1260^4 ??chalenging 1

Question

how any odd and even factors does 1260^4 have??

retro · Answer

1260 = (2^2 * 5 * 7 * 3^2)
1260^4 = (2^8 * 5^4 * 7^4 * 3^8)

This one has (8+1)(4+1)(4+1)(8+1) = 2025 factors.

From this number, all (4+1)(4+1)(8+1) = 225 will be odd since the even part comes from 2^8 alone.

I am slightly unsure of this strategy and I expect someone to correct this if required.

Regards
Rahul

dimri10 · Answer

you are right. it was splendid.
by the way is this a GMATquestion?

sudhir18n · Answer

I dont see any reason for why this cant be...

dimri10 · Answer

it's driving me crazy. i went through kaplan and manhattan and og and never found somthing like that. in a matter of fact most of the hard Q can not be approached in any MGMT/KAPLAN/WHATSEVER except of searching some Arun Sharma's material or other briliant brains from the forums. unless i have seen that Q in some kind of an Indian quantitative book, i surely could'nt solve it.

OldMinolta · Answer

That took a couple of minutes to work through. People struggling with this might like the explanation found for stripping the number down to the prime factors found here. Once you get that part it's pretty intuitive. Thanks for pointing out that you can use this to distinguish the number of odd factors from the even ones!

https://www.mathsisfun.com/prime-factorization.html

dimri10 · Answer

when you learn the concept that:
2 is even and therefore should not be counted. hence you take only 5^4 * 7^4 * 3^8 as factors to dind the odd=225 and the even will be all factors (2025)-225 (odd).

when you know the concept the answer takes no more than 30 sec, but neither kaplan nor MGMT approach this.
i found that at some Indian briliant book named Quantitative workout or something like that.

they make questions like 1010^50+2568987^5+ 345678678^4 look easy to find the reminder when divided by 6.

no kudos for posting the question?

retro · Answer

I'd not expect to find everything everywhere. Some building blocks are given and we may need to construct newer edifices to solve a new problem or a known problem a different way.

In general it is known that any number expressible as a product of primes (or their powers) will have a specific number of factors. Now from within those, the even factors are contributed only by the powers of 2. Hence eliminating the powers of two gives the number of odd factors.

Just remember this: if n = (p1^x1)*(p2^x2)....(pi^xi)...(pn^xn)

where pi is the ith prime and xi is the power to which pi needs to be raised in order to get the original number, then the total number of factors is (x1+1)(x2+2)...(xi+1)...(xn+1)

Regards
Rahul

retro · Answer

Sorry.. forgot adding a couple more of lines:

I also agree that Kaplan's Math workbook does not take you to this level.

Neither does Manhattan. But Manhattan is slightly better in terms of 'providing the building blocks'. They tell you that products of combinations of odd and even numbers and have some exercises to test your agility in that. They also tell you the number of factors (not sure about this last statement) or at least do not leave you at naive a level as Kaplan in this matter. That is one of the reasons I felt my investment in Kaplan books was better spent elsewhere. But the silverlining was that doing Kaplan first pushed me more into questioning myself why I could not take GMAT (by putting very easy questions)

And then when I took the Manhattan stuff, that posed the actual challenges: consistency, conscient answer, application and most important: conquer overconfidence

Regards
Rahul

toughmat · Answer

Good Question. Using prime factorization and odd even rules we can arrive at the solution

1260^4 ??chalenging 1

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