My 6-digit passcode for internet banking is made up of: my house number; followed by half my house number, minus one; followed by
double my house number, plus one. All six digits are different.
The first number of my passcode is 2. What is my full passcode?

The correct answer is 281357

Explanation:

To find the passcode, let's first determine the house number because based on this all the digits of the passcode can be found. Firstly, we know the house number includes two digits and the first digit is 2, this means possible answers to the house number include 20, 21, 22, 23, 24, 25, 26, 27, 28, and 29.

Moreover, the odd numbers (21, 23, 25, 27, 29) can be eliminated because if we divide this by 2 as required to find the second two digits, this would lead to a decimal number, and only whole numbers are allowed in passcodes.

This leaves 20, 22, 24, 26, and 28 as the possible answers. Now, using each of these, let's find the four following digits in each case and determine if there are repeating numbers as the six digits should be different. Also, remember the first number is the house number, then the house number divided by 2 minus 1, and then the house number multiplied by 2 plus one.

20:

20  (first number)    20 ÷ 2 = 10 -1 = 9 (second number)     20 x 2= 40 + 1 =41 (third number)

This is not the correct answer because the second number is only one digit (9) and if it is written with 0 (09) the number 0 repeats

22:

This answer can be immediately eliminated because there is a repeated number (2)

24:

24 (first number)    24 ÷ 2 = 12 -1 = 11 (second number)     24 x 2 = 48 + 1 =49 (third number)

The number 1 repeats, and therefore this is not a valid answer

26:

26 (first number)    26 ÷ 2= 13 - 1 = 12 (second number)    26 x 2 = 52 + 1 = 53 (third number)

There is a repetition of the number 2, this makes this answer invalid

28:

28 (first number)    28 ÷ 2 = 14 - 1 = 13 (second number)  28 x 2 = 56 + 1 = 57 (third number)

In this code (281357) there is no repetition and all requirements are met.

ek = (m·v2) / 2

v - velocity of system after the collision

m = m1 + m2

ek = (m1 + m2) ·v2/2

ek = (200 + 250) ·2,32/2 = 450·5,29/2 = 1190,25 j

haye waan ku helaay.

explanation: