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

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.

• Writing the three numbers as $n-2,n,n+2$ the product is $n^3-4n$. Then solving $n^3-4n=87\times 10^6$ yields $n=443.11$ so you know where you need to start looking ($n=444$ would be the very first number to try). – lulu Jul 1 '18 at 19:37
• @lulu How did you solve that without a calculator? – Topa Jul 1 '18 at 19:39
• I would use THIS and THIS as a starting point. There are various ways to approximate these numbers without a calculator. – Bumblebee Jul 1 '18 at 19:39
• Oh, I missed that requirement. Well, dropping the $4n$ term changes almost nothing and then you get $n\approx 443$ pretty easily. – lulu Jul 1 '18 at 19:41
• Topa, it is easy to find $4.4<\sqrt[3]{87}<4.5$ just by testing values. Then just check manually $n=441,\ldots,449$ – Clayton Jul 1 '18 at 20:09

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

$n = 400 + m + 4$

$n, n-2, n+2 \approx 4.m\times 10^2$ 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.

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.

• "A quick computation" is not allowed because YOU USED A CALCULATOR! Otherwise it is not quick to get $4.4^{3}$ – Topa Jul 1 '18 at 20:31
• I don't want to argue the point. I find $4.4$ extremely easy to cube (well under $30$ seconds with pencil and paper). Same with $4.5$. Perhaps you do not. To stress: I find those computations much, much easier than a lot of fine tuned thinking about the missing terms. Again, it is obvious that complexity is, to some extent, subjective and I can see where people might prefer alternate means. – lulu Jul 1 '18 at 20:40
• $(4 + m)^3 = 64 + 48m + 12m^2 + m^3 \approx 87$ means $48m$ is slightly less than $23$ so $m=.5$ is just too big. $m = .4$ is a bit too small. Which is what we need. – fleablood Jul 1 '18 at 22:42

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.

• Ok your answer is right but can we do it without trial and error. Like some relation between the 7 and the tens digit. These numbers happen to be easy to get from trial and error but if we get a case that takes a lot of time then trial and error is not an option to get the answer in a couple of minutes. – Topa Jul 1 '18 at 20:29
• I don't know the digit-by-digit process for cube roots. If you are supposed to solve problems like this without a calculator they need to be sized so the amount of calculation is reasonable. I don't see anything wrong with trial and error. Here I think it is the fastest route to a solution. I don't think there is a general approach to problems like this. They often depend on knowing arithmetic facts. This one is much easier if you know $4^3=64,5^3=125$ instantly but not everybody does. – Ross Millikan Jul 1 '18 at 20:36
• The course I am taking solves problems without trial and error. So taking trial and error approach would defeat the point of the course. Like there are questions that would take months to solve with trial and error. – Topa Jul 1 '18 at 20:42

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 $$(n-1)^3\leq 11\cdot 10^6\implies n-1< \sqrt[3]{11}\cdot 100 <2,3\cdot 100\implies n \leq 230$$

• There is no idea at all in here. You just bound 2n between 400 and 600. I bounded 2n between 400 and 500 in the OP. This answer doesn't solve anything. I need to get the tens digit without a calculator. – Topa Jul 1 '18 at 19:55
• Perhaps you are right, however it does lead me to a finally answer. – Aqua Jul 1 '18 at 20:11

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.