Product of 3 consecutive even integers is 87*****8. Find missing digits without a calculator?

Question

Q: Product of 3 consecutive even integers is 87,***,**8 where * are missing digits. Find missing digits without a calculator?

I know the answer is 87,526,608 with the even integers 442, 444, 446 but that is with a calculator. I should find it without a calculator.

I found the ones and hundreds digit of the even integer easily. The ones digit can be 0, 2, 4, 6, 8. We try the product of 3 consecutive even digits and only the product of 2, 4, and 6 have 8 as ones. So, the consecutive even integers have to be **2, **4, **6.

Next for the hundreds I checked 400x400x400 and it is 64,000,000 while 500x500x500 is 125,000,000. 87,***,**8 is between these numbers so the hundreds digit of the middle number has to be 4. Since the ones are 2, 4, and 6, the hundreds digit of all three consecutive even integers is 4.

Now how should I proceed to get the tens digit OR is there a way to get all the missing digits without knowing the 3 consecutive even digits.

Edit: Ok I am getting the exact same answers that involve cubing of 2-digit numbers. This is not allowed.

Answer 1

$(n-2)n(n+2) = n^3 - 4n$

$n = 400 + m + 4$

$n, n-2, n+2 \approx 4.m\times 10^2$ and so we need

$4.m^3 \approx 87$

$4.m^3 = 64 + 3*16*\frac 1{10}m + 3*4*\frac 1{100}m^2 + \frac 1{1000}m^3 \approx 87$

$4.8m + .12m^2 + .001m^3 \approx 23$

If $m = 5$ then $4.8m = 24$ is too high but $m =4$ is about right. $4.8m = 19.2$ and $.12*16 = 8*.24\approx 2$ and $0.001 = .065$ is bit too small. But too small is okay as $(400 + 10m)^3 < (400 + 10m +2)(400 + 10m + 4) (400 + 10m +6)$.

So the middle digit is $4$. .... that is if there is any solution.

As for finding the missing digits... well it's tedious to multiply $442*444*446$ on paper... but it's not undoable.

Answer 2

Having gotten that the numbers are $4*2,4*4,4*6$ I would just search. We can ignore the offsets of $2$ and look for $4*4^3\sim 87*****8$ I would start with $5$ because it is the center of the range and because I know that $45^2=2025$. Multiplying by another $45$ gets us $9$ in the leading digit, so that is barely too large. Next I would try $442\cdot 444\cdot 446$ and hit gold.

Answer 3

Idea:

We have $$(2n-2)2n(2n+2) = 87*****8$$

Since $$8(n-1)^3\leq 8n(n^2-1) \leq 8n^3 $$ we have

$$(n-1)^3\leq {87*****8 \over 8} \leq n^3 $$

So $$(*)\;\;\;\;10875001\leq n^3 \implies 8\cdot 10^6<n \implies 200<n$$

(but clearly if we observe (*) we can do better estimate to $n>220$) and $$(n-1)^3\leq 11\cdot 10^6\implies n-1< \sqrt[3]{11}\cdot 100 <2,3\cdot 100\implies n \leq 230$$

Answer 4

Letting the three numbers be $n-2,n,n+2$ we see that the product is $n^3-4n$. Studying $n^3-4n\pmod {10}$ tells us that the last digit must be a $4$ for the product to end in $8$.

Ignoring the $4n$ we see that we must have $$n^3\approx 87\times 10^6\implies n\approx \sqrt[3] {87}\times 100$$

Clearly $4<\sqrt[3] {87}<5$ so we must have $n\in \{404,414,\cdots, 494\}$. A quick computation shows that $$4.4<\sqrt[3] {87}<4.5$$ which implies $$n=444$$ and we are done.

Answer 5

There are many techniques used for solving these kind of problems mentally. Most require practice and skill at using them, coupled with a much better than average mental arithmetic ability.

You had identified the first and last digits fairly easily, with the first two digits being between $40$ and $50$. One technique for identifying the second digit is assessing how close to the result $40$ times $50$ times $45$ is (close to $45^3$) half way between $40$ and $50$. This is easily done mentally and is $90 000$ which is too big by $3000$. $40^3$ is $64 000$, $45^3$ is close to $90 000$ so we have a $5$ digit spread of $26 000$ or about $5000$ per digit. Being too big by only $3000$, the first two digits can't be anything but $1$ less than $45$.

Now, the explanation is $100$ times longer than the actual mental computation. Finding cube roots of $6$ digit numbers with a similar mental approach, acquired with practice, can be done in a few seconds.

The cube root of $681472$ has a last digit of 8 and a solution $> 80 (512000)$ but $< 90 (729000)$ hence $88$. It is based on mental interpolation of fitting a solution between a high and low easily calculated estimate, whereby the margins of difference are big and easily estimated. Whatever can be done mentally, I'm sure could be done by average mathematicians with pencil and paper.