# 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. :(

• Hint 97*93 = (100-3)(100-7) – AD. Aug 26 '10 at 16:19
• @Chiaotzu: You originally used some words "complex," "division-algebra" that have other meanings in mathematics. I removed them so that people won't misunderstand the topic of your question. – Larry Wang Aug 26 '10 at 16:19
• @AD.: I'd be more inclined to try 43*93*47*97=(70-27)(70+23)(70-23)(70+27) because of the pairs that lend themselves to $(a-b)(a+b)=a^2-b^2$ (but I haven't actually done anything with it, just thinking out loud). – Isaac Aug 26 '10 at 16:34
• 3007 = 31*100 - 31*3 – Aryabhata Aug 26 '10 at 16:40
• @Isaac: that is a great way to proceed. Indeed, 43*93*47*97/3007 = 3*43*47 (as others have noted) = 3*(45-2)*(45+2) = 3*(45^2 - 2^2) = 3*((40+5)^2 - 2^2) = 3*(1600+400+25-4) = 3*2000 + 3*25 - 3*4, etc. In general, multiplication can be done quickly by subtracting squares (and factoring can be done by recognizing differences of squares). – whuber Aug 26 '10 at 18:26

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.

• I'm not getting anything. Can you liek "Solve" the question. I know I should not ask someone to solve the problem, but I'm not aware of mod and ≡. The only thing about mod I know is it returns remainder. :\ – Electrifyings Aug 26 '10 at 17:06
• Ok, here is a bit simpler way to state it: since 100A + B = 97A + 3A + B, it follows that 97 divides 100A + B iff 97 divides 3A + B. Now put A,B = 30,7, i.e. the digits of 3007 in radix 100. – Bill Dubuque Aug 26 '10 at 17:18
• chiaotzu: Here's the way one would read the first relation in Bill's answer: saying that 100 is congruent to 3, modulo 97, is the same as saying 100 divided by 97 leaves a remainder of 3. Now, go read the next statements in the same way. – J. M. is a poor mathematician Aug 26 '10 at 17:18
• Whoa! I get it now and its a trickier easiest solution man. Sure it takes a genius to think that way to solve problem quickly, I'll try to learn this technique with some practise. Thanks a lot. And thanks J. Mangaldan for explaining those tricky symbols and bringing it into MY language. :) – Electrifyings Aug 26 '10 at 20:08

(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.

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.

• Thanks man, very well explained. You've applied the same logic that I was looking for. 8 minutes is something I can't afford and I must do it in less than 2 minutes next time. :D Prime factorization is something I missed and didn't checked. 93 is quite tricky number and I never though of 31*3 factors LOL. Thanks once again. :) – Electrifyings Aug 26 '10 at 20:11

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.

• Thanks for answer man, but the question neither gave any options, nor expected approximation as I mentioned in my question. But your method is a perfect workaround for longer calculations and does the dirty work in seconds. :D – Electrifyings Aug 26 '10 at 20:05

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 $X = 6063$.

• Thanks man, didn't get much because I'm new to this "mod" stuff in maths, but I'm sure I'll get used to it soon. :) – Electrifyings Aug 26 '10 at 20:12
• No problem. I'm sort of being facetious here; since you know the quotient is an integer it makes a lot more sense to factor than to do what I did. This is merely by way of illustration of an alternate method. – Qiaochu Yuan Aug 26 '10 at 20:18
• @chiaotzu: I see now that you did not necessarily know that X was an integer. But this is a reasonable guess. – Qiaochu Yuan Aug 26 '10 at 21:08

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

• Appreciate it man. I'm not looking for approximation though. But I'll have a look at your method too. :) – Electrifyings Aug 26 '10 at 20:17
• yes, for exact answer, you can see before doing the aproximation that 9021 = 3 x 3007 so x = 2021*3 = 6063 but this is "luck" for this situation – laurent Aug 26 '10 at 20:48
• Yeah man, I suppose they were asking exact answer because of this "luck" thing. :) – Electrifyings Aug 27 '10 at 3:49

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}