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There are four consecutive prime numbers written in

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There are four consecutive prime numbers written in [#permalink] New post 26 Dec 2005, 13:41
There are four consecutive prime numbers written in ascending order. The product of the first three is 4199 and product of last three is 7429. What is the largest of the four numbers and what is the least of four numbers?
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Re: Number Theory Challenge [#permalink] New post 26 Dec 2005, 15:54
arungovind wrote:
There are four consecutive prime numbers written in ascending order. The product of the first three is 4199 and product of last three is 7429. What is the largest of the four numbers and what is the least of four numbers?


13, 17, 19 and 23.

the cle is 9, the last digits of the product of the primes. we need primes whose unit digits give 9 as unit digit of their products.
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 [#permalink] New post 26 Dec 2005, 15:56
Prime factorise the numbers 4199 & 7429.

We get the prime numbers 13,17,19,23

So the least of 4 numbers is 13
and the largest number is 23
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Re: Number Theory Challenge [#permalink] New post 26 Dec 2005, 16:05
arungovind wrote:
There are four consecutive prime numbers written in ascending order. The product of the first three is 4199 and product of last three is 7429. What is the largest of the four numbers and what is the least of four numbers?


The prime numbers less than 50 are...

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41...etc.

First three numbers must be around approx 20 ish for the product to be around 4000.

13*17*19 = 4,199
17*19*23 = 7,429

So largest of the four ==> 23
Smallest of the four ==> 13

I used a little intuition and brute force :roll:
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Re: Number Theory Challenge [#permalink] New post 26 Dec 2005, 21:13
a*b*c = 4199 , b*c*d= 7429 ---> (b*c) * ( d-a) = 3230 = 323*10
Since b*c can't be a multiple of 10 ---> d-a must be a multiple of 10
since 323 has no factors of 2,3,4,5 ,6, 7 (this we can judge by using divisibility tricks) ---> The best d-a is 10 (coz if d-a = 8 or larger ----> there are A LOT other primes in between ----> violate the "consecutive" characteristics of this sequence)

we have 20*20*20= 8000>> 4199 ----> the least number must be < 20
1) a=2 ---> d=12 , not a prime --> out
2) a=3 ---> d= 13, there're 3 primes in between ---> out
3) a=5 --> d=15 , not a prime ---> out
4) a=7 -->d=17 ---> 7,11,13,17 ...but 7*11*13 has unit digit of 1 instead of 9 ---> out
5) a=11 -->d=21 , not a prime --> out
6) a= 13 ---> d=23 ---> 13,17,19,23 : WOW!
7) a= 17 ---> d=27, not a prime --> out
8) a= 19 ---> d=29 ---> there's only 1 prime in between ---> out

So the best sequence is 13,17,19,23.

It looks redundant but in fact, the reasoning process is very short and quick.
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Re: Number Theory Challenge [#permalink] New post 26 Dec 2005, 21:17
TeHCM wrote:
arungovind wrote:
There are four consecutive prime numbers written in ascending order. The product of the first three is 4199 and product of last three is 7429. What is the largest of the four numbers and what is the least of four numbers?


The prime numbers less than 50 are...

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41...etc.

First three numbers must be around approx 20 ish for the product to be around 4000.

13*17*19 = 4,199
17*19*23 = 7,429

So largest of the four ==> 23
Smallest of the four ==> 13

I used a little intuition and brute force :roll:


WOW, easy and nice way to solve this problem. 8-)
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 [#permalink] New post 27 Dec 2005, 10:02
Did this an old timer way. Got first 13 and the rest just came.

13, 17, 19, 23
  [#permalink] 27 Dec 2005, 10:02
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