the product of 44 and 11 gives 1034 as answer. what is the base of the system??
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Think of how we write numbers. Supposing we have a base of 10, what would the number 35 be? It'd be 3 * 10^1 + 5 * 10^0 => 3 * 10 + 5 * 1 => 3 * 10 + 5
What would it be with a base of 9?
3 * 9^1 + 5 * 10^0 => 3 * 9 + 5
Using that format, we can construct an algebraic equation:
(4x + 4) * (1x + 1) = 1x^3 + 0x^2 + 3x + 4
4 * (x + 1) * (x + 1) = x^3 + 3x + 4
4 * (x^2 + 2x + 1) = x^3 + 3x + 4
4x^2 + 8x + 4 = x^3 + 3x + 4
4x^2 + 8x = x^3 + 3x
0 = x^3 - 4x^2 + 3x - 8x
0 = x^3 - 4x^2 - 5x
0 = x * (x^2 - 4x - 5)
0 = x * (x - 5) * (x + 1)
x = 0 , x = 5 , x = -1
x can't be 0, and a base of -1 doesn't make any sense, but x = 5 is our last option. Let's try it:
4 * 5 + 4 = 24
1 * 5 + 1 = 6
1034 =>
1 * 5^3 + 3 * 5 + 4 =>
125 + 15 + 4 =>
144
24 * 6 => 144
It works
What would it be with a base of 9?
3 * 9^1 + 5 * 10^0 => 3 * 9 + 5
Using that format, we can construct an algebraic equation:
(4x + 4) * (1x + 1) = 1x^3 + 0x^2 + 3x + 4
4 * (x + 1) * (x + 1) = x^3 + 3x + 4
4 * (x^2 + 2x + 1) = x^3 + 3x + 4
4x^2 + 8x + 4 = x^3 + 3x + 4
4x^2 + 8x = x^3 + 3x
0 = x^3 - 4x^2 + 3x - 8x
0 = x^3 - 4x^2 - 5x
0 = x * (x^2 - 4x - 5)
0 = x * (x - 5) * (x + 1)
x = 0 , x = 5 , x = -1
x can't be 0, and a base of -1 doesn't make any sense, but x = 5 is our last option. Let's try it:
4 * 5 + 4 = 24
1 * 5 + 1 = 6
1034 =>
1 * 5^3 + 3 * 5 + 4 =>
125 + 15 + 4 =>
144
24 * 6 => 144
It works
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(4b + 4)(b+1) = b³ + 3b + 4
4b² + 8b + 4 = b³ + 3b + 4
b³ - 4b² - 5b = 0
b(b²- 4b - 5) = 0
b(b-5)(b+1) = 0
b = -1, 0, 5
Check: base = 5
(4*5+4)(5+1) = 5³ + 3*5 + 4
24*6 = 125+15+4
144 = 144
Check: base = -1
[4*(-1) + 4]*(-1+1) = (-1)³ + 3*(-1) + 4
0*0 = -1-3+4
0 = 0
Either -1 or 5 could be the base and satisfies the conditions but number bases are conventionally positive integers so choose 5 as the base.
Base of the system = 5
4b² + 8b + 4 = b³ + 3b + 4
b³ - 4b² - 5b = 0
b(b²- 4b - 5) = 0
b(b-5)(b+1) = 0
b = -1, 0, 5
Check: base = 5
(4*5+4)(5+1) = 5³ + 3*5 + 4
24*6 = 125+15+4
144 = 144
Check: base = -1
[4*(-1) + 4]*(-1+1) = (-1)³ + 3*(-1) + 4
0*0 = -1-3+4
0 = 0
Either -1 or 5 could be the base and satisfies the conditions but number bases are conventionally positive integers so choose 5 as the base.
Base of the system = 5
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Suppose the base is b.
In base 10 the numbers are 4b + 4 and 11 = b + 1 and b³ + 3b + 4
(4b + 4)(b + 1) = b³ + 3b + 4
4b² + 8b + 4 = b³ + 3b + 4
4b² + 8b = b³ + 3b
4b + 8 = b² + 3
b² - 4b - 5 = 0
(b - 5)(b + 1) = 0
b = 5 . . . ignore the negative solution
It's base 5.
In base 10 the numbers are 4b + 4 and 11 = b + 1 and b³ + 3b + 4
(4b + 4)(b + 1) = b³ + 3b + 4
4b² + 8b + 4 = b³ + 3b + 4
4b² + 8b = b³ + 3b
4b + 8 = b² + 3
b² - 4b - 5 = 0
(b - 5)(b + 1) = 0
b = 5 . . . ignore the negative solution
It's base 5.
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hexadecimal