1. What is the remainder when 2^190 is divided by 5?
How do I approach these questions; do I write it down on paper, or use my calculator(how do I see the remainder?)?
2. Figure: A large right triangle, shortest base in on the ground, shorter height is vertical, longest hypotenuse is slanting at the left. This large triangle is labeled A(bottom left)C(top)D(bottom right/90 degree angle). Drawing a line from D(the 90 degree corner) down to segment AC(the hypotenuse of the larger triangle) at point B, perpendicular to AC.So:
In the figure above, the perimeter of triangle ABD equals x, the perimeter of triangle ADC equals 2x, and AD=16(the base, horizontal, of the larger triangle also the shortest side AD). What is the length of line BC?(the larger portion of line AC that is "cut off" by the point B that was drawn from point D(90 degrees) down perpendicular to AC.)
So apparently the length of AC is 32... How? Is it just that I multiply 16 by 2 since the PERIMETER of big triangle is two times(2x) of the smaller triangle ABD(x)?(Is the perimeter proportionate with the sides of a shape?)
I apologize for the abundance of words, but I really need to figure this out- it irks me not knowing how to do these! Thanks in advance!
Note that the powers of 2 end in digit that cycles:
2,4,8,6,2,4,8,6,...
Since 190 = 47*4 + 2, 2^190 ends in 4
So, the remainder is 4 when divided by 5.
If AC=32, the the hypotenuses AC and AD are in the ratio 2:1. So also are sides
AD and AB. So, AB=12, making BC=24
And yes, perimeters are in the same ratio as the sides, since they are just the sum of the sides. The areas scale as the square of the ratio, since they are two sides multiplied.
Steve, how does 190 = 47*4 + 2? And I still don't get where you see that it ends in 4, so 4 is the remainder.
AND my question was how do we find out that AC=32? That was not given in the question.
Also, I meant how did you get 190 = 47*4 + 2
clearly, 190 = 47*4 + 2
That means 47 whole cycles of digits 2,4,8,6 have been used, leaving 2 more: 2 and 4.
2^6 ends in the same digit as 2^2
and so on, till 2^190 = 2^(47*4) + 2
okay, makes sense. But how does it have anything to do with dividing by 5 and knowing that 4 is the remainder?
1. To find the remainder when 2^190 is divided by 5, you can use the concept of modular arithmetic. Here's how you can approach the question:
- Start by writing down the powers of 2:
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
...
- Notice that the units digit of the powers of 2 repeats in a pattern: 2, 4, 8, 6, 2, 4, 8, 6, ...
- Since we're interested in finding the remainder when dividing by 5, we only need to consider the units digit.
- The units digit of 2^1 is 2, 2^2 is 4, 2^3 is 8, and so on. We can see that there is a pattern: 2, 4, 8, 6, ...
- To find the units digit of 2^190, you can divide 190 by 4 (the length of the pattern). The remainder will tell you which power of 2 to use.
- 190 divided by 4 gives a remainder of 2. So, we need to find the units digit of 2^2.
- The units digit of 2^2 is 4. Therefore, the remainder when 2^190 is divided by 5 is 4.
You can write down the powers of 2 and keep dividing to find the pattern, or you can use a calculator to calculate the powers and observe the pattern. In this case, using a calculator would be more efficient.
2. In the given figure, we have a right triangle ABC, where AC is the hypotenuse, AD is the base, and AB is the height. We also have a line segment BD perpendicular to AC, where point D is the right angle.
- It is given that the perimeter of triangle ABD is x and the perimeter of triangle ADC is 2x.
- The perimeter of a triangle is the sum of the lengths of its sides. So, for triangle ABD, the perimeter is AB + BD + AD, which is x.
- For triangle ADC, the perimeter is AC + CD + AD, which is 2x.
- Since AD is given as 16, we can substitute this value into the equations:
x = AB + BD + 16
2x = AC + CD + 16
- It is also given that AC is twice the length of AD. Since AD is 16, AC is 2 * 16 = 32.
- Using this information, we can further simplify the equations:
x = AB + BD + 16
2x = 32 + CD + 16
- Now, we need to calculate the length of line segment BC. To do that, we can subtract AB from AC (since BC is the "cut off" portion of AC by point B):
BC = AC - AB
= 32 - BD
- However, we still need the value of BD. To find that, we can rearrange the equation x = AB + BD + 16 to solve for BD:
BD = x - AB - 16
- Now we can substitute this expression for BD into the equation for BC:
BC = 32 - (x - AB - 16)
= 48 - x + AB
So, the length of line segment BC is 48 - x + AB, where x is the perimeter of triangle ABD and AB is the length of the height of triangle ABD.
To find the length of line segment BC, you need to know the values of x and AB given in the problem statement.