M26-17

Question

What is the largest prime number that can be obtained by adding three 2-digit numbers formed from the list {1, 3, 6, 7, 7, 7}?

A. 97
B. 151
C. 209
D. 211
E. 219

Bunuel · Accepted Answer

Official Solution:

What is the largest prime number that can be obtained by adding three 2-digit numbers formed from the list {1, 3, 6, 7, 7, 7}?

A. 97
B. 151
C. 209
D. 211
E. 219

What is the largest possible sum of three 2-digit numbers that can be formed using the numbers {1, 3, 6, 7, 7, 7}? To find the maximum sum, we start by maximizing the first digit of each number. We try 76, 73, and 71, which gives us a sum of 220, but since it's even, it cannot be a prime number.
 
Next, we try the next largest sum by switching the digits in 76 to get 67, giving us the numbers 67, 73, and 71. We then need to check whether the sum of these three numbers, which is 211, is a prime number.
 
The most efficient way to prove that a large enough positive integer  x  is prime, is to prove that it cannot be divided by any primes up to  \sqrt{x} . Therefore, to prove that 211 is prime, we need to prove that it cannot be divided by any primes up to  \sqrt{211} , which is less than 15 (since  15^2 = 225 ). It is easy to see that 211 is not divisible by 2, 3, or 5. Therefore, we need to check divisibility by 7, 11, and 13. To check whether 211 is divisible by any of the primes 7, 11, or 13, we find easily identifiable multiples of these primes close to 211. We find that 210 = 7*30 is a multiple of 7, so 211 is not divisible by 7. 220 = 11*20 is a multiple of 11, so 211 is not divisible by 11. 260 = 13*20 is a multiple of 13, so is 260 - 13*4 = 208, thus 211 is not divisible by 13 either. Therefore, 211 is not divisible by any primes up to  \sqrt{211} , so it must be a prime number.

Answer: D

Gemmie · Answer

From the answer choices

1. The largest one, 219, is divisible by 3 => Not a prime

2. Check the 2nd largest, 211

211 < 225 = 15^2

Check whether 211 is divisible by each prime smaller than 15 => No
=> 211 is a prime

For the unit digit of the sume of 3 numbers to be 1
=> 1, 3 and 7

=> 71, 73, 67 or 61, 73, 77 or 71, 63, 77
=> We find at least a possible combination for 211 to be sum

=> Answer is 211

NothingComes3asy · Answer

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?

ENGRTOMBA2018 · Answer

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  
eq  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+1-1| = 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.

Aadi01 · Answer

you can use following formula to determine validity of prime
for any prime no>3 it can expressed as p=6n+1 or p=6n-1 ; n is the +ve integer value

.

Bunuel · Answer

First of all there is no known formula of prime numbers.

Next: 
Any prime number p, which is greater than 3, when divided by 6 can only give the remainder of 1 or 5 (remainder cannot be 2 or 4 as in this case p would be even and the remainder cannot be 3 as in this case p would be divisible by 3).

So, any prime number p, which is greater than 3, could be expressed as  p=6n+1  or  p=6n+5  or  p=6n-1 , where n is an integer greater than 1.

But: 
 Not all number which yield a remainder of 1 or 5 upon division by 6 are primes , so vise-versa of the above property is not true. For example 25 yields the remainder of 1 upon division be 6 and it's not a prime number.

Hope it's clear.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

rickyric395 · Answer

Hey  Bunuel , why have you not considered repeated numbers i.e   (cause if all three are same then it won't be prime)? Cause the question didn't mentioned anything about repeated numbers.

Bunuel · Answer

First, notice that we have three 7s, not four. We aim to get the highest total from three 2-digit numbers, so we need to increase the tens place in each number. That's why we're using all three 7s in the tens place.

vn049636 · Answer

I think this is a  high-quality  question and I agree with explanation.

rish_dutton · Answer

I don't think this question is up to the GMAT OG standards from the 100s of OG quant questions I have solved. I would have never had to ask myself whether or not repetition is allowed while building numbers from the set. You could assume either way equally.

I get it that the indication that repetition is not allowed is that there are repeated numbers in the set but even so, it should have made it explicitly clear that repetition is not allowed. This question deviates from testing the logical skills and goes into the territory of confusing question comprehension.

Bunuel · Answer

I don't agree. I think the question is fine. The list {1, 3, 6, 7, 7, 7}, comprising  six digits  where 1, 3, and 6 appear once and 7 appears three times, aligns with the requirement to form three 2-digit numbers, utilizing all  six digits . This setup clearly indicates that 1, 3, and 6 should be used once each and 7 three times. The specific composition of the list is intentional. If it were the case that we could use digits in the way you think, it would be just said '...using digits 1, 3, 6, and 7'. This specific list structure and the matching number of digits required for the solution point directly to how the numbers should be formed and used in the solution.

anish0953 · Answer

good and high quality question , just that i find it hard to find sol in under 2 min

rak08 · Answer

what if this is my approach
the largest number in the option is 219, which is not prime as divisible by 3
then i jump to 211 ~ not divisble by 3,5,2,7,9,11,23 so i take it as prime
and then i recheck if time permits
211-73= 138 - 71 = 67

Bunuel · Answer

Makes sense.

But you’d still need to check whether 211 can actually be formed as the sum, because there is one more prime in the options, 151.

ZaraBee · Answer

Hi there, I assumed that all options (except) E are prime numbers using the divisibility methods. E.g. D (211) is not divisible by 2, 3, 4, 5, 6, 7, 8, 9, 10 and then worked backwards to find 3 digit combination to arrive at 211. Is my method correct?

Bunuel · Answer

If you first check which choices are prime, then take the largest prime among them and check whether you can form valid numbers that add up to it, then OK. However, I find that approach a bit odd.

P.S. C and E are not prime.

Vivek1707 · Answer

I did not quite understand the question. The quesiton needs to clearly specify that repetitions are not allowed. I started testing values such s 77, 77 73 and then 77 77 67 etc ... ended up using up alot of time.

Bunuel · Answer

Your doubt is addressed above. Please review.

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