How to mentally compute the cube root of the cube of a two digit number.

Thread (1/10)

In base 10, the last digits of the cubes of digits are distinct.

If d = 0, 1, 4, 5, 6, or 9, then d^3 ends in d.

If d = 2, 3, 7, or 8, then d^3 ends in 10-d.

(2/10)
So if you're given the cube of an integer, you can figure out the last digit of the cube root.

For example, if 50653 is the cube of an integer, it's the cube of a number ending in 7.

(3/10)
Let n = 10a + b be a two-digit number, with a and b single digit numbers.

Then 10a <= n <= 10(a+1), and so n^3 is between (10a)^3 and (10(a+1))^3, i.e. between 1000 a^3 and 1000 (a+1)^3.

(4/10)
So suppose x is the cube of a two-digit number n = 10a + b.

Chop off the last three digits of x.

Then a, the first digit of n, is the largest number whose cube is no greater than what's left of x.

(5/10)
This means that if you know the cubes of the digits, you can find the first digit of the cube of a two-digit number.

(6/10)
For example, suppose we know 50653 is a cube.

Lopping off the last three digits gives 50.

3^3 < 50 < 4^3, so the first digit of the cube root of 50653 is 3.

(7/10)
At the beginning of this thread we showed how to find the last digit of the cube root, and now we've shown how to find the first digit.

(8/10)
Another example. Suppose you know that 438976 is a cube.

The last digit of the cube root must be 6.

438 is between 7^3 and 8^3 and so the first digit of the cube root must be 7.

438976 = 76^3.

(9/10)
Final example. Suppose you know x = 103823 is a cube, x = n^3.

Then the last digit of n is 7 because 7^3 ends in 3.

103 is between 4^3 = 64 and 5^3 = 125, so the first digit of n is 4.

n = 47.

(10/10)

More from Maths

In light of my tweet thread about the category of finite sets and commutative monoids (https://t.co/jnY0wZZbxq), I thought I might try to say what the analogue is for braided monoidal things (although much of this is still somewhat hypothetical).


It's also just kind of a cool combinatorial structure! I've been talking to @CreeepyJoe about this lately, as well as @grassmannian.

The first thing you have to know is that, in a braided monoidal category you can still have commutative monoids. Since a braided monoidal category C has a "twist" map for every object β(x):x⊗x→x⊗x, if x is a monoid you can ask for the following diagram to commute:


Remember that being symmetric monoidal just means that if you take the twist map above and do it twice, you get the identity map, but braided monoidal doesn't mean that. But it's okay! You can still define commutative monoids here.

But so anyway, we can talk about commutative monoids in braided monoidal categories.
It is trying when mathematicians declare condescendingly that there is no point doing things because their models tell them so. Well maybe some of the assumptions don't hold up. How did that work out for the no additional risk from large events and no point in border controls...


During wave 1 cases fell very fast, faster than I think most people were expecting. Particularly in Scotland. Rt was probably ~0.5 until we started easing off.

This was despite a constant leak of cases coming out of hospitals and LTC facilities as we were rationing PPE and are policies were nowhere near ideal. There was insistence from infection control that droplet protections were sufficient. We have all learned a lot since then.

Not to mention we have learned to avoid the shit show of actively importing cases into care homes. We've learned not to repeat that. Other sectors have learned too.

We've learned a lot and there's no reason we can't control this new variant. But we will not manage if we don't try and act with clarity of purpose.

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