At Mathnasium of Friendswood, our instructors know our students thrive in a fun and engaging environment. In particular, riddles can be a fun way to challenge your math student. See if you can work with your math student to solve the two riddles provided below!

Math Riddle #1:

I have only red, blue, and yellow candies. They are all red except for 5 of them. They are all blue except for 11 of them. They are all yellow except for 12 of them. How many of each color candy do I have?

The solution to Riddle #1:

How can we solve this riddle? Let’s start by thinking about what the riddle is telling us. If all the candies are red except 5, I have r + 5 total. We also know that the remaining 5 must be blue and yellow, so b + y = 5. If all the candies are blue except 11, then the total is also b + 11. The 11 must be yellow and red, so y + r = 11. And finally y + 12 = total and r + b = 12.

We can use b + y = 5 and y + r = 11. Let’s solve for y in each equation to get:

y = 5 - b and y = 11 - r

Now, set them both equal:

5 - b = 11 - r

Combine like terms by adding r to both sides and subtracting 5 from both sides:

- b + r = 6

We also know that r + b = 12. Solve both this equation and -b + r = 6 for r:

r = 6 + b and r = 12 - b. Now, set them equal to each other:

6 + b = 12 – b

Combine like terms:

2b = 6

Divide by the coefficient:

b = 3

We now know that there are 3 blue candies and can solve for the other values by plugging in our 3 for b where it appears with y and r. So:

b + y = 5 is actually 3 + y = 5 knowing what we know about blue. We solve and find that y = 2. We have 2 yellow candies!

Do the same with r + b = 12. r + 3 = 12, so r = 9. There are 9 red candies!

There are 9 red, 3 blue, and 2 yellow candies.

Math Riddle #2:

I made a batch of 3 dozen cookies and want to sort them into bags to sell at a bake sale. I get hungry and eat some of the cookies but don’t pay attention to how many I eat. I try to make bags of 3 cookies each. Oops! I end up with not enough cookies for my final bag – I have two leftover cookies. I decided to start over. This time, I placed 4 cookies in each bag. This time I can evenly distribute the cookies into bags, with none leftover. If I know I ate fewer than 10 cookies, how many cookies did I eat?

Answer to Riddle #2:

I know that whatever number of cookies I have has to be divisible by 4 with no remainders, so the first thing to do is to figure out what the possibilities are:

28 and 32 are the only numbers in the 26-36 range (remember, I know I ate fewer than 10 cookies) that are evenly divisible by 4.

Now, we need to figure out which of these two numbers gives us a remainder of 2 when we divide by 3.

Three goes into 28 nine times, with a remainder of one. After all, 3 x 9 = 27, and 28 - 27 = 1. 28 is not how many cookies we have because we need a remainder of 2.

Three goes into 32 ten times, with a remainder of two. Again, the math tells us that 3 x 10 = 30, and 32-30 is 2. This IS the number we want. The answer to the riddle is that I ate 4 cookies. We started with 36 and ended up with 32.

These riddles were inspired by similar riddles found here. Try the other riddles at the link if you are up for some more challenging fun! We know at Mathnasium that students are more engaged with math when it is fun! We are always trying to incorporate fun ways to challenge our students, just like we can with these riddles. You can trust that your math student is being stimulated and enjoying their time spent at Mathnasium of Friendswood! Give us a call today at (832) 569-5073 to see if our programs are right for you!