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

b³ = 3375 

=> b = \(\sqrt[3]{3375}=15\)

Because a,b,c are three consecutive odd number  and b = 15 so

a = 13 ; b = 15 ; c = 17

ac = 15 . 17 = 221 

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

 1

1 is odd,so the 3rd number is even

=> The 4th number is odd and so is the 5th number

=> The 6th number is even 

=> The 7th number is odd and so is the 8th number

Hence,the 3rd number and every third number after that are even.

So the numerical orders of the even numbers are divisible by 3

In the first 1000 numbers,there are :

(999 - 3) : 3 + 1 = 333 (even numbers) and :

1000 - 333 = 667 (odd numbers)

1 is odd,so the 3^rd number is even

=> The 4^th number is odd and so is the 5^th number

=> The 6^th number is even 

=> The 7^th number is odd and so is the 8^th number

Hence,the 3^rd number and every third number after that are even.

So the numerical orders of the even numbers are divisible by 3

In the first 1000 numbers,there are :

(999 - 3) : 3 + 1 = 333 (even numbers) and :

1000 - 333 = 667 (odd numbers)

The odd number between the six remaining numbers is:

1337 : 7 = 191

So 7 consecutive numbers are 185 ; 187 ; 189 ; 191 ; 193 ; 195 ; 197

The odd number between the six remaining numbers is:

1337 : 7 = 191

So 7 consecutive numbers are 185 ; 187 ; 189 ; 191 ; 193 ; 195 ; 197

The odd number between the six remaining numbers is:

1337 : 7 = 191

So 7 consecutive numbers are 185 ; 187 ; 189 ; 191 ; 193 ; 195 ; 197

We have :

22009 is even number 

32009 = 34.502 . 3 = .....1 . 3 = ......3 (odd number)

72009 = 74.502 . 7 = .....1 . 7 = .....7(odd number)

 = 94.502 . 9 = ......1 . 9 = .......9(odd number)

So : 22009 + 32009 + 72009 + 92009 =  .....2 + .....3 + ....7 + .....9 = ......21 (odd number)

We have :

2²⁰⁰⁹ is even number 

3²⁰⁰⁹ = 3^4.502 . 3 = .....1 . 3 = ......3 (odd number)

7²⁰⁰⁹ = 7^4.502 . 7 = .....1 . 7 = .....7(odd number)

 = 9^4.502 . 9 = ......1 . 9 = .......9(odd number)

So : 2²⁰⁰⁹ + 3²⁰⁰⁹ + 7²⁰⁰⁹ + 9²⁰⁰⁹ =  .....2 + .....3 + ....7 + .....9 = ......21 (odd number)

22009 is even number 

32009 = 34.502 . 3 = .....1 . 3 = ......3 (odd number)

72009 = 74.502 . 7 = .....1 . 7 = .....7(odd number)

 = 94.502 . 9 = ......1 . 9 = .......9(odd number)

So : 22009 + 32009 + 72009 + 92009 =  .....2 + .....3 + ....7 + .....9 = ......21 (odd number)

We have : abcabc = abc x 1001 = abc x 7 x 11 x 13

Because abc is a prime number,the number of divisors of abcabc is :

(1 + 1)4 = 16

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

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

It's 23,54

it's 23,54

Let n is number of pages. I is the page which was removed.

1 + ... + n - i = 3030

3030 + i = ( n +1 ) * n : 2

Let n is number of pages. I is the page which was removed.

1 + ... + n - i = 3030

3030 + i = ( n +1 ) * n : 2 

Let n is number of pages. i is the page which was  removed.

1+...+n -i =3030

3030+i=(n+1) * n :2 

If a is odd,then a³ and a² are also odd,so a³ + a² + 1 is odd

If a is even,then a³ and a² are also even,so a³ + a² + 1 is odd

So a³ + a² + 1 is always odd with any integer a

If a is odd,then a3 and a2 are also odd,so a3 + a2 + 1 is odd

If a is even,then a3 and a2 are also even,so a3 + a2 + 1 is odd

So a3 + a2 + 1 is always odd with any integer a

If a is odd,then a3 and a2 are also odd,so a3 + a2 + 1 is odd

If a is even,then a3 and a2 are also even,so a3 + a2 + 1 is odd

So a3 + a2 + 1 is always odd with any integer a

Put \(\left(2k-1\right)\left(2k+1\right)=123476543\)

\(\Leftrightarrow4k^2-1-123476543=0\)

\(\Leftrightarrow4k^2-123476544=0\)

\(\Leftrightarrow4\left(k^2-30869136\right)=0\)

\(\Leftrightarrow k^2=30869136=5556^2\)

\(\Leftrightarrow k=5556\)

\(\Rightarrow\left\{{}\begin{matrix}2k-1=11111\\2k+1=11113\end{matrix}\right.\)

Ans : 11111 & 11113.

Put (2k−1)(2k+1)=123476543

⇔4k2−1−123476543=0

⇔4k2−123476544=0

⇔4(k2−30869136)=0

⇔k2=30869136=55562

⇔k=5556

⇒{2k−1=111112k+1=11113

Ans : 11111 & 11113.

123476543 = 11111 x 11113

a + b + c is odd ; 2b is even,so a + b + c - 2b is odd

=> a - b + c is odd => (a + b + c)(a - b + c) is odd

a + b + c is odd ; 2b is even,so a + b + c - 2b is odd

=> a - b + c is odd => (a + b + c)(a - b + c) is odd