The number of terms is:
(200 - 1): 1 + 1 = 200 (number)

We have:
From 1 to 9 have a two digit number
From 10 to 100 has 19 digits two
From 101 to 200 there are 20 digits two
So 1 to 200 are: 1 + 19 + 20 = 40 digits 2
=> Yes: 200 - 40 = 160 with no 2 digits

Answer: 160

số kẹo còn lại trong hộp là:100-4=96(viên)

số bạn bè có là:96:24=4(bạn)

ĐS:4 bạn bè

123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14 123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14 123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14123+133+...+1n3<14123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)123+133+...+1n3<11.2.3+12.3.4+...+1(n−1)n(n+2)=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(21.2.3+22.3.4+...+2(n−1)n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(11.2−12.3+12.3−13.4+...+1(n−1)n−1n(n+2))=12(12−1n(n+2))=12(12−1n(n+2)) =14.12n(n+2)=14.12n(n+2)⇒123+133+...+1n3<14−12n(n+2)<14⇒123+133+...+1n3<14−12n(n+2)<14 ⇒123+133+...+1n3<14⇒123+133+...+1n3<14 So 123+133+...+1n3<14

Có:2=12+12

5=22+12 

29=52+22

............................

Tương tự: số thứ 6=292+52=866

=>Số thứ 7=8662+292=750797

The number of pencils Mary has is:

       9 x 5 = 45(pencils)

The number of pencils John and Mary have is:

       45 + 16 = 61(pencils)

                     Answer: 61 pencils

Given a sum :  2 + 4 + 6 + ... + 2k

The number of terms is : 2k−22+1=k2k−22+1=k ( terms )

So, the result of this sum is : k.(2k+2)2=k.(k+1)k.(2k+2)2=k.(k+1)

Example : 2 + 4 + 6 = 2 + 4 + 2.3 ; so k = 3; then 2 + 4 +6 = k.(k+1)=3.4 .

ab+¯¯¯¯¯ba=10a+b+10b+a=11(a+b)=n2ab¯+ba¯=10a+b+10b+a=11(a+b)=n2

⇒⇒Put a+b=11.k2a+b=11.k2

We have 0<a+b≤180<a+b≤18

⇒a+b=11=2+9=3+8=4+7=5+6⇒a+b=11=2+9=3+8=4+7=5+6

So, (a;b)∈{(2;9);(9;2);(3;8);(8;3);(7;4);(4;7);(5;6);(6;5)}

(a) Assume that the square number 16 = 42 is equal ton3n3,so n = 48

(b) Assume that the cubic number 27 = 33 is equal ton5n5,so n = 135

 1 Selected by MathYou

5n8 is divisible by 9 => (5 + n + 8) ⋮⋮ 9 means (13 + n) ⋮⋮ 9.

=> n = 5.

=> 3m9 = 558 - 189 = 369 => m = 6.

So, m + n = 5 + 6 = 11.

a) There are : 7 x 7 x 6 x 5 = 1470 numbers

b) There are : 4 x 3 x 2 x 1 = 24 old numbers 

Nhiều người trả lời vãi chật cả chỗ

Let a,b,c be the number of the correctly answered questions , unanswered questions , incorrectly answered questions of a participant respectively

Then the participant's total score is 5a + b - c with a + b + c = 30

If a is odd,then 5a and b + c are odd,so b - c is odd and 5a + b - c is even

If a is even,then 5a and b + c are even,so b - c is even and 5a + b - c is even

From 2 cases,we know that the total score of all participants is always an even number

Because the lowest common multiple of a and b is 50 . So all possible values of a and b is :

(a;b) = {1 ; 2 ; 5 ; 10 ; 25 ; 50}

20 = 2^2 x 5 

20 = 4.5 

The smallest number such that the ptoduct of all it is digit is 20 is : 45 

Good ! ^^

b3 = 3375 = 153 => b = 15 => a = 13 ; c = 17 => ac = 221

b3 = 3375=> b = 15 => a = 13 ; c = 17 => ac = 221

There are 162 numbers from 1 to 200 do not have the digit 2

 1

Tom : 20 candies

Tony : 80 candies

The sum of the equal parts is:
3 + 8 = 11 (part)
Baby number is:
198: 11 x 3 = 54
Large numbers are:
198 - 54 = 144
Answer: SB: 54
                 SL: 144

Oh,this Problem is very difficult