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2 years ago

Round of 389,422 in 1,000

1 answer:

4 0

So,

To round the nearest thousand, look at the number in the hundredths place.
If n ≥ 5, round up
If n < 5, round down

4 < 5, therefore round down.

389,422 --> 389,000

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Determine the rate of change of the following coordinates

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2 years ago

What is 35 percent of 135

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Which power has a greater value? -3^2 or 2^3

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3 years ago

Angles X and Y are supplementary. Angle X is 3 times the measure of angle Y. What is the measure of angle X?

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2 years ago

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Determine n between 0 and 19 such that (2311)(3912) ≡ n mod 20.

sleet_krkn [62]

You can write 2311 and 3912 in the form $20q+r$ :

$2311=115\cdot20+11$

$3912=125\cdot20+12$

Then

$2311\cdot3912=(115\cdot20+11)(125\cdot20+12)$

$2311\cdot3912=115\cdot125\cdot20^2+(11\cdot125+12\cdot115)\cdot20+11\cdot12$

Taken modulo 20, the terms containing powers of 20 vanish and you're left with

$2311\cdot3912\equiv11\cdot12\equiv132\pmod{20}$

We further have

$132=6\cdot20+12$

so we end up with

$2311\cdot3912\equiv12\pmod{20}$

and so $n=12$ .

###

If instead you're trying to find $2311^{3912}\pmod{20}$ , you can apply Euler's theorem. We can show that $\mathrm{gcd}(2311,20)=1$ using the Euclidean algorithm. Then since $\varphi(20)=8$ , and 8 divides 3912, we have

$2311^{3912}\equiv2311^{489\cdot8}\equiv(2311^{489})^8\equiv1\pmod{20}$

To show 2311 and 20 are coprime:

2311 = 115*20 + 11

20 = 1*11 + 9

11 = 1*9 + 2

9 = 4*2 + 1 => gcd(2311, 20) = 1

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3 years ago