Sir ... is the answer ... 250501 ... ???
1002004008016032 has a prime factor p>250000. Find it.
Source(David A Santos)
Hint: (a^3-b^3)(a^3+b^3)
Even though I have put it in the class ix - x level.. it needs a bit of application which is not generally known at ix-x level.
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9 Answers
My logic ....
Consider the first 7 digits of the no. , i.e., 1002004
If we multiply 1002004 by 8 ... it gives a result equal to the 9 left out digits of the no., i.e., 008016032. Hence 1002004 must be a factor of the no., i.e., 1002004008016032 [ factorisation gives 1002004 and 1002 as factors ... ]
Now ... 1002004 can be still factorised further ... into 4 and 250501 ... 4 can be further factorised ... but 250501 can not be further factorised ... Hence 250501 is the reqd. PRIME FACTOR of the no. > 250000
very good work arka..
i had a slightly different proof in mind... but yours is better..
Sir ... please tell ur proof too ... would love to learn that too ...
1002004008016032=1x1015+2x1012+4x109+8x106+16x103+32
1002004008016032=20x1015+2x1012+22x109+23x106+24x103+25
=25x5006-1500-1
Now there is 5006-1 which we factorize further... to get to where arka has reached..
* I think this is the same proof as given in the book too!
Wow .. sir ... the proof is wonderful .... pata nehi mere bheje me kyun nahi aya ...