Cause \(2\cdot\left(3+4\right)>2\cdot3+4\)

So the positive difference between the value of 2 × (3 + 4) and the value of 2 × 3 + 4 is :

\(2\cdot\left(3+4\right)-\left(2\cdot3+4\right)=14-10=4\)

Answer : 4.

The whole angles of the square is \(360^0\)

The pizza is cut into 12 congruent slices => Each slice has the angle of \(\dfrac{360}{12}=30^0\)

So the sum of the central angles of the slices that were not eaten after ate 2 slices is :

\(360-30\cdot2=360-60=300^0\)

Answer : \(300^0\)

Call  the width is x.

We have \(78=2x+6\)

So \(x=36\)  <=> the width is 36 ft.

Answer : 36 ft.

We have : The ratio : Charity A : Charity B = 2 : 3.

The money Sam contributes to Charity B is : \(\dfrac{2500}{2+3}\cdot3=1500\)$.

Answer : 1500$.

A day the man floats \(3-2=1\) (mi north)

So until 47 days the man can go 47 mis north ( Cause 1 mi / day).

In the last day (day 48) the man floats 3 mi north so the distance he went is 47 + 3 = 50 (mi north).

So it will take him 48 days to reach the location.

Let's take the frist 20 positive intergers 1 to 20.

The number of odd prime numbers : 3;5;7;11;13;17;19.

The probability is : \(\dfrac{7}{20}\cdot100\%=35\%\)

\(5\cdot\left(11+4:4\right)\)

\(=5\cdot\left(11+1\right)\)

\(=5\cdot12=60\)

Cut in half so 72/2=36 is the diameters of two circles.

Sum of the areas : \(2S=\dfrac{d^2}{4}\cdot3,14\cdot2=2034,72\left(cm^2\right)\)

Answer : 2034,72 \(cm^2\).

We have \(\overline{a0b}⋮9\Rightarrow a+0+b\in\left\{9;18;27;36;...\right\}\)

But \(a+0+b\le18\)

So \(a+b=\left\{9;18\right\}\)

We have pair (a;b) = (1;8) , (8;1) , (2;7) , (7;2) , (6;3) , (3;6) , (5;4) , (4;5) , (9;0).

But there is only the pair : (4;5) match the condition \(\overline{a0b}=\overline{ab}\cdot9\).

So the answer of the two digit number \(\overline{ab}\) is 45.

We have \(7^{2017}=\left(7^4\right)^{504}\cdot7\)

We have \(7^4\equiv1\left(mod12\right)\)

\(\Rightarrow\left(7^4\right)^{504}\equiv7^{2016}\equiv1\left(mod12\right)\)

\(\Rightarrow7^{2017}\equiv1\cdot7\equiv7\left(mod12\right)\)

So the remainder is 7.

Call : \(\dfrac{\left(n+1\right)^2}{n+23}=Q_{\left(n\right)}+\dfrac{R}{n+23}\)

n + 23 = 0 => n = -23.

\(R=\left(-23+1\right)^2=\left(-22\right)^2=484\)

\(\Rightarrow\dfrac{\left(n+1\right)^2}{n+23}=Q_{\left(n\right)}+\dfrac{484}{n+3}\in Z\)

\(\Rightarrow n+23\inƯ_{\left(484\right)}=\left\{\pm1;\pm2;\pm4;\pm11;\pm22;\mp44;\pm121;\pm242;\pm484\right\}\)

\(\Rightarrow n+23_{MAX}=484\Rightarrow n=484-23=461\)

Answer : 461.

Hey guy (LeQuocTranAnh), the question asked the sum of the digits not numbers.

I think you've wrong in reading thread :<

The sum of the digits 0-9 : 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9, which equals 45.

Next, the sum of the digits of the numbers 10-19 : \(\left(10\cdot1\right)+\left(0+1+...+9\right)\)

Then, the sum of the digits of the numbers 20-29 : \(\left(10\cdot2\right)+\left(0+1+2+...+9\right)\)

So we have this expression :

(0 + 1 +...+ 9) - The numbers 0-9

\(10\cdot1+\left(0+1+...+9\right)\)- The numbers 10-19

\(10\cdot2+\left(0+1+...+9\right)\)- The numbers 20-29

\(10\cdot3+\left(0+1+...+9\right)\)- The numbers 30-39

\(10\cdot4+\left(0+1+...+9\right)\) -The numbers 40-49

\(10\cdot5+\left(0+1+...+9\right)\) -The numbers 50-59

\(10\cdot6+\left(0+1+...+9\right)\) -The numbers 60-69

\(10\cdot7+\left(0+1+...+9\right)\) -The numbers 70-79

\(10\cdot8+\left(0+1+...+9\right)\) -The numbers 80-89

\(10\cdot9+\left(0+1+...+9\right)\)  - The numbers 90-99.

Calculate all of these expressions we have the sum of the digits from 1-99 is 900.

So the sum of the digits from 1 - 100 is 900 + 0 + 0 + 1 = 901.

Answer : 901.

The sum of 1 + 7 + 13 + 19 +...+253 + 259 :

\(S=\left(259+1\right)\cdot\dfrac{\left(\dfrac{259-1}{6}+1\right)}{2}=5720\)

So the remainder is : \(\dfrac{5720}{6}=Q=953,R=2\)

Answer : 2.

\(P=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)

Condition : \(x\ge1\)

a) We have \(x+2\sqrt{x-1}=\left(x-1\right)+2+1\sqrt{x-1}+1=\left(\sqrt{x-1}+1\right)^2\)

So P = \(\sqrt{\left(1+\sqrt{x-1}\right)^2}+\sqrt{\left(1-\sqrt{x-1}\right)^2}\)

\(\Rightarrow\left|1+\sqrt{x-1}\right|+\left|1-\sqrt{x-1}\right|\)

+ With \(1-\sqrt{x-1}\ge0\Rightarrow x\le2:\)

\(P=1+\sqrt{x-1}+1-\sqrt{x-1}=2\)

+ With \(1-\sqrt{x-1}< 0\Rightarrow x>2:\)

\(P=1+\sqrt{x-1}+\sqrt{x-1}-1=2\sqrt{x-1}\)

Sum up : \(\left[{}\begin{matrix}1\le x\le2\Rightarrow P=2\\x>2\Rightarrow P=2\sqrt{x-1}\end{matrix}\right.\)

b) We have inequality \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)

So : \(P=\left|1+\sqrt{x-1}\right|+\left|1-\sqrt{x-1}\right|\ge\left|\sqrt{\left(1-\sqrt{x-1}\right)+\left(1-\sqrt{x-1}\right)}\right|\)

\(\Rightarrow P\ge2\Leftrightarrow1\le x\le2\)

So Pmin = 2 when \(1\le x\le2\).

Factoring : \(2^{58}+1=\left(2^{29}+2^{15}+1\right)\cdot\left(2^{29}-2^{15}+1\right)\)

Because : \(\left\{{}\begin{matrix}2^{29}+2^{15}+1=536903681\\2^{29}-2^{15}+1=536838145=5\cdot107367629\end{matrix}\right.\)

So \(2^{58}+1=\left(2^{29}+2^{15}+1\right)\cdot\left(2^{29}-2^{15}+1\right)=536903681\cdot107367629\cdot5\)

\(\rightarrow\dfrac{2^{58}+1}{5}=\dfrac{536903681\cdot107367629\cdot5}{5}=536903681\cdot107367629\)

So the number \(\dfrac{2^{58}+1}{5}\) is composite.

How about this ?

\(A=1+9+9^2+...+9^{50}\)

\(\Rightarrow9A=9+9^2+9^3+...+9^{50}+9^{51}\)

\(\rightarrow9A-A=8A=\left(9+9^2+9^3+...+9^{50}+9^{51}\right)-\left(1+9+9^2+...+9^{50}\right)\)

\(\Rightarrow8A=9^{51}-1\Rightarrow A=\dfrac{9^{51}-1}{8}\)

We find that \(9^{2k},9^{2k+1},...\)always have the last digit is 1 or 9, or we can simply understand that whether 9 has index number is odd => the last digit is 9, and whether 9 has index number is even => the last number is 1.

\(9^{51}\) has the last digit is 9 (Cause 51 is odd number)

So : \(A=\dfrac{\overline{B9}-1}{8}=\dfrac{\overline{B8}}{8}=\overline{C1}\)

Therefore, the last digit of \(1+9+...+9^{50}\) is 1.

Answer : 1.

Also, we use Venn to this.

Pretend there are 100 americans.

Call x is the number of Americans have type a and type b in their blood

Similarly, we have : \(43+15-x+46=100\Rightarrow x=4\)

So the percent is \(\dfrac{4}{100}\cdot100\%=4\%\).

Answer : 4%.

Call the price is n.

We have : In the first year : The price is : \(n\cdot150\%\)

In the second year : The price is \(\dfrac{n\cdot150\%}{50\%}=\dfrac{3}{2}n\cdot\dfrac{1}{2}=\dfrac{3}{4}n=0,75n=n\cdot75\%\)

So 75% is the correct answer.

Using Venn to solve it.

Pretend there are 100 mathletes.

So there will be 80 own computers and 40 in band and 10 people are not in band and not own computers.

Call the number of all mathletes both own computers and are in band is n.

We have \(80+40-n+10=100\) \(\Rightarrow n=30\)

So the percent : \(\dfrac{30}{100}\cdot100\%=30\%\)

Answer : 30%