Largest 3-Digit Divisor

At first, we notice that $639+385 = 1024$ , so

$N = 1024^3 - 639^3- 385^3 = (639+385)^3 - 639^3- 385^3$ $= 3 . 639 .385 (639+385) = 3.639.385.1024$

Now, $1024 = 2^{10}, 639 = 3^2.71, 385 = 5.7.11$ , so $N = 2^{10}.3^3.5.7.11.71$

Now the largest multiple of $71$ below $1000$ is $994 = 2.7 . 71$ , and thus $994$ is a divisor of $N$ . Now if we check the 3 digit numbers greater than $994$ , i.e. $995-999$ , then we are done.

$995 = 5.199$

$996= 2^2.3.83$

$997$ is a prime.

$998 = 2.499$

$999 = 3^3. 37$

and thus none of them is a divisor of $N$ , and thus the answer is $994$ .

I quite liked this question, because although it can be solved by spamming the large number to get its prime factorisation, which comes out as: $2^{10} \times 3^3\times 5 \times 7 \times 11 \times 71$ , it is not a nice approach. Now, to get prime divisors etc. of an extremely large number, more often than not, we can factorise . Let us recall some useful factorisation regarding cubes , Now, recall that

$a^3 +b^3 + c^3 = 3 a^2 b+3 a^2 c+3 a b^2+3 a c^2+3 b^2 c+3 b c^2 +6abc$

$= 3abc + 3(a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2)$

$= 3abc + (a+b+c) (a^2+b^2+c^2-a b-a c-b c)$ .

Although this factorisation seems pretty convoluted, the main idea is to factorise out $3abc$ then subbing in $a,b,c = 1$ we can get that $(a+b+c)$ is a factor of the really symmetric expression. Next it's just comparing coefficients by trying out $(a+b+c)(ab+bc+ca)$ and trying to match up the terms.

Now, $1024, 639, 285$ must be chosen for some reason. It is quite motivated to consider $639 + 385$ (hoping that since math is beautiful everything should work out) which voila gives $1024$ . By the above factorisation, if $a+b+c = 0$ then $a^3 + b^3 + c^3 = 3abc$ . In this case, we get:

$N = 1024^3 - 639^3 - 385^3 = 3(1024)(-639)(-385) = 3 \times1024 \times 639 \times 385$

The beauty of factorisations is that it leaves us with smaller expressions/numbers to work with. By general methods (such as sum of digits, etc.) we can determine the factorisations of $639$ and $385$ as $3^2 \times 71$ respectively $5 \times 7 \times 11$ . Obviously, $1024 = 2^{10}$ . So we are back with the clean factorisation noted at the beginning of the solution.

Now, to find the largest (3 -digit) divisor, consider taking out the largest prime divisor of $N$ and multiplying smaller prime divisors. This guarantees a larger factor than, say, $2^9 = 512$ , because exponents increase faster, in other words, make the number bigger (to an extent $> 1000$ which should be avoided). By simple trial and error, we can get the answer is $71 \times 7 \times 2 = \fbox{994}$ .

First notice that $1024 = 639 + 385$ . This suggests that we may want to factor the expression using differences and sums of cubes to get a representation of $N$ which is much easier to factor into primes.

For easier manipulation, let $a = 1024, b = 639, c = 385$ . $a = b + c$ and $a - c = b$ . Then $N = a^3 - b^3 - c^3 = a^3 - (b^3 + c^3) =$ $a^3 - (b + c)(b^2 - bc + c^2) = a^3 - a(b^2 - bc +c^2) =$ $a(a^2 - (b^2 - bc + c^2)) = a(a^2 - c^2 + bc - b^2) =$ $a((a + c)(a - c) + bc - b^2) = ab(a + c + c - b) = ab(3c) =$ $3(1024)(639)(385) = 3(2^{10})(3^2\cdot71)(5\cdot7\cdot11) =$ $2^{10}\cdot3^3\cdot5\cdot7\cdot11\cdot71$ .

Beginning with $999$ and working backwards, we have:

$999 = 3^3\cdot37$

$998 = 2\cdot449$

$997 = 997$

$996 = 2^2\cdot3\cdot83$

$995 = 5\cdot199$

$994 = 2\cdot7\cdot71 | N$

Therefore, $994$ is the largest 3-digit integer that divides $N$ .

$a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc$

Now we just substitute $a = 1024, b = -639, c = -385$ , and find that $a + b + c$ quite conveniently reduces to 0, leaving us with $3 \times 1024 \times -639 \times -385$

We prime factorize this to get

$2^{10} \times 3^3 \times 5 \times 7 \times 11 \times 71$

Now we start backwards from 999, which is the largest 3 digit number

$$ $999 = 9 \times 37 \times 3$ $998 = 2 \times 499$ $997 = ?$ $996 = 4 \times 3 \times 83$ $995 = 5 \times 199$ $994 = 2 \times 7 \times 71$

Now since we do have the number to make up 994, our answer is 994

Use the identity: $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$ .

And re-write the given expression for $N$ as: $N = (1024)^3 + (-639)^3 + (-385)^3$ .

By considering $a = 1024$ , $b = -693$ , and $c = -385$ , it can easily be seen that $a + b + c = 0$ , so that $N = 3 \cdot 1024 \cdot (-639) \cdot (-385)$ .

Factorizing the above expression for $N$ gives: $N = 2^{10} \times 3^3 \times 5 \times 7 \times 11 \times 71$ .

Now, the best way to find the largest 3-digit divisor of $N$ (theoretically) is to find the prime factorization of every number starting from 999 (998, 997, 996, and so on) and compare it with the prime factorization of $N$ . But, it is practically better to do some hit and trial first.

Dividing $999$ by the largest prime factor of $N$ , $71$ , gives about $14.07$ . Which means the largest three digit number which is divisible by $71$ (and is a divisor of $N$ ) is $71 \times 14 = 994$ .

To make sure there isn't any greater number than $994$ which is a divisor of $N$ , it is a good idea to check with the help of the prime factorization of the last five $3$ - digit numbers:

$995 = 5 \times 199$

$996 = 2^2 \times 3 \times 83$

$997 = 997$ ( $997$ is a prime number)

$998 = 2 \times 499$

$999 = 3^3 \times 37$

Comparing these with the prime factorization of $N$ reveals that none of the them are divisors of $N$ , as they have one or the other prime factor uncommon with $N$ .

Which means we can safely say that the largest $3$ - digit divisor of $N$ is $994$ .

$N = 1024^3 - (639 + 385)(639^2 + 385^2 - 385 * 639)$

$= 1024 * (1024^2 - 639^2 + 385 * 639 - 385^2)$

$= 1024 * (385 * 1663 + 385 * 254)$

$= 1024 * 385 * 1917$

$= 1024 * 27 * 5 * 7 * 11 * 71$

$994 = 2 * 7 * 71$ is a divisor of N. Other 3 digit numbers larger than 994 can't be factorized into prime factors of N. Hence, the answer is 994.

We take the given equation and rearrange it to make $N = 1024^3 - (639^3 + 385^3)$ Recall that the sum of cubes equation is $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ . We simplify our expression

$1024^3 - (639+385)(639^2 - (639*385) + 385^2)$

$1024^3 - (1024)(310531)$

$1024*(1024^2 - 310531)$

$1024 * (738045)$

The problem asks for a specific divisor, so prime factorization would be useful. Brute force subtraction is not the best path because if would be hard to factorize such a big number. However, with the above method, we have already factored out $2^{10}$ . Factoring even more, we get

$2^{10} \times 3^3 \times 5 \times 11 \times 71$ . With a few educated guesses, we see that $71 \times 14 = \boxed{994}$

Using a^3 + b^3 + c^3 = 3abc + (a +b +c)(a^2 + b^2 + c^2 - ab - bc - ca)

as a +b +c = 0 here

a^3 + b^3 + c^3 = 3abc

and 1024^3−639^3−385^3 = 3 x 1024 x 639 x 385

                                           = 3^3 x 2^10 x 71 x 11 x 7 x 5

which can be rearranged to get 71 x 2 x 7 as the largest 3 digit number. Hence ans = 994

First, we try to factor this strange expression. There isn't a neat factorization for $a^3-b^3-c^3$ , so we try something else.

Notice that since $639+385=1024$ , $639^3+385^3=(639+385)(639^3-639(385)+385^2)=1024(639^2-639(385)+385^2)$ Then, $N=1024^3-639^3-385^3=1024(1024^2-(639^2-639(385)+385^2))$ Also using $639+385=1024$ , $639^2-639(385)+385^2=(639+385)^2-3(639)(385)=1024^2-3(639)(385)$ and so $N=1024(1024^2-(1024^2-3(639)(385)))=1024(3(639)(385))$ This expression is easy to factor, since we can find $639=3^2\cdot 71$ and $385=5\cdot 7\cdot 11$ . Now, $N=2^{10}\cdot 3^3\cdot 5\cdot 7\cdot 11$ By trial and error, we find that $2\cdot 7\cdot 71=\boxed{994}$ is the largest 3 digit divisor of $N$ .

Motivation: When I noticed that $639+385=1024$ , I realized that I could factor sum of cubes, and then simplify the second factor. The rest was trial and error, trying all possible combinations of factors. In hindsight, it makes some sense to try to see that $7\cdot 71=497$ is very close to 500, and multiplying by 2 gives 994.

First, I used difference of cubes to factor $1024^3-639^3$ into $(1024-639)(1024^2+1024\times 639+639^2)$ . Now, since $1024-639=385$ , I can pull out a 385 so I am left with $1024^2+1024\times 639+639^2-385^2$ . To simplify this inner expression, I factor $1024^2-385^2$ using difference of squares to get $639*1409$ . Now I can pull out a 639 so the total expression at this point is $385\times 639(1409+1024+639)$ . $385\times 639$ factors into $3\times 3\times 5\times 7\times 11\times 71$ and the inner expression add up to $3072$ which factors into $(2^{10})\times 3$ . To achieve the greatest 3 digit divisor, take $2\times 7\times 71=994$ .

[Latex Edits - Calvin]

a^3+b^3+c^3=3abc+(a+b+c)(something) as a+b+c is zero a^3+b^3+c^3=3abc if we write the prime factors of these three numbers we shall find that the highest possible 3 digit number is 2 7 71=994

Factoring the number we get that it is equal to:

(2^10)(3^3)(5)(7)(11)(71)

The largest possible 3-digit number made by these factors is 7x71x2 = 994

CONSIDER a = 639; b =385; a+b= 1024; (a+b)^3-a^3-b^3= 3ab(a+b) = 3 639 385 1024;= (2^10) (3^3) 5 7 11 71 by taking all possible divisors nearer to 1000 we get 71 7 2 =994.

N = $2^{10} \times 3^{3} \times 5 \times 7 \times 11 \times 71$

So the largest 3 digit divisor is less than 1000.

$71 \times 7 \times 2$ = $\boxed{994}$

We note that $1024-639=385$ . Therefore, we have

$1024^3-639^3-385^3=385(1024^2+639^2+1024.639)-385^3$

$=385(1024^2-385^2+639^2+1024.639)$

$=385.639[(1024+385)+639+1024]=385.639.3072$

We use prime factorization to get $385.639.3072=2^{10}.3^3.5.7.11.71.$ . Notice that $2.7.71=994$ and a quick check shows that no 3-digit number bigger than 994 has divisors as subset of divisors of N. Hence 994 is the answer.

since $1024-639-385=0$ , $1024^3-639^3-385^3=3(1024)(-639)(-385)=2^{10}.3^3.5.7.11.71$ . $994=71(14)$ , and checking the numbers $995-999$ shows that $994$ is the max

$(a+b)^3-a^3-b^3=a^3+3a^2b+3ab^2+b^3-a^3-b^3=3a^2b+3ab^2=3ab(a+b)$

N=1024³-(639³+385³)=1024³-(639+385)×(639²-639×385+385²)= 1024×(1024²-639²+385(639-385))= 1024×(385×1663+385×254)= 1024×385×1917= 2^10×3³×5×7×11×71 So we try 999=3³×37 not 998=2×499,997,996=2²×3×83 ,995=5×199 So 994=71×7×2 is solution

Note that $639+385=1024$ .

Let $639$ be $a$ and $385$ be $b$ . Then we have: $N=(a+b)^3-a^3-b^3$

Expanding $(a+b)^3$ using Binomial Theorem : $N=a^3+3a^2b+3ab^2+b^3-a^3-b^3$

Cancelling: $N=3a^2b+3ab^2$

Take out common factor: $N=3ab(a+b)$

Substituting them back: $N=3\times385\times639\times1024$

Note that:

• $385=5\times7\times11$
• $639=3^2\times71$
• $1024=2^{10}$

Therefore: $N=3\times5\times7\times11\times3^2\times71\times2^{10}$

Rearranging: $N=2^{10}\times3^3\times5\times7\times11\times71$

Let us prime factorize every number from $999$ to see if the prime factors is included in the prime factorization of $N$ .

• $999=3^3\times37\ \color{#D61F06}✘$
• $998=2\times499\ \color{#D61F06}✘$
• $997=997\ \color{#D61F06}✘$
• $996=2^2\times3\times83\ \color{#D61F06}✘$
• $995=5\times199\ \color{#D61F06}✘$
• $994=2\times7\times71\ \color{#20A900}✔$

Therefore, the answer is $\fbox{994}$ .

using computer to solve (javascript):

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var n=755758080;
var result;
for(var i=1;i<1000;i++)if(n%i==0)result=i;
console.log(result);

Let $A = 639$ and $B = 385$ , now notice that $A+B=1024$ , thus $N=(a+b)^3 -a^3-b^3$ $N=a^3 + 3a^2b +3ab^2 +b^3 -a^3-b^3$ $N=3ab(a+b)$ $N=3\times 639 \times 385 \times 1024$

By factorising, we have $N=2^{10} \times 3^3 \times 5\times 7\times 11\times 71$ Since we have to find the largest 3 digit divisor, let the largest divisor $P$ . We have to find, make sure we have $P|N$ . Since we have prime $2, 3, 5, 7, 11, 71$ , by a little trial and error, choose any primes that makes easier to continue. For me, I choose $71$ as the first prime, we have $\frac{1000}{7} \leq 14$ . Now, find another factor greater or less than 14.

Tips: do backwards For another factor $14$ , we have $14 | N$

Since $GCF(14, 71) = 1$ , we conclude that

$(2\times 7 \times 71) | N$

$994 | N$

Which satisfis the problem

The interesting thing is that 994 and 9940 are both factors of this number.

Getting to N=3 * 1024 * 639 * 385 is easy i guess (lot of solutions explaining it) But getting to 994 must be kind of trial and error as far as i saw in the solutions i reached the solution this way (i found an interval where to do the trial and error) N= 2^10 * 3^3 * 5 * 7 *11 * 71 [ Factorized format]

Largest possible 3 digit no. = 999 [ which is not a divisor; 9 * 3 * 37 ]

9 * 111 = 999 so closest we can get is 9*110 (which is a divisor of N because 9 * 10 * 11)

So the interval is [990,999) is our place where we can find our solution

To check if any number > 990 is a factor

If it is a divisor D then it should be divisible by some of primes in the factored format set S={ 2,3,5,7,11,71 } Other way: all the divisor's prime factors should be present in N

First way: [Eliminating the primes which cannot be common factors of N,D]

5 cannot be a common factor of N,D because only 995 is possible and also it is not divisible by any other prime factors

995=5 * 199 (199 is a prime)

11 can also not be a common factor of N,D because 990 is a multiple of 11 and the next one will be 1001

3^2 can't be a factor because of the similar reason 990,999 are the factors no other number will have two 3s

Numbers with 3 as factor 993=3*331(a prime),

996 - 2 * 2 * 3 * 83 - X [can see without factorizing it also. only 2s and 3s can be there in the divisors from set S]

So 3 cannot be a factor Only prime factors left are N`=2^8 * 7 * 71 [ 2^10 can't be and 2^9 can't be (for obvious reasons 1024 is too be and 512 cant be multiplied be anything smaller than 2) ] => only 2 possiblities 2^7 * 7=896 and 2 * 7 * 71=994 (Y)

Second way: [Reverse analysis-takes time if the interval is big]

999 - 9 * 3 * 37 -X

998 - 2 * 499(prime) -X

997 - prime - X

996 - 2 * 2 * 3* 83 - X [can see without factorizing it also. only 2s and 3s can be there in the divisors from set S]

995 - (it's not divisible by any prime factor other than 5 from set S)

994 - divisible by 71,2,7

994=2 *7 * 71 [Y Solution]

$Let~~ a=639,~~b=385,~~a+b=1024. \\639=3*3*71=3^2*71,~~385=5*7*11,~~1024=2^{10}...( well~ known.)\\(a+b)^3-a^3-b^3=3ab(a+b)=\large 3*3^2*71*5*7*11*2^{10}\\\text {The two out standing options are }\\99 * 10 =990~~0r~~since~7 * 7=49:- 71 * 7 * 2=\color{#D61F06}{994.}\\ \text{The factors of the devisor must be factors of N}\\995=5 * 199 {\huge~~~~~~~ X},\\996=4 * 3 * 83 {\huge ~~~~~~~X},~~\\997:--3/5/7/11\nmid 997,~~~~~ \dfrac{,997}{71}=14.04,~~but13\nmid 997,{\huge~~~~~~~ X},~~\\998=2 * 499:-3/5/7/11\nmid 499,~~~~~\dfrac{499}{71}=7,{\huge~~~~~~~ X},\\999=27 * 37{\huge~~~~~~~ X},~~~$

Simplified, N = 755,758,080, which has prime factorization of $2^{10} \cdot 3^3 \cdot 5 \cdot 7 \cdot 11 \cdot 71$ . Starting at 999 and working backwards, I looked for the first number which could be made by multiplying a combination of these prime factors. 994 is the first number which meets these conditions as it equals $2 \cdot 7 \cdot 71$

1024×1024×1024-639×639×639-385×385×385=755758080

755758080÷999=756514.59 755758080÷998=757272.63 755758080÷997=758032.18 755758080÷996=758792.25 755758080÷995=759555.86 755758080÷994=760320

So 994 is the answer