Division with 4 digit number in denominator I've got a question in my task sheet. The question is as follows.
$$
\frac{43\cdot93\cdot47\cdot97}{3007}=X
$$
Find the exact value of $X$. I've tried a lot, but couldn't find easier way to do it without calculator, which of course, is not allowed in exam. There are no options, they're just asking the value of $X$.
Would love if someone could help to give Method to solve the problem. As I said, I know how to solve above problem with the help of calculators and I've already found the factorization with help of calc, but no luck in manual mode. :(
Thanks in advance. :)
 A: HINT $\ \rm mod \; 97\!: \: 100 \;\equiv\; 3 $
Hence $\; 3007 \;\equiv\; 30 \cdot 100 + 7$
$\quad\quad\quad\quad\quad\quad\;\; \;\equiv\; 30 \:\;\cdot\;\: 3 \;\: + \; 7 \;$
$\quad\quad\quad\quad\quad\quad\;\; \;\equiv\; 0$
I.e. cast out 97's in analogy to cast out nines. See also here where I discuss casting out 91's.
A: This must be a multiple choice question that must be done under a time limit... so approximate!
Here we go:
$$\frac{43\cdot 93\cdot 47 \cdot 97}{3007} \approx \frac{50 \cdot 100 \cdot 50 \cdot 100}{3000}$$
We can easily reduce this to $\frac{5 \cdot 100 \cdot5 \cdot 10}{3}$ which is simply $\frac{25000}{3} = 8333 \frac{1}{3}$.  This is not all that satisfactory since the answer is $6063$.
Let's be a little finer with our approximation (while being no more painful):
$$\frac{43\cdot 93\cdot 47 \cdot 97}{3007} \approx \frac{40 \cdot 100 \cdot 50 \cdot 100}{3000}.$$
So, we are now looking at $\frac{40 \cdot 100 \cdot 5}{3}$ which is simply $\frac{20000}{3} = 6666 \frac{2}{3}$.  That should be close enough, but we already have more information than you may think, because the true answer is less than both of these estimates.  
A: (20:30)
First we notice that 43, 47 and 97 are all prime numbers and 93=3x31. Clearly 3 is not a factor of 3007, let's see about the rest using the good ol' Euclidean algorithm.


*

*43 is clearly not (as 43x7=280+21=301 therefore 300=43x6+42, and 427 is clearly not divisible by 43)

*in 47 case, 300=6x(47+3) therefore the reminder is 47-18=29, and again 297 is clearly not divisible by 47

*31 gives us 310=31x10 so 300=9x31+21, and 217 equals to 31x7. So we have one divisor, that is 97x31 = 3007


Therefore the result is 43x47x3 which should be a simple calculation.
(20:38)
8 minutes, would have taken 5 if I'd began with 31 :)
I tried to describe my thoughts into words and be not very formal. I hope it's clear and it gave you some insight about how I solved it.
Addendum:
Say that it was the case where no factors are common, what's then? Then we have two options either use approximations by noticing how 97x93 is (95+2)(95-2) and same with 43x47=(45+2)(45-2) so it'd be simpler to try approximating 45x45x95x95/3000
Or you can take the numbers you've calculated in with the Euclidean algorithm when you checked for common factors and you'd have 3007/43, 3007/47, etc. So you can just multiply their inverses for the result.
A: The numerator is about $16000000$, so the quotient is about $5300$, say plus or minus a thousand.  Now I am going to play some tricks.  
The numerator is $-9 \bmod 50$ and the denominator is $7 \bmod 50$, so $X \equiv 13 \bmod 50$.
The numerator is $6 \bmod 9$ and the denominator is $2 \bmod 9$, so $X \equiv 3 \bmod 9$.  Hence $X \equiv 213 \bmod 450$.  So $X$ can really only be one of $4263, 4713, 5163, 5613, 6063$, and maybe $6513$.
The numerator is $8 \bmod 11$ and the denominator is $4 \bmod 11$, so $X \equiv 2 \bmod 11$.  We now know the value of $X \bmod 4950$, which is more than enough.  Looking through the above list this gives $X = 6063$.
A: You can estimate it this way very quicly:
43*93*47*97 = (50-7).(100-7).(50-3).(100-3) = (50-7).(50-3).(100-7).(100-3)
= (100²/4 - 10.100/2 + 21) * (100² - 10.100 + 21) =
= (100.(25-5)+21) * (100.(100-10)+21) = 
= 2021 * 9021 ≈ 18 000 000 (if needed for aprox: greater error around 20/2000 => 1%

using 3000 instead of 3007 (error 7/3007 around 0,2% so less than 1% previous)

18 000 000 / 3000 = 6000 if needed, you can reduce aproximation 
correcting denominator error (1%) => 6000 + 60 = 6060

A: This can be done the hard way without too much effort.
There is a trick for squaring numbers that end in $5$. Basically
$(10n+5)^2 = 100n(n+1)+25$. A bit more "symbolically", this can be written as 
$[n,5]^2 = [n(n+1),2,5]$.
So $95^2=[9 \cdot 10,2,5]=9025$
and $45^2 = [4 \cdot 5,2,5] = 2025$
Then 
\begin{align}
   43\cdot 93\cdot 47 \cdot 97
   &= (45-2)(45+2) \cdot(95-2)(95+2) \\
   &= (2025-4)(9025-4) \\
   &= 2021 \cdot 9021 \\
   &= (2K + 21)(9K+21) \\
   &= 18K^2 + (9K+2K) \cdot 21 + 441 \\
   &= 18K^2 + 231K + 441 \\
   &= 18,231,441
\end{align}
It isn't that hard to find that $18,231,441 \div 3007 = 6063$

$$\frac{43\cdot93\cdot47\cdot97}{3007}$$
Seeing as how this is a test question, there most probably will be some cancellation going on. The prime factors in the numerator are
$$3,31,43,47,97$$
A quick computation shows that $3007 \div 31 = 97$.
So
\begin{align}
   \frac{43\cdot93\cdot47\cdot97}{3007}
   &= 43 \cdot 3 \cdot 47 \\
   &= 3 \cdot (45-2)(45+2) \\
   &= 3 \cdot 2021 \\
   &= 6063
\end{align}
