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hram777 [196]
2 years ago
11

The difference btwn increasing a number by 20% and decreasing the same number by15% is 14 what is the number?​

Mathematics
1 answer:
mote1985 [20]2 years ago
4 0

Answer:

The number is:

  • n = 40

Step-by-step explanation:

Let 'n' be the number

  • Increasing a number by 20% = (n + n×20/100)
  • Decreasing a number by 14% = (n - n×15/100)

As the difference between increasing a number by 20% and decreasing the same number by15% is 14.

Thus, the expression becomes

\left(n\:+\:\frac{n}{5}\right)\:-\:\left(n\:-\:\frac{3n}{20}\right)\:=\:14

n+\frac{n}{5}-n+\frac{3n}{20}=14

\frac{7n}{20}=14

Multiply both sides by 20

\frac{7n}{20}\cdot \:20=14\cdot \:20

7n=280

Divide both sides by 7

\frac{7n}{7}=\frac{280}{7}

n=40

Thus the number is:

  • n = 40
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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

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###

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}

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