1
$\begingroup$

The series of Find a factor puzzle is started by Culver Kwan, and asks the solver to identify a factor of a certain large number within a certain range using some mathematical identities. This should be tagged with , and .


Find a factor of $104060405$ within $10100$ and $11000$, using various mathematical identities. You should not use a computer.

$\endgroup$
2
  • $\begingroup$ Does seeing a factorization count as using an identity? This factors easily if you rot13(oernx vg va gur zvqqyr), so to speak. $\endgroup$
    – msh210
    May 25, 2020 at 3:45
  • $\begingroup$ @msh210 I do not understand what are you saying, but the identity is a Wikipedia page. $\endgroup$ May 25, 2020 at 4:10

2 Answers 2

5
$\begingroup$

To be honest, like msh210 said in the comments, I just saw the factorisation:

$104060405 = 104050000 + 10405 = 10405*(10000+1) = 10405*10001$

You could find it by recognising

most of the fourth row of Pascal's triangle $1 4 6 4 1$, and using
$(x+1)^4=x^4+4x^3+6x^2+4x+1$
with $x=100$ to write the number as
$104060405=101^4+2^2=10201^2+2^2$

And then applying the identity

for the product of sums of two squares:
$10201^2+2^2 = (102*100+1*1)^2+(102*1-100*1)^2 = (102^2+1^2)(100^2+1^2)$

but that seems a bit convoluted to me.

Edit:

The intended trick was to use:

Sophie Germain's identity which is
$x^4+4y^4 = (x^2+2xy+2y^2)(x^2-2xy+2y^2)$
In this case we have $x=101$ and $y=1$, so the factors are
$101^2\pm 202 + 2 = 10203\pm 202$

$\endgroup$
2
  • $\begingroup$ Do you know the Sophie Germain's identity? It is in a small section of a Wikipedia page, but you can see the explanation in other sites. I used this identity to create this problem. Still, your answer is good and a check mark will be rewarded. I will try not to make a question which can be done with very simple factorisation next time. $\endgroup$ May 25, 2020 at 13:20
  • $\begingroup$ @CulverKwan I was not really aware of that identity. I have no doubt seen it before, but it is not one that I have ever needed so it is not in my arsenal as it were. $\endgroup$ May 25, 2020 at 14:10
1
$\begingroup$

The number trivially splits into

$5 \times 20812081 = 5 \times 10001 \times 2081 = 10001 \times 10405$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.