We know that the number 7 is the prime followed by a cube. Which next number is also a prime followed by a cube ?
There is no number other than 7 that is a prime followed by a cube.
To prove, let us assume that n3-1 is a prime for n.
N2-1 = (n^1)(n2+n+1)
Now, n-1 divides n3-1.
If n-1>1, then we are done as it is a contradiction to n3-1 being a prime.
In the figure given with this question, place the numbers from 1 to 19 in the circles in a manner that each side of the triangle sums up to 17.
If we roll two dices (six sided normal dice) together.
what is the probability that the first one comes up with a 2 and the second one comes up with a 5?
The probability will be 1/36.
For the first dice, there can be six possibilities. Similarly, for the second dice as well, there can be six possibilities.
Thus the total possibilities is 6 * 6 = 36.
The outcome we need is that the first comes up with a 2 and the second comes up with a 5. That is possible only in one possibility.
Therefore the required probability is 1/36.
How high will you have to count before you use the letter A?
One Thousand
Consider the situation that there is a pond where some flowers have grown up and some bees are hovering over the flowers. Now read the following statements carefully:
1. If each of the bees lands on a flower, then one bee does not get a flower.
2. If two bees share each flower, then there is one flower left.
Now, can you calculate the number of flowers in the pong and the number of bees hovering over them?
There are three flowers in the pond and four bees are hovering over it.
If you remove one from eleven, it becomes ten. If you remove one from nine, it becomes ten.
How is this possible ?
It is possible if you are using Roman Numerals.
9 = IX
10 = X
11 = XI
Now it is all possible.
A devotee visits 9 temples when he visits India. All these nine temples have one thing in common - there are 100 steps in every temple. The devotee puts Re.1 coin after climbing up every step. He does the same while climbing down every step. At each temple, the devotee offers half of his money from his pocket to god. In this way, his pocket becomes empty after his visit to 9th temple.
Can you calculate the total amount he had initially ?
146900
In such cases, we calculate from the last thing. Thus in this case, let us start counting from the 9th temple. Since, he must have offered 100 coins while climbing down stairs, it means that he must have offered Rs. 100 to god on the ninth temple and offered Rs. 100 while climbing up steps. Thus it is clear that he had Rs. 300 in his pocket before climbing the steps of the 9th temple.
Similarly, we will calculate for all the temples backwards.
Before eight temple: (300+100)*2 + 100 = 900
Before seventh temple: (900+100)*2 + 100 = 2100
Before Sixth temple: (2100+100)*2 + 100 = 4300
Before fifth temple: (4300+100)*2 + 100 = 8900
Before fourth temple: (8900+100)*2 + 100 = 18100
Before third temple: (18100+100)*2 + 100 = 36,500
Before second temple: (36500+100)*2 + 100 = 73300
Before first temple: (73300+100)*2 + 100 = 146900
Therefore he had Rs. 146900 initially.
Can you find out the number of the parking space in which the car is parked?
The parking number is 87. You will know why if you invert the image and then see.
There was a blind man. He had four socks in his drawer either black or white. He opened it and took out two socks. Now the probability that it was a pair of white socks is 1/2.
Can you find out the probability that he had taken out a pair of black socks ?
This question can be a tricky one. The probability that is asked will be zero.
If the probability off taking out a pair of white socks is 1/2, it means that there is no black pair. This is because if there had been, there would have been three cases – white pair, black pair and mixed pair. But since in the question it is clear that there were just two chances (1/2), we can fairly deduce that there were only one black sock and the rest were white.
Can you find out the remainder when 3^300 is divided by 5?
3
It is obvious that it is not feasible to calculate 3^300 as it will take too much of time. So we will use a trick to solve the question. We will calculate the remainder of each power till we find a pattern.
3^1 divided by 5 leaves the remainder 3.
3^2 divided by 5 leaves the remainder 4.
3^3 divided by 5 leaves the remainder 2.
3^4 divided by 5 leaves the remainder 1.
3^5 divided by 5 leaves the remainder 3.
3^6 divided by 5 leaves the remainder 4.
As you can see that the pattern is now repeating itself and it will go on like this till 3^300 and beyond. Since every fourth remainder is same as the first, we will look for the power of 4 only. 300 is divisible by 4. Therefore at the power of 300, the first remainder will repeat itself and the remainder will be 3.