Hard math puzzles, only attempt if you dare!

1. Alright, let's give this a shot. First off, there are 4 prime numbers between 1 and 10. 

We have \(2,3,5,7\)

However, since the product of these two prime numbers must be less than 10 (single digit number), the only two numbers that qualify is \(2 \cdot 3 = 6\)

So now, we have two different numbers to work with. \(263, 362\)

Since their the reverses, take the larger number, 362 is the original. 

We find that \(362-263=99\), meaning we found our number. 

Adding the digits up, we get

\(2+6+3 = 11\)

So 11 is our answer. 

Thanks! :)

2. Note that if Harry Puttar (nice name LOL) refills 2 litres and drinks 3 litres, he will drink 1 litre of milk every single day. 

Thus, he should run out of milk by the 40th day. 

The number of litres he drank can be calculated with the sequence

\(1+2+3+4+5+6+...+39+40 = 820\)

Of these total 820 litres, there WAS 40 litres of milk,  meaning he drank

\(820-40 = 780\)litres of water. 

So 780 is our answer. 

Thanks! :)

3. My logic for this problem might be flawed, but let me give it my best shot. 

If One side of the bottom layer of a triangular pyramid has 11 balls, 

The first layer has 1 ball

The Second layer will have \(3 = (1 + 2) \)balls

The Third layer will have \(6=(1 + 2 + 3)\) balls.

The Fourth layer has\( 10=(1 + 2 + 3 + 4)\) balls.

The Fifth layer will have \(15=(1 + 2 + 3 + 4 + 5)\) balls

The sixth layer there are\( 21=(1 + 2 + 3 + 4 + 5 + 6)\) balls

So there will be \(28,36,45,55 \quad \text{and} \quad 66 \) balls in the remaining layers.

So, adding the total number of balls up, we get

\(1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 = 286 balls.\)

So 286 is the answer. 

Again, not sure about this one, should be right. 

Thanks! :)

I will give you that you got the three right! Nice job, and the method for number 3 was my strategy too. Great job NotThatSmart!

3. First, let's note that If any number LQ is multiplied by 9, right most digit of the product will be (10 - Q).

So, from JKL/9 = LQ, we get, L = 10 - Q

LQ can be mathematically represented as\(10L + Q = 10(10 - Q) + Q = 100 - 9Q \)

Since \(PQR/6 = LQ\), \(PQR = 6LQ = 6(100 - 9Q) = 600 - 54Q\)

We now try various values for Q such as 1, 2, 3, .... , etc, till the middle digit of the product turns out to be the same as the value of Q used. We get the correct answer for Q at the very third attempt, namely 3. Hence,

\(L= 10 - 3 = 7\)

So we finally have

\(219/3 = 73 \\ 438/6 = 73 \\ 657/9 = 73\)

Thanks! :)

5. I THINK this is the solution. Unless,umm..idk

The old man temporarily added his dog to the 17, making a total of 18 dogs.

First son got 1/2 of 18 = \(1/2\cdot 18 = 9\)

Second son got 1/3 of 18 = \(1/3 \cdot 18=6\)

Third son got 1/9 of 18  = \(1/9 \cdot 18 = 2\) for a total of 17.

Then I think the old man stole his dog back...so yay?

Nice questions! I might not answer them every week, but it sure was a nice little brain teaser!

Thanks! ::)

Alright, here are the solutions that I got. 

1. It is given that the two digits of the required number are prime numbers i.e. 2, 3, 5 or 7. Note that 1 is neither prime nor composite. Also, the third digit is the multiplication of the first two digits. Thus, hundreds digit and units digit must be either 2 or 3 i.e. 2_2, 2_3, 3_2 or 3_3 which means that there are four possible numbers - 242, 263, 362 and 393.

Now, it is also given that - the difference between its reverse and it is 99. So 263 and 362 satisfy this condition. Hence, the sum of the three digits is 11 in each case.

2. It is given that the man has 40 litres container of milk. Also, he will drink 1 litre on the first day and refill the container with water, will drink 2 litres on the second day and refill the container, will drink 3 litres on the third day and refill the container, and so on till 40th day. Thus at the end of 40 days, he must have drunk (1 + 2 + 3 + 4 + ..... +38 + 39 + 40) = 820 litres of liquid.

Out of those 820 litres, 40 litres is the milk which he had initially. Hence, he must have drunk 780 litres of water.

3. As there are 11 balls along one side, it means that there are 11 layers of balls. The top most layer has 1 ball. The second layer has 3 (1+2) balls. The third layer has 6 (1+2+3) balls. The fourth layer has 10 (1+2+3+4) balls. The fifth layer has 15 (1+2+3+4+5) balls. Similarly, there are 21, 28, 36, 45, 55 and 66 balls in the remaining layers.

Hence, the total number of balls is = 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 = 286 balls

4. Let's start with JKL = 9 * LQ. Note that L appear on both the side. Also, after multiplying LQ by 9 the answer should have L at the unit's place. The possible values of LQ are 19, 28, 37, 46, 55, 64, 73, 82 and 91; out of which only 64, 73 and 82 satisfies the condition. (As all alphabets should represent different digits)

Now, consider PQR = 6 * LQ. Out of three short-listed values, only 73 satisfies the equation.

Also, ZYX = 3 * LQ is satisfied by 73.

Hence, Z=2, Y=1, X=9, P=4, Q=3, R=8, J=6, K=5, L=7

219/3 = 438/6 = 657/9 = 73

5. The old man temporarily added his dog to the 17, making a total of 18 dogs.

First son got 1/2 of it = 9

Second son got 1/3 of it = 6

Third son got 1/9 of it = 2 for a total of 17.

He then steals his dog back and goes away......