There are: 100 \(\times\) 1% = 1 (litre of salt)

After evaporating, 1 litre of salt become 5% of solution, which means that the solution now contains 1 \(\div\) 5% = 20 litres 

Hence, there is: 100 - 20 = 80 litres of water which has been evaporated

Duncan has subtracted: 14 x 6 = 84 \(\Rightarrow\) Duncan's starting number: 25 + 84 = 109

Taz has added: 14 x 8 = 112 \(\Rightarrow\) Taz's starting number: 25 - 112 = -87

Hence, the sum of Duncan's and Taz's starting number: 109 + (-87) = 22

(Or if you are having a calculator in your hand, you can add these 24 numbers without thinking)

There are 24 numbers of them in total

\(\Rightarrow\) Each number 1, 2, 3, 4 exists 6 times in each colunm (thousands, hundreds, tens, and units)

\(\Rightarrow\) The sum of the digits in each column is: (1 + 2 + 3 + 4) x 6 = 60

\(\Rightarrow\) The sum of all of the four-digit numbers whose digits are permutations of 1, 2, 3 and 4 is:

 60 x 1000 + 60 x 100 + 60 x 10 + 60 x 1 = 66660

66 mi/h = 1.1 mi/h

\(\Rightarrow\) n_min = 10 

The largest three-digit base 5 number is: 444₅ =  \(4\cdot5^2+4\cdot5^1+4\cdot5^0\) = 124₁₀

The smallest four-digit base 4 number is: 1000₄ = \(1\cdot4^3+0\cdot4^2+0\cdot4^1+0\cdot4^0\) = 64₁₀

_{\(\Rightarrow\)} the absolute difference, expressed in base 10, between the largest three-digit base 5 number and the smallest four-digit base 4 number is: 124 - 64 = 60

b = -c

\(\Rightarrow\) bc = (-c)c = -c² = -9 = -3²

^{\(\Rightarrow\)} c = \(\pm\)3

Case 1: c = 3 \(\Rightarrow\) b = -3

\(\Rightarrow\) ab = a(-3) = 1

\(\Rightarrow\) a = \(\dfrac{-1}{3}\) (unsatisfy c < -a)

Case 2: c = -3 \(\Rightarrow\) b = 3

 ab = 3a = 1

 a =  \(\dfrac{1}{3}\) (satisfy c < -a)

Hence, we have a + b = \(\dfrac{1}{3}\) + 3 = \(3\dfrac{1}{3}\)

HAVE YOU EVER TRIED: 2, 3, 5, 6?

But there is only one value for a² + b² + c² + d², can you find that?

12 = 5 + 5 + \(\dfrac{6}{3}\)

12 = \(6\cdot5-6\cdot3\)

12 = \(3\cdot5-3\)

12 = \(6\cdot3-6\)

...

3a² = 2a³

=> \(\dfrac{a^3}{a^2}\) = \(\dfrac{3}{2}\) => a = \(\dfrac{3}{2}\)

(1 + 2 + 3 + ... + 99) + 10

= 4950 + 10 = 4960

We have:

49 = 7 . 7 ( from the number 77)

36 = 4 . 9 (from the number 49)

18 = 3 . 6 (from the number 36)

SO THE NEXT NUMBER IS: 1 . 8 = 8

We have:

2 = 1 + 1

4 = (1 + 2) + 1

7 = (1 + 2 + 3) + 1

11 = (1 + 2 + 3 + 4) + 1

16 = (1 + 2 + 3 + 4 + 5) + 1

SO THE NEXT NUMBER IS: (1 + 2 + 3 + 4 + 5 + 6) + 1 = 22

P/s: Your question is incorrect

It must be: "FIND THE MISSING NUMBER" or "FIND THE NUBER THAT IS MISSING", not number missing

This is easy. Each number is formed by reading out the digits, grouping digits together and saying the number of identical digits followed by the digit.

1 is read as "one one" (one number 1) -> 11

11 is read as "two ones" (two number 1) -> 21

21 is read as "one two, one one" (one number 2 then one number 1) -> 1211

1211 is read as "one one, one two, two ones" (one number 1, then one number 2, then two number 1 -> 111221.

SO THE NEXT NUMBER IS: "three one, two two, one one" -> 312211

Hey don't cheat man! You should do your homework YOURSELF, not DEPENDENTLY

\(\dfrac{6}{3}\times\dfrac{8}{4}=2\times2=4\)

\(\dfrac{3}{8}\times\dfrac{5}{6}=\dfrac{15}{48}=\dfrac{5}{16}\)

\(\dfrac{108}{12}\times\dfrac{8}{81}=9\times\dfrac{8}{81}=\dfrac{8}{9}\)

\(123\times\dfrac{133}{688}=\dfrac{16,359}{688}\)

\(\dfrac{4}{10}=\dfrac{40}{100}=40\%\)

\(\dfrac{23}{100}=23\%\)

\(\dfrac{254}{1000}=\dfrac{25.4}{100}=25.4\%\)

\(\dfrac{7890}{10000}=\dfrac{78.9}{100}=78.9\%\)

Find the sum of the digits?

Hey! Correct explanation but you calculate incorrectly. Try again!