"All of this probably seems pretty obvious to you. All right then. Let’s use a base other than ten and see how you do. Let’s write out a number in base 7. We have seven symbols at our disposal: 0, 1, 2, 3, 4, 5, and 6. Wait, you ask — why not 7? Because there is no digit for seven in a base 7 system, just like there is no digit for ten in a base 10 system. Ten is the point where we need two digits in a decimal system, and analogously, seven is the point where we’ll need two digits in our base 7 system. How will we write the value seven? Just like this: 10. Now stare at those two digits and practice saying “seven” as you look at them. All your life you’ve been trained to say the number “ten” when you see the digits 1 and 0 printed like that. But those two digits only represent the number ten if you’re using a base 10 system. If you’re using a base 34 system, “10” is how you write “thirty-four.” Very well, we have our seven symbols. Now how do we interpret a number like 6153₇? It’s this: 6153₇ = 6 × 7³ + 1 × 7² + 5 × 7¹ + 3 × 7⁰. That doesn’t look so strange: it’s very parallel to the decimal string we expanded, above. It looks weirder when we actually multiply out the place values: 6153₇ = 6 × 343 + 1 × 49 + 5 × 7 + 3 × 1. So in base 7, we have a “one’s place,” a “seven’s place,” a “forty-nine’s place,” and a “three hundred forty-three’s place.” This seems unbelievably bizarre — how could a number system possibly hold together with such place values? — but I’ll bet it wouldn’t look funny at all if we had been born with 7 fingers. Keep in mind that in the equation above, we wrote out the place values as decimal numbers! Had we written them as base-7 numbers (as we certainly would have if base 7 was our natural numbering system), we would have written: 6153₇ = 6 × 1000₇ + 1 × 100₇ + 5 × 10₇ + 3 × 1₇. This is exactly equivalent numerically. Because after all, 1000₇ is 343₁₀. A quantity that looks like an oddball in one base system looks like the roundest possible number in another."