# How many six-digit numbers are there where the third digit is equal to the second last digit, ...

How many six-digit numbers are there where the third digit is equal to the second last digit, the digit in the ten-thousands place is equal to the digit in the hundreds place, and the product of all the digits is equal to the square of a natural number?

Attempt:

We are looking for numbers of the form $$abc bcf$$, where $$ab^2c^2f=n^2$$ for some $$n \in \mathbb{N}$$. This means that $$af = m^2$$ for some $$m \in \mathbb{N}$$. Now, I have separated it into three cases:

1. Case: $$a = f$$, there are $$10^3$$ such numbers,
2. Case: $$a = x^2$$ and $$f = y^2$$ and $$a \neq f$$, there are $$10^2 \cdot 6$$ such numbers,
3. Case: $$a = 2$$ and $$f = 8$$ or $$f = 2$$ and $$a = 8$$, there are $$2 \cdot 10^2$$ such numbers.

Thus, I found that there are 1800 such numbers (but the solutions state 1458). Where did I go wrong?

Note: $$0$$ is not a natural number here.

• Zero is a square, is it not? So maybe some $10$s rather than $9$s? Commented Jun 19 at 17:38
• Do you know whether or not the problem considers $0$ to be a natural number? This makes a significant difference re: the problem statement that "... the product of all the digits is equal to the square of a natural number". Commented Jun 19 at 17:43
• $0$ is not a natural number here ...
– user1316790
Commented Jun 19 at 17:44
• If you knew $0$ is not a natural number, why did you edit it to replace the $9$s with $10$s? If zero is not a natural number my comment wasn't valid and you shouldn't have changed anything. Commented Jun 19 at 21:38
• I think the deleted answer gives a hint to how the may have double counted. In the $a,f$ are not square but have a square product the only way to do that is with odd powers of $2$. $(2,2),(2,8)$ and $(8,8)$. But $(2,2)$ and $(8,8)$ were included in case $1$ so double counting. And maybe a failure to count $(2,8)$ separately from $(8,2)$. So they get $18\times 81$ rather than your (and I believe it to be correct) $17\times 81$. (I wouldn't mind if someone could confirm that.) Commented Jun 19 at 22:14

If, as indicated in the edit, $$0$$ is not a natural number, that means the product of digits can't be zero and there are only $$9$$ possibilities for each digit. Consequently every $$10$$ in your calculation should be a $$9$$. This leads to an answer of $$1377$$.

A quick python script confirms that there are exactly $$1377$$ integers with this property, so the answer you were given is wrong. (The largest possible product of digits is $$9^6=729^2$$.)

squares=[n*n for n in range(1,730)]
r=0
for n in range(100000,1000000):
x=str(n)
y=1
for z in x:
y*=int(z)
if y in squares and x[1]==x[3] and x[2]==x[4]:
r+=1
print(r)


Incidentally, if $$0$$ is treated as a natural number, the answer is not $$2600$$, despite two previous posts. This is because numbers where $$af$$ is not a square but $$b$$ or $$c$$ is $$0$$ still give a square product. There are $$19$$ choices for $$b,c$$ with at least one $$0$$. There are $$90$$ overall possibilities for $$a,f$$ and of these $$26$$ have $$af$$ being a square (nine with $$a=f$$, nine with $$f=0$$, six combinations of $$1,4,9$$ and two of $$2,8$$) so there are $$19\times 64=1216$$ cases missing from the other answers, giving a total of $$3816$$.

EDIT: Especially Lime correctly pointed out in their answer that my math didn't properly account for all the $$abcbcf=0$$ cases, so here is a fixed version.

We are looking for numbers of the form $$abcbcf$$, where $$ab^2c^2f=n^2$$ for some $$n∈N$$.
This means that $$af=m^2$$ for some $$m∈N$$.

This looks good so far, but the issue of whether $$a=0$$ creates a valid "six digit number" seems like an important detail. (As well as the question of whether $$n\in N$$ includes $$n=0$$ or not.)

If we start by counting only the non-zero-$$n$$ solutions to $$ab^2c^2f=n^2$$ then $$af=m^2$$ applies and that is solved most simply when:

Case 1: $$a=f \rightarrow af=a^2=m^2$$, there are 9 non-zero choices for $$a$$, giving $$9\cdot9^2$$ such numbers.

That leaves the $$a\neq f$$ cases, for which we can see that $$af=m^2\rightarrow af=p^2q^2$$ for which $$a$$ can only be factored out in one of two ways:

Case 2a: $$a=p^2 \rightarrow af=p^2(f)=p^2q^2 \rightarrow f=q^2$$, there are three choices for $$a$$ leaving two remaining choices for $$b\neq a$$, giving $$3\cdot 2\cdot9^2$$ such numbers.

Case 2b: $$a\neq p^2 \rightarrow m^2=p^2q^2=p(pq^2)$$, which gives two choices for $$a$$ leaving one choice for $$b\neq a$$, giving $$2\cdot 1\cdot9^2$$ such numbers.

That all together that gives: $$9\cdot9^2+3\cdot 2\cdot9^2+2\cdot 1\cdot9^2=(17)\cdot9^2=1377$$ solutions to $$abcbcf=n^2$$ for non-zero $$n\in N$$.

That (potentially) leaves the $$n=0$$ cases, which are much easier to account for (since there are no further requirements containing the digits make squares or whatnot). If we assume that $$012123$$ does not count as a "six-digit number" then $$a\neq 0$$ applies and:

Case 3a: $$b=0$$, leaves 9 choices each for $$a$$, $$c$$, and $$f$$, giving $$9^3$$ such numbers.

Case 3b: $$c=0$$, leaves 9 choices each for $$a$$, $$b$$, and $$f$$, giving $$9^3$$ such numbers.

Case 3c: $$f=0$$, leaves 9 choices each for $$a$$, $$b$$, and $$c$$, giving $$9^3$$ such numbers.

Case 3d: $$b=c=0$$, leaves 9 choices each for $$a$$ and $$f$$, giving $$9^2$$ such numbers.

Case 3e: $$b=f=0$$, leaves 9 choices each for $$a$$ and $$c$$, giving $$9^2$$ such numbers.

Case 3f: $$c=f=0$$, leaves 9 choices each for $$a$$ and $$b$$, giving $$9^2$$ such numbers.

Case 3g: $$b=c=f=0$$, leaves 9 choices for $$a$$, giving $$9^1$$ such numbers.

That all together that gives: $$(3)\cdot9^3+(3)\cdot9^2+(1)\cdot9^1=2439$$ solutions to $$abcbcf=0$$ when $$a\neq 0$$.

This makes the grand total: $$1377+2439=3816$$ as pointed out by Especially Lime

Also, it should be noted somewhere that this all assumes the decimal base.

So $$a\in\{1,2,3,4,5,6,7,8,9\}$$ and similarly for $$b$$, $$c$$, and $$f$$ in Case 1.

For Case 2a, only $$a\in\{1,4,9\}$$ will satisfy the conditions of $$a=p^2\leq9$$ (for $$a\neq0$$).

For Case 2b, only $$p=2=q$$ can satisfy $$p\cdot q^2\leq 9$$. This leaves only $$a\in\{2,8\}$$.