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# What is the largest prime number that can divide evenly into

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What is the largest prime number that can divide evenly into [#permalink]

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16 Oct 2005, 01:02
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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

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16 Oct 2005, 01:33
# of integers = 109-11 + 1 = 99
Sum = 99/2 (11 + 109) = 5940

Prime Factorizatoin:
= 594 * 10 = 297 * 2 * 5 * 2
= 3 * 99 * 2^2 * 5
= 3^2 * 33 * 2^2 *5
= 3^3 * 11 * 2^2 *5

Biggest Prime Factor = 11

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Current Student
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16 Oct 2005, 01:51
Wilfred> you broke the beast down by dividing by 10 and then 99. Didn`t see that. Cool.

What if the same problem read:

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

Any similiar sort of formula we can apply?

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16 Oct 2005, 01:54
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

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.

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Current Student
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16 Oct 2005, 02:05
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??

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16 Oct 2005, 02:16
GMATT73 wrote:
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

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16 Oct 2005, 03:11
GMATT73 wrote:
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....

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16 Oct 2005, 03:24
Some extra common divisibility tests:

Divisible by 3 if the sum of the numbers is divisble by 3.
Ex: 342 (3+4+2=9)

Divisible by 4 if the last two digits are divisible by 4.
Ex: 1748 (48/4=12)

Divisible by 6 if it is divisible by both 2 and 3.
Ex: 4326

Divisible by 9 if the sum of the digits is divisible by 9.
Ex: 22428 (2+2+4+2+8=18)

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16 Oct 2005, 04:01
Divisible by 7 if the last digit can be doubled and subtracted from the rest of the number to get a factor of 7.

Ex: 203----->3*2=6 (20-6=14)

Divisible by 12 if it is divisible by both 3 and 4.

Ex: 324 (3+2+4=9 and 24/4)

Divisible by 13 if the last digit can be deleted, multiplied by 9, and subtracted from the rest of the number to get a factor of 13.

Ex. 325----->9*5=45 (45-32=13)

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16 Oct 2005, 07:26
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.............

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Re: Arithmetic progression divisor   [#permalink] 16 Oct 2005, 07:26
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# What is the largest prime number that can divide evenly into

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