We have :

5x1=5, 5x5=25, 5x5x5=25x5=125, 5x5x5x5=25x5x5=125x5=625

So the missing number is 625

you have a wrong question

it's divisible when it's remander 0.

      If group 1 has 108 : 4 = 27 (students), 27 is not half of the number of students in group 2.

=> If group 1 has 28 students, group 2 has : 28 : 2 = 14 ( students )

=> Group 3 has : 28 + 2 = 30 (students)

=> Group 4 has : 28 - 2 = 26 ( students )

Try : 28 + 14 + 30 + 26 = 42 + 30 + 26 = 98 (impossible)

=> Group 1 has : 30 students

=> Group 2 has : 30 :2 = 15 ( students)

=>30+15 = 45

=> Group 3 has: 30 +2 = 32 ( students)

=>45+32=77

=> Group 4 has: 30 - 2 = 28 ( students)

=>77+28=105 (impossible)

=> Group 1 has 32 students

=> Group 2 has 32:2=16(students)

=> Group 3 has : 32+ 2 = 34(students)

=> Group 4 has : 32 -2= 30 (students)

Try:32+16+34+30=48+34+30=82+30=112 (impossible)

=> Wrong question

hey Ton That Khac Trinh, you are wrong . 90  cannot divisable 28.

List the number have exactly 1 number 8 from 100 to 400 : 108, 118, 128, 138, 148, 158, 168, 178, 180, 181, 182, 183, 184, 185, 186, 187, 189, 198, 208, 218, 228, 238, 248, 258, 268, 278, 280, 281, 282, 283, 284, 285, 286, 287, 289, 298, 308, 318, 328, 338, 348, 358, 368, 378, 380, 381, 382, 383, 384, 385, 386, 387, 389, 398 ( 188, 288, 388 is elminate because have 2 8s)

So there are 54 number have exactly 1 number 8 from 100 to 400

\(\sqrt{1+2+3+...+n}\)=\(\sqrt{6+...+n}\)

Because n is the smallest and n > 1 so n = 2

\(\Rightarrow\sqrt{6+2}\)=\(\sqrt{8}\)=\(\sqrt[\sqrt{2}]{2}\) (elminate)

=>n=3

=>\(\sqrt{6+3}\)=\(\sqrt{9}=3\)(possible)

So n=3

Error : Wrong Question

that be answered !

oh sorry i mistaped. that will be 2025-2015 not 2025+2015. sorry about that. :( 

If n and m are smallest number

=>2015 + n = 2025       2015 - m = 1936

                 n = 2025 + 2015      m = 2015 - 1936

                 n =        10               m =         79

=>n + m = 10 + 79 = 89

@3@=6x3x2x1=36

(1+2+3)+(2+3+4)+(3+4+5)+...+(38+39+40)+(39+40+41)

=6+9+12+...+117+120

=3x(2+3+...+39+40)

=3x819

=2457

Because 72 and 36 \(⋮\)2 and 4 so we have 2 cases :

Case 1: 72 and 36 \(⋮4\)

We have 36:4=9 

->72:9=\(8\) =>\(\dfrac{1}{8}\) (elminate)

Case 2: 72 and 36\(⋮2\)

We have :36:2=18

->72:18=4=>\(\dfrac{1}{4}\) (possible)

=>Cos 72⁰ x Cos 36⁰ =\(\dfrac{1}{4}\)

error:wrong question

We have:

1+1=2

2+1=3

3+2=5

5+2=7

7+3=10

10+3=13

13+4=17

17+4=21

21+5=26

...

So the 25th number is:26+5+6+6+7+7+8+8+9+9+10+10+11+11+12+12=157

Answer: 157

a, b, c, d \(\ne\)0

a,b,c,d=1

 \(\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)

\(=\dfrac{1+1+1}{1}+\dfrac{1+1+1}{1}+\dfrac{1+1+1}{1}=3\)

\(\Rightarrow A=\dfrac{1}{1+1+1}+\dfrac{1}{1+1+1}+\dfrac{1}{1+1+1}+\dfrac{1}{1+1+1}=\dfrac{1}{3}\times4=\dfrac{4}{3}\)