Brain teaser hard maths puzzles with answers

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In these Maths Puzzles Brain Teasers, four sets of numbers are given. These numbers are related to each other Mathematical and Logical by a sequence or pattern. Your challenge i… More

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2+3=8,
3+7=27,
4+5=32,
5+8=60,
6+7=72,
7+8=??

Solve it?

98

2+3=2*[3+(2-1)]=8
3+7=3*[7+(3-1)]=27
4+5=4*[5+(4-1)]=32
5+8=5*[8+(5-1)]=60
6+7=6*[7+(6-1)]=72
therefore
7+8=7*[8+(7-1)]=98
x+y=x[y+(x-1)]=x^2+xy-x

You can place weights on both side of weighing balance and you need to measure all weights between 1 and 1000. For example if you have weights 1 and 3,now you can measure 1,3 and 4 like earlier case, and also you can measure 2,by placing 3 on one side and 1 on the side which contain the substance to be weighed. So question again is how many minimum weights and of what denominations you need to measure all weights from 1kg to 1000kg.

For this answer is 3^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729.

A high school has a strange principal. On the first day, he has his students perform an odd opening day ceremony:

There are one thousand lockers and one thousand students in the school. The principal asks the first student to go to every locker and open it. Then he has the second student go to every second locker and close it. The third goes to every third locker and, if it is closed, he opens it, and if it is open, he closes it. The fourth student does this to every fourth locker, and so on. After the process is completed with the thousandth student, how many lockers are open?

The only lockers that remain open are perfect squares (1, 4, 9, 16, etc) because they are the only numbers divisible by an odd number of whole numbers; every factor other than the number's square root is paired up with another. Thus, these lockers will be 'changed' an odd number of times, which means they will be left open. All the other numbers are divisible by an even number of factors and will consequently end up closed.

So the number of open lockers is the number of perfect squares less than or equal to one thousand. These numbers are one squared, two squared, three squared, four squared, and so on, up to thirty one squared. (Thirty two squared is greater than one thousand, and therefore out of range.) So the answer is thirty one.

How can I get the answer 24 by only using the numbers 8,8,3,3.

You can use the main signs add, subtract multiply and divide.

1) 8/(3-(8/3))
= 8/(1/3)
= 24

2)((8 x 3!)/3)+8
= ((8 × 3 × 2 × 1)/3)+8
= (48/3)+8
= (16)+8
= 24

3)(3!/)*8

4)(8-3)!/(8-3)
( × )!
( + )!
√(8×8×3×3)
8+(8×(3!/3))
((√(8+8) × (3/3))!
√(8+8) × (3+3)
(log base(3!/3) of 8) × 8
((log base(3!/3) of (8+8))!

I am thinking of a 6-digit number. The sum of the digits is 43.

And only two of the following three statements about the number are true:

(1) it's a square number,
(2) it's a cube number, and
(3) the number is under 500000.

Only Statements 1 and 3 are true. the number is 499849.

7072 = 499849
499849<500000

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