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.
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!
6 answers
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
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?
0 answers