Calculate a six digit number that is smaller than 500000 such that the sum of its digits is 43 exactly.
499849
Let us begin with the maximum sum of a digit under 500000. Of course it will be the sum of 499999.
4 + 9 + 9 + 9 + 9 + 9 = 49.
Now, the required number must be close to it.
Rooting 499999, we get 707.106
Taking the square of 707, we get 499849.
4 + 9 + 9 + 8 + 4 + 9 = 43
Thus, the required number is 499849.
Below, you can see three complete equations and one incomplete. Based on the three complete ones, can you complete the incomplete one?
5 $ 4 $ 3 $ 9 = 4215
6 $ 9 $ 2 $ 6 = 3816
4 $ 7 $ 3 $ 3 = 1122
7 $ 2 $ 7 $ 4 = ____
2418 is the answer.
The algo that has been used here is
A $ B $ C $ D = [first digit of (A * D)] [last digit of (B * C)] [first digit of (B * C)] [last digit of (A * D)]
Using the same algo,
7 $ 2 $ 7 $ 4 = [first digit of (7 * 4)] [last digit of (2 * 7)] [first digit of (2 * 7)] [last digit of (7 * 4)]
= 2418
Can you think of three numbers that gives a prime number as their product and also the difference between the second and first is equal to the difference between third and second?
The three numbers are: -3, -1 and 1.
Remember that we never asked for positive numbers.
You along with your friend are standing in front of two houses. Each of those houses inhabits a family with two children.
Your friend tells you the below two facts:
1) On your left is a family that has a boy who likes accounts but the other child loves science.
2) On the right is a family with a seven year old boy and a new born baby.
You ask him, "Does either of the family have a girl?"
To this he replies, "I am not quite sure. But can you guess that? If you are right, I will give you $200."
Which family do you think is likely to have a girl ?
1/2
In the house on the left, there are three possibilities:
Younger Older
Girl Boy
Boy Girl
Boy Boy
We cant have a girl-girl option because it has been mentioned that there is at least one boy in that house. Now all these outcomes are equally likely, and we have two events with girls, the chances of having a girl in the left house is 2/3.
In the house on the right, there are two possibilities because we already know that the older child is a boy:
Younger Older
Girl Boy
Boy Boy
Here the chances are 1/2.
Thus you must choose the house on the left for better chances at winning.
I engaged in a strange activity. My birthday was approaching and I decided to collect money for my birthday bash. On the first day of the month, I kept a dollar in my piggy bank, on the second, I kept two dollars and on the third, I kept three and so on.
On my birthday, I had a total of 276 dollars in my piggy bank. Can you find out on which day of the month was my birthday?
23rd
The easiest way to find out without engaging in any formula would be to simply add them:
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 = 276
Thus, the birthday falls on the 23rd day of the month.
You have $100 with you and you have to buy 100 balls with it. 100 is the exact figure and you cant go below or above the numbers and you have to use the entire $100. If there is no kind of tax applied how many of each of the following balls will you be able to buy:
Green Balls costing $6
Yellow Balls costing $3
Black Balls costing $0.10
Now, how many of each must you buy to fulfill the condition given?
1 green ball = $6
29 yellow balls = $87
70 black balls = $7
Detailed Explanation :
Let us denote the number of balls to be brought with a variable
p = green sponges
q = yellow sponges
r = black sponges
Now using a simple expression:
p + q + r = 100
6p + 3q + 0.1r = 100
Multiplying the first equation by 6 then 3 and subtracting the two equations, we get:
6p + 6q + 6r = 600
3p + 3q + 3r = 300
3q = 500 - 5.9r
3p = 2.9r - 200
Usually, we have three equations to solve three variables. However, we will manage with just two here because we know that p and q are non-negative integers.
If 3q >/= 0, then 500 - 5.9r >/= 0.
Using this, r
Also, if 3p >/= 0, then 2.9r - 200 > 0. This implies that r >/= 68.97.
Now, the number of black balls must be an even dollar amount or we will land up with a fraction. Considering the expressions, you know that there can be just two integers that can satisfy the condition while lying in between 68.97 and 84.75.
If we take r as 80 and solve the equations (the last two equations), we will get p = 10.67 and q = 9.33. But p and q must be integer and not fraction. Thus you can cancel out 80.
If we take r as 70, we will get p = 1 and b = 29.
Therefore following are the number of balls that we must buy:
1 green ball = $6
29 yellow balls = $87
70 black balls = $7
Below, you will find the mathematical proof that 10 equals 9.99999…. But is that possible or there is something wrong about it? Can you find the error?
x = 9.999999...
10x = 99.999999...
10x - x = 90
9x = 90
x = 10
No error is present in this proof and 10 is actually equal to 9.99999…. The proof is absolutely correct.
Can you find a number that lies one third of the distance between 1/3 and 2/3?
Let us find a bigger multiple first. Lets multiply the numerator and denominator by 9 in both the numbers.
Now, 1/3 = 9/27
And, 2/3 = 18/27
Clearly, 12/27 lies one third on the distance between the two numbers now.
Alex and Martin are planning on a vacation. Alex says, "It will be better if we take the train to the hotel. We will reach faster."
To this, Martin says, "No. You are wrong. The train stops at halfway to the hotel and then we will have to walk the rest of the way. We should take the bikes to the hotel directly."
Alex says, "You are wrong."
Now, Alex takes the train while Martin takes the bike. The train travels with four times the speed of bike. The bike travels at twice the speed of walking.
Who do you think will reach the hotel first?
Martin
S = total distance to the hotel.
V = walking speed.
2V = bike speed.
8V = train speed.
A = time taken by Alex
M = time taken by Martin
A = (S/2)/8V + (S/2)/V = S/16V + S/2V
M = S/2V
Now if you notice, A exceeds M by S/16V. Therefore, Martin will reach the hotel first.
One night, I thought of ways that can be used for creating a palindrome. So I decided that I will turn into a larger number by adding the reversed digits to the original number and keep doing it till I finally obtained a palindrome.
I am not sure if this process will always result in a palindrome eventually but I was able to produce a four digit palindrome. Can you guess my starting number?
My starting number was 192.
192 + 291 = 483
483 + 384 = 867
867 + 768 = 1635
1635 + 5361 = 6996