The "prime sum" of an integer n greater than 1 is the sum of all the

Question

The "prime sum" of an integer n greater than 1 is the sum of all the prime factors of n, including repetitions. For example , the prime sum of 12 is 7, since 12 = 2 x 2 x 3 and 2 +2 + 3 = 7. For which of the following integers is the prime sum greater than 35 ?

(A) 440
(b) 512
(C) 620
(D) 700
(E) 750

(A) 440
(b) 512
(C) 620
(D) 700
(E) 750

Bunuel · Accepted Answer

Good question. +1.

On the GMAT, answer choices are always in ascending/descending order, so trying option C firsts gives an idea which direction to go next if C is not correct.

WoundedTiger · Answer

The "prime sum" of an integer n greater than 1 is the sum of all the prime factors of n, including repetitions. For example , the prime sum of 12 is 7, since 12 = 2 x 2 x 3 and 2 +2 + 3 = 7. For which of the following integers is the prime sum greater than 35 ?

(A) 440
(b) 512
(C) 620
(D) 700
(E) 750

(A) 440
(b) 512
(C) 620
(D) 700
(E) 750

Before getting down to solving on careful observation B(2^9, So sum is 18),D(7*100----> 7*2^2*5^2) and E(250*3---->5^3*2*3) can be ruled out

C looked good cause 620 is multiple of 31 so ideally should be the closest one.Always start with C and then decide to move on.

Ans is C

Bunuel · Answer

SOLUTION

The "prime sum" of an integer n greater than 1 is the sum of all the prime factors of n, including repetitions. For example , the prime sum of 12 is 7, since 12 = 2 x 2 x 3 and 2 +2 + 3 = 7. For which of the following integers is the prime sum greater than 35 ?

(A) 440
(b) 512
(C) 620
(D) 700
(E) 750

(A) 440
(B) 512
(C) 620
(D) 700
(E) 750

Start by testing the middle option:

(C) 620 = 2*2*5*31, hence the "prime sum" of 620 is 2 + 2 + 5 + 31 = 40 > 35. Since there can be only one correct answer, then it must be C.

Answer: C.

AKG1593 · Answer

Ans C

Solved it the prime factorization way
440= 2*2*2*5*11 Sum=22
512= 2^6 Sum=2*6=12
620=2*2*5*31 Sum=40>35
Hence correct.

KarishmaB · Answer

A few ideas that could help you tackle such a question quickly:

Large numbers are made in two ways:

Either by taking small prime numbers and raising them to higher powers or taking large prime numbers in the first place.

31*2 = 62 (large prime number)
2^6 = 64 (many small prime numbers)

To get a sum as large as 35 or more, you would need some large prime numbers. So options such as 512, 700 and 750 which have small prime numbers as factors should be ignored first. 440 also splits into 11, 4( which is 2*2), 2, 5 - relatively small prime numbers.

You are left with 620 which is 31*2*2*5 (sum greater than 35)

Answer (C)

Ergenekon · Answer

Bunuel, you wrote that we shoud start by middle option. Can you explain the reason? Thanks.

BrentGMATPrepNow · Answer

This question requires us to find the  prime factorization  of the answer choices

A. 440 = (2)(2)(2)(5)(11). 
PRIME SUM = 2 + 2 + 2 + 5 + 11 = 22

B. 512 = (2)(2)(2)(2)(2)(2)(2)(2)(2)
PRIME SUM = 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 18
 
C. 620 = (2)(2)(5)(31)
PRIME SUM = 2 + 2 + 5 + 31 =   40

STOP

We've found the number that has a prime sum that's greater than 35.

Answer: C

RELATED VIDEO
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ScottTargetTestPrep · Answer

Scanning our answer choices we want to find the number that contains a large prime factor. Thus, considering answer choice C, we have:

620 = 62 x 10 = 31 x 2 x 2 x 5

The sum is 31 + 5 + 2 + 2 = 40.

Answer: C

IanStewart · Answer

Two things about this - in a question like this one, there's no way to know, if you test an answer choice and find that it produces an answer that is too small, whether you should try out a larger or smaller answer choice. There isn't a strong relationship, when numbers are fairly similar in size, between the size of a number and the size of its 'prime sum'. The prime sum of 1,000,000,000, for example, is smaller than the prime sum of 67.

But when you  would  know, after testing a wrong answer choice, whether the right answer is smaller or larger, you also should not test C first. I wouldn't test answer choices at all for a question like the one below (I'd use a quick estimate and units digits), but it illustrates how best to proceed if that's the strategy you decide to use:

If x is positive, and x^2 + x = 5112, what is the value of x?
A) 68 
B) 70
C) 71
D) 72
E) 74

In this question, if x > 0, the larger the value of x, the larger the value of x^2 + x. So if we test an answer choice, and x^2 + x is smaller than 5112, we'll know the right answer is larger than the one we tested. If you decided then to test answer C first:

• 1/5 of the time you'll get lucky, and C will be correct
• 4/5 of the time, C will be wrong, and then you will need to test precisely one other answer choice (e.g. if C is too large, you could test A, say, and that will either be right, or if it's wrong, you'll know B is right)

But if you decide to test B or D first -- say we test B:

• 1/5 of the time you'll get lucky, and B will be correct
• 1/5 of the time, B will be too large, and you will know A is correct without testing any other answer choice
• 3/5 of the time, B will be too small, and then you only need to test D. If that's right, you're done, if D is too large then C is right, and if D is too small, then E is right

So in questions like this, you never need to test more than two answer choices, but you if you test B or D first, rather than C, you need to test only one choice twice as often (2/5 of the time instead of 1/5). 
 
Testing answers isn't a very useful strategy anyway on the GMAT, at least for higher-level test takers, so this isn't as useful to understand as it might seem, but if someone is going to do it, it probably makes sense to know how to do it optimally.

And I'd add that there is a myth I've seen repeated in some prep materials that on "which of the following" questions, the GMAT typically makes the right answer D or E more often, so test takers working in order from A through E spend more time on the question. If you actually look at this empirically -- look at published official questions of this type to see how often each answer choice is correct -- you'll find there's no truth to that myth.

MHIKER · Answer

Back solving starting from option C
620 = 10*62 = 2*5*31*2: sum = 40 could be answer. 
Let's move up.
B = 512 =  2^9 ; sum of the 9 2's =18 less than 35

So,620; 2+2+5+31=40

The answer is C

Nipungupta9081 · Answer

Since all the options are Increasing Order Let's start with C

620 = 2*2*5*31 
Sum of all prime factors = 2+2+5+31 = 40

Well this fits Answer - C

totaltestprepNick · Answer

A) 440 --> the prime factorization of 440 is 2*2*2*5*11, and the sum of those primes is 2+2+2+5+11 = 22, which is less than 35, so A is out
B) 512 --> the prime factorization of 512 is 2^9, and the sum of those primes is 2*9 = 18, which is less than 35, so B is out
C) 620 --> the prime factorization of 620 is 2*2*5*31 and the sum of those primes is 2+2+5+31 = 40, which is greater than 35, so C is our answer and we can stop here

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