# Some mind boggling Math problems to challenge your friends

Category : EDUCATION Author : Krishna Mishra Date : Wed Jan 17 2018 Views : 15

I have here for you some Math problems which are basically for VIII grade and above grade students. You can use these problems to stump your friends.

**1. **__n ^ 2 + 45 = a perfect square__

The goal here is to tell as many as possible integers which are possible so that when 45 is added to their square, a perfect square is obtained. After they had surrendered, tell them the solution.

**SOLUTION: **

Let's say the square obtained is x. Then....

n ^ 2 + 45 = x ^ 2

=> 45 = x ^ 2 - n ^ 2

This can be factored out as

45 = (x + n)(x - n)

Now we have to make pairs of numbers whose pair sums up to 45. This will be easy.

(7 + 2)(7 - 2) = 45

(23 + 22)(23 - 22) = 45

(9 + 6)(9 - 6) = 45

Thus, **n = 2, x = 7**

n = 22, x = 23

n = 6, x = 9

**2. **__AA + BB + CC = ABC__

You have to find out the values for variables A, B, and C such that

AA

BB

+ CC

-------

ABC

-------

Now let's try to solve it. Even if you take the maximum possible values of A, B, and C, you will always get the hundreds place as 2. So, A has to be 1 or 2. Now in the ones place, A + B +** **C = a unit digit C.

So A + B = 10.

If A = 1 then B = 9 and if A = 2 then B = 8

11 + 99 + CC = 19C

So 110 + CC = 19C

This means that 1 + C = 9 (tens), so C = 8.

Let's check it.

110 + 88 = 198 (HENCE VERIFIED)

You can verify that the other statement is false.

Using this method, you can solve as many of these combinations as you want.

**Thank you!**

I have here for you some Math problems which are basically for VIII grade and above grade students. You can use these problems to stump your friends.

**1. **__n ^ 2 + 45 = a perfect square__

The goal here is to tell as many as possible integers which are possible so that when 45 is added to their square, a perfect square is obtained. After they had surrendered, tell them the solution.

**SOLUTION: **

Let's say the square obtained is x. Then....

n ^ 2 + 45 = x ^ 2

=> 45 = x ^ 2 - n ^ 2

This can be factored out as

45 = (x + n)(x - n)

Now we have to make pairs of numbers whose pair sums up to 45. This will be easy.

(7 + 2)(7 - 2) = 45

(23 + 22)(23 - 22) = 45

(9 + 6)(9 - 6) = 45

Thus, **n = 2, x = 7**

n = 22, x = 23

n = 6, x = 9

**2. **__AA + BB + CC = ABC__

You have to find out the values for variables A, B, and C such that

AA

BB

+ CC

-------

ABC

-------

Now let's try to solve it. Even if you take the maximum possible values of A, B, and C, you will always get the hundreds place as 2. So, A has to be 1 or 2. Now in the ones place, A + B +** **C = a unit digit C.

So A + B = 10.

If A = 1 then B = 9 and if A = 2 then B = 8

11 + 99 + CC = 19C

So 110 + CC = 19C

This means that 1 + C = 9 (tens), so C = 8.

Let's check it.

110 + 88 = 198 (HENCE VERIFIED)

You can verify that the other statement is false.

Using this method, you can solve as many of these combinations as you want.

**Thank you!**

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