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Any quick way to solve this besides brute force (keep dividing by 2)??

Sorry, no OA as I just made up this problem. Trying to hammer in the concept

for this specific case, 5940 , take out 0 ---> 594 , notice that 9=5+4 ----> it must be divisible by 11 , quickly do that , get 11*54 ....54 is divisible by 2 ---> 27 left ...we no for sure, no other prime can occur ---> confirm it's 11.

Laxie> Thanx, I will remember that time saving tip on how to divide by 11.

What if the problem were stated this way:

What is the largest prime number that can divide evenly into the product of all integers 11 to 19, inclusive?

Any tricks of the trade to share??

That is to find the prime which is the closest to the bigger end of the product. In this example 11 to 19, 19 itself is a prime ,thus the answer is 19 . Consider other example: 11 to 48, then the answer is 47.
I think to find prime promply , you should remember some signs of divisibility:
11: abc --> b= a+c
8 : abcde---->cde is divisible by 8
there're still some, my head is heavy now, i can't recall, i'll let you know soon

Laxie> Thanx, I will remember that time saving tip on how to divide by 11.

What if the problem were stated this way:

What is the largest prime number that can divide evenly into the product of all integers 11 to 19, inclusive?

Any tricks of the trade to share??

Hi GMATT73, I hope I got your question right. I suppose you're asking what is the largest prime number that can divide evenly into this product:

11 * 12 * 13 * 14 * 15 * 16 * 17 * 18 * 19.

We know for 12,14,15,16,18, they can be broken down into smaller prime numbers like 2, 3, 5 etc. So forget about them. The remaining are all prime numbers 11,13,17 and 19. The largest prime number that can therefore divide into the product is 19.

Sorry I couldn't reply earlier... I'm trying to make myself interested in working today. I just feel so tired....

Re: Arithmetic progression divisor [#permalink]
16 Oct 2005, 06:26

GMATT73 wrote:

What is the largest prime number that can divide evenly into the sum of all integers 11 to 109, inclusive? From arithmetic progression, I get: (First+Last)*(Last-First+1)/2--------> (11+109)*(109-11+1)/2 simplified to 120*99/2------->99*60= 5940 Any quick way to solve this besides brute force (keep dividing by 2)?? Sorry, no OA as I just made up this problem. Trying to hammer in the concept

i think factorization is the best approach but the question is how to do factorization fast?

lets try sum is 5940. here we donot need to factorize 5940. we can take out 10 (which has 1, 2 and 5 as factors) as a factor of 5940 because it makes easy to factorize 594. now lets get the factors of 594.
594=1x2x3x3x3x11.

5940=1x2x2x3x3x3x5x11.

so 11 is the largest prime number.............

gmatclubot

Re: Arithmetic progression divisor
[#permalink]
16 Oct 2005, 06:26

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...