So, we end up with

3415₇ × 432₇ = 22045543₇

and thus

3415₇ × (-432₇) = -22045543₇

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