These math riddles will make you think a little bit

• It costs $77 to buy B apples, B bananas, and A+B cherries.
• The price of an apple is $a, of a banana is $b, and of a cherry is $c.
• What do you get if you put together what Einstein and Pythagoras found?

Raise your hand if you liked math the most in school. If you ask this question in a room with more than 100 people, only 7-8 might raise their hands.

Why? Should we blame mathematicians like Euclid and Pythagoras? Well, they are not to blame, even though many people do. Instead, these and other mathematicians from all over the world have changed the way we look at numbers in big ways.

Even though numbers are hard to work with on their own, things get even harder when you add letters to them. Was the day busy? Do you want to take a break and clear your mind? Well, this blog post might be the best thing you read all day.

Math Riddle 1:

What do you get if you put together what Einstein and Pythagoras found?

Math Riddle 2:

The numbers a, b, and c are all positive integers.

The price of an apple is $a, of a banana is $b, and of a cherry is $c.

It costs $77 to buy B apples, B bananas, and A+B cherries.

How much would an apple, two bananas, and a cherry cost?

Math Riddle 3:

Prove that 2 + 1 = 1?

Do you want to check to see if you got it right?Move down!

Math Riddle 1:

What do you get if you put together what Einstein and Pythagoras found?

Answer:

m (a2 + b2) = E m c2)

Math Riddle 2:

All three of these numbers are positive integers.

The price of an apple is $a, of a banana is $b, and of a cherry is $c.

It costs $77 to buy B apples, B bananas, and A+B cherries.

How much would an apple, two bananas, and a cherry cost?

Answer:

Let’s try to find a solution…

b*a + b*b + (a+b)*c = 77

77 = b(a+b) + c(a+b).

(a+b)(b+c) = 77

Now look at the question: The numbers a, b, and c are all whole numbers that add up to 1.

We know that it is a whole number that needs to be multiplied by itself to get 77. Simply multiply 77 by the factors.You will see that this is an easy composite number to break down into its parts. – 1, 7, 11, and 77.

The number 77 can be made by adding 1 to 77 or 11 to 7. This is important to know.

Problem solved. Since neither A+B nor B+C can equal 1, it must be 7 and 11.

So a + 2b + c = (a + b) + (b + c) = 7 + 11 = 18.

Math Riddle 3:

Show that 2 + 1 = 1?

Answer:

Assume that a = b.

Then we multiply each side by a.

Following that, we’ll see that a2 = ab [a2 = ab].

when put next to ab

There is no difference. Remove b2. [a2 b2 = ab b2]

We will look at both sides.

Now, there is a b on each side. [(a + b) (a b) (a b) = b (a b)]

We’ll split things up by a

olé and less than b

a + b = b. [a + b = b]

But I stated that a = b, so

Do you agree?[b + b = b]

So if b = 1,

Then I did this sum. [1 + 1 = 1]

proves that 2 equals 1

We know you frowned when you saw the letters written between the numbers. But, hey, you can’t deny that the last 5–10 minutes you spent reading this blog were fun.

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