00:01
We have two pots here.
00:03
In part a, we determine the value of the base x, knowing that the number 211 in base x is equal to the number 152 in base 8.
00:18
The second part b, we convert the x -dessimal number f3a7c2 to octal to base 8.
00:29
So in part a, the condition is that the number 211 base x is equal to the number 152 base 8 or octal.
00:48
So now we are going to write what each number means in the corresponding base to have an equation in decimal base.
00:58
So this number means the number two here multiplies the power of the base corresponding to exponent 0 1 2.
01:16
Because it's the second, the third position from the right to left.
01:20
The first position on the right corresponds to power 0 of the base.
01:25
Second on the left or to the left corresponds to a position.
01:33
Power 1 of the base and this third one corresponds to power 2.
01:38
So the number 2, the digit 2, multiplies the power 2 of the base plus and then number 1 here corresponds to or multiplies power 1 of the base plus and this number 1 here multiplies the power 0 of the base.
02:00
That's got to be equal to and we do the same on the right side.
02:08
So simplies this and the right side is the same ideas but we know the base in this case it's not a variable but a number is 8 so the 1 over here multiplies power 2 of 8 plus number 5 here multiplies power 1 of 8 and the last digit to the right the number 2 multiplies 8 to the 0 and that's expression we want to solve for x which is the base we want to discover here okay let's simplify a little bit we get 2x square plus x plus 1 because x2 series 1 equal 64 plus this is 5 times 8 it's 40 plus 8 to the series 1 we get 2 and this means that 2x square plus x plus 1 is equal to 106.
03:37
And now we put this 106 to the left or equivalently we subtract 106 both sides of the equation to get 2x square plus x minus 105 equals 0.
03:56
And so this is a base 10 equation.
04:00
We know it's base 10 because we have transformed each number on this equation to base 10 by doing the polynomial representation is in the corresponding basis.
04:12
So this is a base 10 equation and it's a polynomial degree 2 so we know how to solve this.
04:21
We get negative 1 more or less square root of 1 minus 4 times 2 times negative 105 over 4 which is 2 times 2 get 901 more less squared off 1 plus and this product over here is 8 times 105 is 400 or 840 over 4 this is 1 more less square of 841 over 4 and 1 more less square root of 841 is 29.
05:14
So we get x is equal to negative 1 more or less 29 over 4.
05:25
So we have first solution negative 1 plus 29 over 4, which is equal to 28 over 4 is 7.
05:42
Let's see the other root x2 is negative 1 minus 29 over 4 and that is negative 30 over 4, which divided by 2 both sides and both numbers, we get negative 15 over 2.
05:59
So it's negative 4 .5.
06:03
So it means this is a solution because the base got to be an integer number.
06:09
So the base x is 7.
06:18
So we have sold the first part.
06:20
And to verify that it's very easy because if we put now x equal to 7 and calculate this side here, we got to get 106, which is the value of the right hand side.
06:34
Okay, so that's part a, so in per p we want to transform the number, the xadecimal number, f3a7c2, f3a7c2, base 16 in a number in base 8.
07:11
So we get to transform to base 8.
07:14
And the idea here is, of course, we can transform this number to base 10 and then go to from base 10 to base 2.
07:24
That would be a lot of work nevertheless.
07:28
So it's better to use base 2 as the intermediate base.
07:32
That if it goes each four digits in base 2 that is in binary, corresponds to one digit in the exodecimal.
07:42
And three binary digits corresponds to one digit in octal.
07:47
So what we got to do is to put this number into base 2, which is quite straightforward.
07:55
And then group the digits in base 2 we get from this transformation in groups of 3 from right to left and then put any group of 3 into a digit which will be in base 8 so let's do that so the the procedure will be that given number we put it in base 2 binary then we put binary into octal base okay, so let's do that.
08:37
So f, which is, let's say, is corresponds to 15, digit 15 x -decimal base.
08:47
In base 2 in binary will be for 1s because it's 1, 2 times 1 times 2 to 1, that is 2 plus 2 to the 3 is 8...