# Operations in the Klasies Caves

July 13, 2012 9:00 AM Subscribe

I was helping my 4th grade age sister-in-law with her math homework, and came across a question that I couldn't figure out. Can you do it?

Picture Of The Problem

It's been a few months since I last glanced at this but I gave it my best for about an hour's time and I was stumped.

Picture Of The Problem

It's been a few months since I last glanced at this but I gave it my best for about an hour's time and I was stumped.

(3^2*4)+6 = 42

(12/6)^3=8

60/4=90-(10*6)-15

On preview, what etrigan said.

posted by duckstab at 9:08 AM on July 13, 2012

(12/6)^3=8

60/4=90-(10*6)-15

On preview, what etrigan said.

posted by duckstab at 9:08 AM on July 13, 2012

I'm not sure about a rigorously analytic way to solve it but I got it pretty quickly by inspection.

The circle with a dot is ^ and the raindrop is /. So equation 2 is (12 / 6) ^ 3 = 8.

So then moving up to the first equation, the bull-horn thing is X and the square thing is +. So (3^2 * 4) + 6 = 42.

Then the circle with slash is -, so 60 / 4 = 90 - (10 X 6) - 15.

posted by Perplexity at 9:11 AM on July 13, 2012

The circle with a dot is ^ and the raindrop is /. So equation 2 is (12 / 6) ^ 3 = 8.

So then moving up to the first equation, the bull-horn thing is X and the square thing is +. So (3^2 * 4) + 6 = 42.

Then the circle with slash is -, so 60 / 4 = 90 - (10 X 6) - 15.

posted by Perplexity at 9:11 AM on July 13, 2012

Your are probably not thinking about the exponent (power of) operator.

First one is (3^2 * 4) + 6 = 42

posted by gus at 9:12 AM on July 13, 2012

First one is (3^2 * 4) + 6 = 42

posted by gus at 9:12 AM on July 13, 2012

You may be wondering how to get the solution.

There are five symbols. So a reasonable guess is that they're addition, subtraction, multiplication, division, and exponentiation (powers).

I'm going to rewrite the equations with letters replacing the symbols:

(3 A 2 B 4) C 6 = 42

(12 D 6) A 3 = 8

60 D 4 = 90 E (10 B 6) E 15

Let's focus on the simplest equation first, the second one. what are A and D? Well, first we look at what 12 D 6 could be: it's either 18, 6, 72, 2, or 12^6 = (big number), depending on what D means. We want to somehow combine one of these numbers with 3 to get 8; the only way to do that is 2^3. So D is division, A is exponentiation, and we have

(3^2 B 4) C 6 = 42

60/4 = 90 E (10 B 6) E 15

or, doing the arithmetic,

(9 B 4) C 6 = 42

15 = 90 E (10 B 6) E 15.

Pretty much the same procedure applies to the first equation. C could be -, +, or *, in which case we need 9 B 4 to be 48, 36, or 7 respectively. The only one of these that works is to have B = *, C = +. That leaves

15 = 90 E 60 E 15

and fortunately - is the only symbol left, and 15 = 90 - 60 - 15.

posted by madcaptenor at 9:14 AM on July 13, 2012 [8 favorites]

There are five symbols. So a reasonable guess is that they're addition, subtraction, multiplication, division, and exponentiation (powers).

I'm going to rewrite the equations with letters replacing the symbols:

(3 A 2 B 4) C 6 = 42

(12 D 6) A 3 = 8

60 D 4 = 90 E (10 B 6) E 15

Let's focus on the simplest equation first, the second one. what are A and D? Well, first we look at what 12 D 6 could be: it's either 18, 6, 72, 2, or 12^6 = (big number), depending on what D means. We want to somehow combine one of these numbers with 3 to get 8; the only way to do that is 2^3. So D is division, A is exponentiation, and we have

(3^2 B 4) C 6 = 42

60/4 = 90 E (10 B 6) E 15

or, doing the arithmetic,

(9 B 4) C 6 = 42

15 = 90 E (10 B 6) E 15.

Pretty much the same procedure applies to the first equation. C could be -, +, or *, in which case we need 9 B 4 to be 48, 36, or 7 respectively. The only one of these that works is to have B = *, C = +. That leaves

15 = 90 E 60 E 15

and fortunately - is the only symbol left, and 15 = 90 - 60 - 15.

posted by madcaptenor at 9:14 AM on July 13, 2012 [8 favorites]

Yeah. My first clue here was looking and seeing there were five operators. Once I knew that, this means it had to be something more than just addition, subtraction, division, and multiplication. Given the 4th grade math curriculum (very little multivariable calculus) the power operator is the next logical one.

posted by Betelgeuse at 9:16 AM on July 13, 2012

posted by Betelgeuse at 9:16 AM on July 13, 2012

Wow. It would never in a thousand years occur to me that fourth graders would be learning about exponents. Is your sister-in-law in an accelerated/gifted program?

posted by croutonsupafreak at 9:17 AM on July 13, 2012 [1 favorite]

posted by croutonsupafreak at 9:17 AM on July 13, 2012 [1 favorite]

As for how to solve it:

I started with the bullseye symbol in the second equation. We've got either

posted by nebulawindphone at 9:17 AM on July 13, 2012 [1 favorite]

I started with the bullseye symbol in the second equation. We've got either

- x*3 = 8
- x+3 = 8
- x-3 = 8
- x/3 = 8
- x^3 = 8

- If it's #1, then x needs to be 2.66666666666..., and there's nothing we can do with 12 and 6 to make 2.666666666666....
- If it's #2, then x needs to be 5, and there's nothing we can do with 12 and 6 to make 5.
- If it's #3, then x needs to be 11, and there's nothing we can do with 12 and 6 to make 11.
- If it's #4, then x needs to be 24, and there's nothing we can do with 12 and 6 to make 24.
- If it's #5, then x needs to be 2. We can divide 12 by 6 to make 2.

posted by nebulawindphone at 9:17 AM on July 13, 2012 [1 favorite]

Thanks all :) Great explanation madcaptenor!

posted by symbollocks at 9:18 AM on July 13, 2012

posted by symbollocks at 9:18 AM on July 13, 2012

croutonsupafreak: Nope, just regular class as far as I know. I know, it's surprising to me too.

posted by symbollocks at 9:19 AM on July 13, 2012

posted by symbollocks at 9:19 AM on July 13, 2012

Fun midday math teaser, thanks :)

Did it about the same way as others. Started in the middle. and worked up then down.

posted by effigy at 9:19 AM on July 13, 2012

Did it about the same way as others. Started in the middle. and worked up then down.

posted by effigy at 9:19 AM on July 13, 2012

What everyone else said. I looked at the symbols, and realized there were 5, but wasn't sure yet what the extra one was. Looking at the middle problem, I wrote out the equation 4 ways, assuming that the teardrop symbol was not whatever the extra one was. So, I had

(12 + 6) O 3 = 8

(12 - 6) O 3 = 8

(12 x 6) O 3 = 8

(12 / 6) O 3 = 8

and then I simplified. The only one that was possible was the last one, if the O symbol was an exponent operator. With those translated, I attacked the first one and realized that V must mean times since (9 x 4) + 6 was the only way to make that work. After that, the last symbol was power of elimination.

Thanks for this; I'll share it with the 4th grade teacher I know!

posted by booksherpa at 9:21 AM on July 13, 2012

(12 + 6) O 3 = 8

(12 - 6) O 3 = 8

(12 x 6) O 3 = 8

(12 / 6) O 3 = 8

and then I simplified. The only one that was possible was the last one, if the O symbol was an exponent operator. With those translated, I attacked the first one and realized that V must mean times since (9 x 4) + 6 was the only way to make that work. After that, the last symbol was power of elimination.

Thanks for this; I'll share it with the 4th grade teacher I know!

posted by booksherpa at 9:21 AM on July 13, 2012

*Wow. It would never in a thousand years occur to me that fourth graders would be learning about exponents. Is your sister-in-law in an accelerated/gifted program?*

My thought too - fourth grade is, what, age 10? I stopped maths when I was sixteen but even then I had to think about what 'operations symbols' were.

posted by mippy at 10:06 AM on July 13, 2012 [1 favorite]

This thread is closed to new comments.

Dot in circle = to the X power, where X is the number after the dot in circle

Capricorn / double-geyser symbol = times

Hollow square = plus

Not sign / circle with slash = minus

posted by Etrigan at 9:08 AM on July 13, 2012