Find a 9-digit number, which you will gradually round off starting with units, then tenth, hundred etc., until you get to the last numeral, which you do not round off. The rounding alternates (up, down, up ...). After rounding off 8 times, the final number is 500000000. The original number is commensurable by 6 and 7, all the numbers from 1 to 9 are used, and after rounding four times the sum of the not rounded numerals equals 24.

Can you make the number 24 by utilizing the numbers 1, 3, 4 and 6? You must use one number only one time and you can use mathematical operation symbols anytime anywhere.

Find the number of squares in the below-given picture.

Can you solve below mathematical equation?

2^1234 - 2^1233

A and B have a certain number of chocolates with them. If B gives one chocolate to A, they will have an equal number of chocolates. But if A gives one chocolate to B, then A will be left with half the number of chocolates that B has.

Can you find out the number of chocolates they have right now?

John's five friends Rachel, Jacob, Phoebe, Joey and Chandler live on the same road.

Rachel lives in the house: A, Jacob lives in the house: B, Phoebe lives in the house: C, Joey lives in the house:D, Chandler lives in the house: E

Mathematically
B*C*D = 1260
B + C + D = 2E
2A = E

The road numbers run from 2 to 222.

Can you tell me the house number of each of John's friends?

A man had five children. He had $100 with him to give to his children. He decided to start with the youngest child and then give $2 more than each younger child to his next elder child.

For example, if he gives $x to the youngest child, he will give $(x+2) to the next one, $[(x+2) + 2] to the next one and so on.

Can you find out how much did the youngest one receive?

John needs to purchase 100 chocolates from three different shops and he has exactly 100 rupees to do that which he must spend entirely. He must buy at least 1 Chocolate from each shop.

The first shop is selling each chocolate at 5 paise, the second is selling at 1 rupee and the third is selling at 5 rupees.

How many chocolates should he buy from each shop?

By using all numbers, i.e. 123456789 and subtraction/addition, operators number 100 can be formed in many ways.
Example: 98 + 7 + 6 - 5 - 4 - 3 + 2 - 1 = 100

But if we add a condition use of the number 32 is a must. Then there are limited solutions.
One of such solution is: 9 - 8 + 76 + 54 - 32 + 1 = 100

Can you tell me any other solution?

Can you write down eight eights so that they add up to one thousand?