# Which number can I erase?

All positive integers greater than $$2$$ are written on a board. First we erase number $$3$$ and $$5$$.

With 4 positive integers $$a,b,c,d$$ satisfying $$a+b=c+d$$, if $$ab$$ is erased, then $$cd$$ can be erased, otherwise $$cd$$ cannot be erased.

For example, $$3=3 \times 1$$, $$3+1=4=2+2$$ , then $$2 \times 2 = 4$$ is erased.

a. What are the conditions of a number that can be erased ?

b. If not only $$3$$ and $$5$$, but every prime number is erased at the beginning, can all other numbers be erased as well? If not, what are the conditions of a number to be erased ?

(Sorry for my last question, English is my second language)

• What makes you say you can't erase $7$? Surely you could erase $7=7\times1$, then say $7+1=6+2$, so you can erase $12$. I don't understand. I also think the question is poorly worded and is missing other relevant information. You haven't specified that $a,b\neq c,d$. Can we choose any number to erase if it has the specified property? Or are we given a starting number, $ab$, to erase and have to find all other numbers, $cd$, that we are then allowed to erase? – Jam Oct 16 '18 at 12:28
• @Jam : thanks for your advice. I will edit my question. – apple Oct 16 '18 at 12:31
• Thanks for the edit, it makes a lot more sense :) – Jam Oct 16 '18 at 14:49

A prime $$p>5$$ can be erased if $$m=2(p-1)$$ has been erased, as $$2+(p-1)=p+1$$. Note that $$m=2\times(p-1)=4\times\frac{p-1}{2},$$ where $$p-1$$ is even because $$p>5$$ is prime. This factorization of $$m$$ shows that it can be erased if $$\frac{p+5}{2}$$ has been erased, because $$\frac{p-1}{2}+4=\frac{p+5}{2}+1,$$ where clearly $$\frac{p+5}{2} becase $$p>5$$.

A composite number $$n=uv$$ with $$u,v>1$$ can be erased if $$m=u+v-1$$ has been erased, as $$1+(u+v-1)=u+v.$$ Of course $$m because $$u,v>1$$.

In particular, an integer $$n>5$$ can be erased if all integers less than $$n$$ have been erased, so you can use induction.

Somewhat a long comment (not a complete answer).

The first thing that I would suggest is that you should work through some examples and see what happens. I'll help with some of the initial steps.

To think about what you're erasing, let $$n$$ be a number that has been erased. Then, for each pair of factors $$ab$$ so that $$ab=n$$, compute $$a+b=m$$. Next, you find all other ways to write $$m$$ as a sum $$m=c+d$$ and then delete each $$cd$$ found in this way. Then, you continue until there isn't anything more to delete (or you see the pattern).

In your set-up, you start with $$3$$ and $$5$$ deleted.

Deleted: $$3$$ and $$5$$.

• Using the deleted $$3$$, you know that $$3$$ factors as $$3=1\cdot 3$$. Therefore, $$a+b=1+3=4$$. The only sums equal to $$4$$ are $$1+3$$ and $$2+2$$. Taking $$c=2$$ and $$d=2$$ gives $$cd=4$$, so $$4$$ is deleted. Since we've considered all pairs of factors of $$3$$, there is nothing more that $$3$$ can tell us.

Deleted: $$3$$, $$4$$, and $$5$$.

• Using the deleted $$4$$, you know that $$4$$ factors as $$4=1\cdot 4$$ or $$2\cdot 2$$. In the first case, $$1+4=5$$, and $$5$$ can be written as $$5=1+4$$ or $$5=2+3$$. The first case leads to $$1\cdot 4=4$$, which is already deleted, but $$2\cdot 3=6$$, which can now be deleted. On the other hand, using $$2\cdot 2=4$$, we have two ways to write $$4$$ as a sum, $$4=1+3$$ or $$4=2+2$$. In either case, the product is $$1\cdot 3=3$$ or $$2\cdot 2=4$$, both of which have already been deleted. Therefore, there is nothing more that $$4$$ can tell us.

Deleted: $$3$$, $$4$$, $$5$$, and $$6$$.

• Using the deleted $$5$$, you know that $$5$$ factors as $$1\cdot 5=5$$. Therefore, the sum of these two is $$1+5=6$$. There are three ways to write $$6$$ as a sum, $$1+5=6$$, $$2+4=6$$, and $$3+3=6$$. The first pair has product $$1\cdot 5=5$$, which has already been deleted. The second pair has product $$2\cdot 4=8$$, which can now be deleted. The third pair has product $$3\cdot 3=9$$, which can also be deleted.

Deleted: $$3$$, $$4$$, $$5$$, $$6$$, $$8$$, and $$9$$.

• Using the deleted $$6$$, you know that $$6$$ factors as $$1\cdot 6=6$$ and $$2\cdot 3=6$$. Therefore, the corresponding sums are $$1+6=7$$ or $$2+3=5$$. There are three sums to get $$7$$, $$1+6=7$$, $$2+5=7$$, and $$3+4=7$$. There are two sums to get $$5$$, $$1+4=5$$ and $$2+3=5$$. These, combined allow one to delete $$6$$, $$10$$, $$12$$, $$4$$, and 6\$.

Deleted: $$3$$, $$4$$, $$5$$, $$6$$, $$8$$, $$9$$, $$10$$, and $$12$$.

Continue this until you find a pattern.