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Re M2617 [#permalink]
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16 Sep 2014, 01:25
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Official Solution:The numbers {1, 3, 6, 7, 7, 7} are used to form three 2digit numbers. If the sum of these three numbers is a prime number \(p\), what is the largest possible value of \(p\)? A. 97 B. 151 C. 209 D. 211 E. 219 What is the largest possible sum of these three numbers that we can form? Maximize the first digit: \(76+73+71=220=even\), so not a prime. Let's try next largest sum, switch digits in 76 and we'll get: \(67+73+71=211=prime\). Answer: D
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Re: M2617 [#permalink]
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18 Mar 2015, 10:20
When we say that we will make three 2 digit numbers from (1,3,6,7,7,7), it is clear that we have to use these numbers. Besides, notice carefully that 7 is already repeating. So, it is clear that we have to make three numbers by using each and every given number once because there are 6 numbers.
Actually we are not doing hit and trial. Our target is to make highest possible numbers. So we are carefully picking digits to make largest possible numbers.



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Re: M2617 [#permalink]
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27 Jun 2015, 14:08
Is there a better way to find out how 3 numbers will add to a prime? Obviously 1 of 3 or 3 of 3 have to be odd, but anything else?



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Re M2617 [#permalink]
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09 Mar 2016, 11:13
why can't any no be used repetitively?



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Re: M2617 [#permalink]
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09 Mar 2016, 11:29
pksam wrote: why can't any no be used repetitively? We can use numbers from the given set only, which is {1, 3, 6, 7, 7, 7}. So, we can use 1, 3, and 6 only once and 7 thrice.
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Re: M2617 [#permalink]
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11 Apr 2016, 13:11
How are you supposed to know 211 is prime is 2 min? I have no idea. Yes you can check numbers up to 20, but how do we know?



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12 Apr 2016, 06:57



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12 Apr 2016, 07:04
Thank you so much for your reply. I want to start off by thanking you for the incredible content you have posted here for eveyone to use. Your solutions are simple and witty. I really appreciate your style and I will you all the best!!
I did take a look at the post that you referred to. I do understand how we typically carry out prime factorizations, but in this case, the factor could 29 or 57 for example, in that case, it is extremely difficult to know if it is a prime number or not. My question really is: prime number don't have a formula, so how do we know if such a large number is a prime without checking if it's divisible by a bunch of numbers, which makes the questions impossible.
Thank you in advance and keep rocking!!



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NothingComes3asy wrote: How are you supposed to know 211 is prime is 2 min? I have no idea. Yes you can check numbers up to 20, but how do we know? For primality, all you need to do is to check for divisibility of the number in question by prime numbers from 2 to \(\sqrt{number}\) Example, for checking primality of 211 > \(\sqrt {211}\) \(\approx\) 15 (but less than 15, you need to remember square of numbers 1 to 20 and cubes of 1 to 10 for GMAT). Thus you need to check divisibility of 211 by prime numbers from 2 to 14 (=2,3,5,7,11,13) For this, remember the divisibility tests > 211 is NOT even > eliminate 2,4,6,8,10,12,14 211 by 3 > sum of the digits \(\neq\) divisible by 3 > eliminate 3,6,9,12 211 by 5> does not end in 5 or 0 > eliminate 5 and 10 211 by 7 > 210 is clearly divisible by 7 > 211 can not be divisible by 7 > eliminate 7 and 14. 211 by 11 > sum of odd place digits  sum of even place digits = 2+11 = 2 = NOT a multiple of 11 > eliminate 11 211 by 13 > regular division is the only way > 130 is a multiple of 13 and 211 is 81 more than 13 > as 81 is NOT a multiple of 13 > eliminate 13. Thus, 211 IS a prime number. Hope this helps.



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Re: M2617 [#permalink]
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12 Apr 2016, 07:09
NothingComes3asy wrote: Thank you so much for your reply. I want to start off by thanking you for the incredible content you have posted here for eveyone to use. Your solutions are simple and witty. I really appreciate your style and I will you all the best!!
I did take a look at the post that you referred to. I do understand how we typically carry out prime factorizations, but in this case, the factor could 29 or 57 for example, in that case, it is extremely difficult to know if it is a prime number or not. My question really is: prime number don't have a formula, so how do we know if such a large number is a prime without checking if it's divisible by a bunch of numbers, which makes the questions impossible.
Thank you in advance and keep rocking!! To check whether 211 is prime you should try dividing by prime numbers up to \(\sqrt{211}\approx{14.5}\), so you should try dividing 211 by 2, 3, 5, 7, 11 and 13.
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Re: M2617 [#permalink]
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27 Jul 2016, 00:37
An alternate way to solve The numbers {1, 3, 6, 7, 7, 7} are to be used to form three 2digit numbers. Let the numbers be AB, CD, EF so their sum can be expressed as (A+C+E)*10 + (B+D+F). In the sum (A+C+E)*10 + (B+D+F), the first term ends with zero (multiplying by "0"), so the units digit is not impacted by the first term. As we want to maximize the number, we should maximize the total for (A+C+E). So we can use the numbers {7, 7, 7}. This leaves out {1, 3, 6}. The total of units digits {1, 3, 6} ends in zero, hence cannot be a prime [A number > 10, ending with 0, 5, or even cannot be prime]. So lets switch a 7 from 10's with the next largest number i.e., 6. The sets are {6, 7, 7} and {1, 3, 7}. The units digit are now {1, 3, 7} = 11, ends with 1. So can be prime. The total now is {6 + 7 + 7} *10 + {1 + 3 + 7} = 20*10 + 11 = 211, which is prime. Thanks
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Re: M2617 [#permalink]
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01 Aug 2016, 15:05
In case of prime number with number of digits > 1, I think unit digit of number will have to be 1,3,7 or 9. It cannot end with 5 and (no chance for 2 as well). We can use this point as well while selecting the numbers.
Please correct me if we cannot use this point.



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Re: M2617 [#permalink]
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02 Aug 2016, 00:23



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Re: M2617 [#permalink]
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02 Aug 2016, 11:18
Bunuel wrote: PerseveranceWins wrote: In case of prime number with number of digits > 1, I think unit digit of number will have to be 1,3,7 or 9. It cannot end with 5 and (no chance for 2 as well). We can use this point as well while selecting the numbers.
Please correct me if we cannot use this point. ______________ Yes, that's correct. Thanks for confirming.



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Re: M2617 [#permalink]
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14 Dec 2016, 03:37
Hi Buenel,
The numbers {1, 3, 6, 7, 7, 7} are used to form three 2digit numbers. If the sum of these three numbers is a prime number pp, what is the largest possible value of pp?
Do such type of questions mean that we have to use all the numbers given?
I know this is not prime, but if 77+77+77 =prime, could not that be the correct answer?
Many Thanks.



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Re: M2617 [#permalink]
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14 Dec 2016, 04:18
dsheth7 wrote: Hi Buenel,
The numbers {1, 3, 6, 7, 7, 7} are used to form three 2digit numbers. If the sum of these three numbers is a prime number pp, what is the largest possible value of pp?
Do such type of questions mean that we have to use all the numbers given?
I know this is not prime, but if 77+77+77 =prime, could not that be the correct answer?
Many Thanks. I understand what you mean. The answer is no, you have only three 7's, so that's the maximum number of 7's you can use to form a number.
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Re: M2617 [#permalink]
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02 Feb 2017, 22:23
I'm not sure if this approach is correct but I will explain my reasoning.
The sum of all the number in the set add up to 31. Now 3+1=4 (sum of digits) For every possible combination of different 2 digit numbers using the numbers from the given set, the sum of digits when reduced to a single digit must be 4.
A; 97 > 9+7 = 16 > 1+6=7 B: 151 > 1+5+1 = 7 C: 209 > 2+0+9 = 11 > 1+1 = 2 D: 211  > 2+1+1 = 4 (matches) E: 219 > 2+1+9 = 12 > 1+2
Of course, if there were two options which summed up to 4, you'd have to make a guess. But I suppose an educated guess is better than a random one.



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Re: M2617 [#permalink]
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27 Mar 2017, 10:50
My alternative approach:
{1, 3, 6, 7, 7, 7} > we need a prime number > each of the 3 numbers has to be odd (as there is only one even digit in the set), thus, the biggest odd numbers are:
77 73 61 ===== = 211 (prime)



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Re: M2617 [#permalink]
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01 May 2017, 20:50
The Unit digit has to be 1,7, or 9 (from the answer choice) the only combo that will give you one of them is 7,3,1 . Just take it from there. so you have 7 3 1
only make sense to put 7 with 7 7 with 3 and 6 with 1 in order to get the max.







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