What is the value of the positive integer a?

Hi shriramvelamuri,

Yes, there is a pattern involving prime factorization that can help speed you up on this question. While it's a relatively rare pattern, the GMAT might slip in into your Test in 1 question (when you're scoring at a sufficiently high level).

For a positive integer to have EXACTLY 6 factors, the prime-factorization of that number will yield 2 DIFFERENT primes, one of which shows up TWICE.

For example:
12 = (2)(2)(3) = 2 different primes, one of which shows up twice.

Factors of 12: 1,12, 2,6, 3,4

We can use this pattern to quickly find the various integers that fit this pattern (and 'leapfrog' all of the integers that don't). Since the two Facts "hint" at the fact that we're going to be dealing with 2-digit numbers, I'm going to limit my work to those.....

Start with the prime that shows up twice:
(2^2)(3) = 12
(2^2)(5) = 20
x7 = 28
x11 = 44
x13 = 52
x17 = 68
Etc.

(3^3)(2) = 18
(3^3)(5) = 45
x7 = 63
x11 = 99

(5^2)(2) = 50
(5^2)(3) = 75

(7^2)(2) = 98

With this much smaller list of numbers, you should be able to pinpoint the possibilities that "fit" the Facts.

GMAT assassins aren't born, they're made,
Rich

VERITAS PREP OFFICIAL SOLUTION:

If you look first at statement 2, you can see that it's pretty useless, as there are 76 positive integers less than 77 so statement 2 is obviously insufficient. So as you start to look at statement 1, you should begin thinking in that range of statement 2 so that you're efficient in your work.

Knowing Data Sufficiency, you may want to try the number at the exact end of the range in statement 2 - they don't pick those inequality ranges by accident! If a is 75 and a + 1 is 76, then a has six factors (1 and 75; 3 and 25; and 5 and 15), and 76, a + 1, has six factors (1 and 76, 2 and 28, 4 and 19).

In seeing those factors and thinking systematically, you might recognize that each of those factorizations includes a square (75 has 25; 76 has 4) and start looking for similar numbers further down the scale. One more such pairing is 44 and 45: 44 has a square in its factorization (4) and so does 45 (9). And each does have six factors (44: 1 and 44; 2 and 22; 4 and 11. 45: 1 and 45; 3 and 15; 5 and 9). So a could be either 44 or 75, and therefore the answer is E.
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ans E..
1) it can give us various values insufficient
2) it can again give us more than 1 value for eg a=44(44 and 45 have 6 factors) and a=75(75 and 76 have 6 factors)...


A detailed procedure
the criteria is that a and a+1 have 6 factors each.
Now what can constitute 6 factors-
1) 6=1*6=(0+1)(5+1)....so we are looking for the FIFTH power of a prime number, hence 2^5=32, and 3^5>76.
2) Next 6=2*3=(1+1)(2+1)....so we are looking at two different prime numbers, one of which is used twice or has a square.
a)Let us start if the square is that of 2, so 2^2=4. Now we have to multiply 4 with different prime numbers
4*3=12; 4*5=20; 4*7=28; 4*11=44; 4*13=52; 4*17=68 and 4*19=76
b) Now check with 3^2 or 9...
9*2=18; 9*5=45; 9*7=63
c) Now is square is of 5..
25*2=50; 25*3=75
No other combinations will fit in as next is 7^2*2=98

Now you can write these in ascending order
12, 18,20,28,32,44,45,50,52,63,68,75,76..
Check for value of a .... 44,45 and 75,76
So a can be 44 or 75

E
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HI Chetan,

Is there a logical short cut or a logical way for solving this question.

Here you go:

a is a positive integer (given)

St1: a and (a+1) each have exactly 6 factors.

No limit defined on the value of a as a can have any value between 1 and infinity.

Not sufficient

St2: a < 76

a can have any value between 1 and 76
Not sufficient

Combining :

a and (a+1) each have exactly 6 factors and a < 76.

key here is : a and (a+1) are co-prime integer.

Consider p, q, r, and s be 4 different prime numbers.

for number of factors to be 6

a = p^2 * q
(a+1) = r^2 * s

as we have to find the value less than 76... substituting values will not take much of a time....

the only value comes up is

a = 44 = 2^2 * 11
(a+1) = 45 = 3^2 * 5

Hence C is sufficient.

Note: We can also use the following property :
GCD of two co-prime integers = 1

Hi, the criteria is that a and a+1 have 6 factors each.
Now what can constitute 6 factors-
1) 6=1*6=(0+1)(5+1)....so we are looking for the FIFTH power of a prime number, hence 2^5=32, and 3^5>76.
2) Next 6=2*3=(1+1)(2+1)....so we are looking at two different prime numbers, one of which is used twice or has a square.
a)Let us start if the square is that of 2, so 2^2=4. Now we have to multiply 4 with different prime numbers
4*3=12; 4*5=20; 4*7=28; 4*11=44; 4*13=52; 4*17=68 and 4*19=76
b) Now check with 3^2 or 9...
9*2=18; 9*5=45; 9*7=63
c) Now is square is of 5..
25*2=50; 25*3=75
No other combinations will fit in as next is 7^2*2=98

Now you can write these in ascending order
12, 18,20,28,32,44,45,50,52,63,68,75,76..
Check for value of a .... 44,45 and 75,76
So a can be 44 or 75

E
_________________

Hi chetan2u,

You should double-check your work: 64 does NOT have 6 factors (it has 7).

GMAT assassins aren't born, they're made,
Rich

Hi Empower,

Is there a shortcut to solve this question. Going after every number will waste our time. please educate me.


quote="EMPOWERgmatRichC"]Hi chetan2u,

You should double-check your work: 64 does NOT have 6 factors (it has 7).

GMAT assassins aren't born, they're made,
Rich[/quote]

hi thanks buddy...
I should have been careful as 64 is a perfect square so cannot have even number of factors...
+1 for you
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the answer is E

statement 1 - not sufficient - multiple answers

statement 2 - not sufficient - 2 answers 44 and 75

chetan2u

Hello,
Can you guide me how you picked numbers?

Let's say I tried 2^2*3 which is 12 and has 6 factors but the next is a prime number

So I tried a perfect square (25) but didn't work.

So how to figure out what numbers will satisfy the given criteria. (I know perfect squares have odd factors)

Hi TheNightKing,

I provide a lengthier explanation of how to use TEST IT in my post above, but here is the 'key' to this question: while it's a relatively rare pattern, the GMAT might slip in into your Test in 1 question (when you're scoring at a sufficiently high level). For a positive integer to have EXACTLY 6 factors, the prime-factorization of that number will yield 2 DIFFERENT primes, one of which shows up TWICE.

For example:
12 = (2)(2)(3) = 2 different primes, one of which shows up twice.

Factors of 12: 1,12, 2,6, 3,4

We can use this pattern to quickly find the various integers that fit this pattern (and 'leapfrog' all of the integers that don't). Since the two Facts "hint" at the fact that we're going to be dealing with 2-digit numbers, I'm going to limit my work to those.....

Start with the prime that shows up twice:
(2^2)(3) = 12
(2^2)(5) = 20
x7 = 28
x11 = 44
Etc.

(3^3)(2) = 18
(3^3)(5) = 45
x7 = 63
x11 = 99

(5^2)(2) = 50
(5^2)(3) = 75

(7^2)(2) = 98

With this much smaller list of numbers, you should be able to pinpoint the possibilities that "fit" the two Facts.

GMAT assassins aren't born, they're made,
Rich

Nice trick Rich. Thank you! and Yes, it is a rare pattern but one should be prepared for the worst

In DS questions like these, it becomes evident from the looks of one of the statements that it will not be sufficient when taken alone. In this question, for example, Statement II is insufficient when taken alone since it gives us a range for a whereas we are required to find a definite value for a.

This gives us the opportunity to eliminate some options quickly and you should always be on the lookout for such chances. We can eliminate options B and D. The possible answers are A, C or E.

Let us now analyse statement I. Statement I talks about two numbers: a and (a+1). These are two consecutive numbers. Any two consecutive numbers will always be co-prime and will not have any other factor in common except 1.
Also, of two consecutive numbers, one will always be odd and the other even. This is a lot of data. Statement I also says that a and (a+1) have exactly 6 factors each. This means that both a and (a+1) are composite numbers.

The general form of a composite number is N = \(a^p * b^q*c^r*\)…… For such a number, the number of factors is given by the expression (p+1) (q+1) (r+1)…..

Applying this concept to our numbers a and (a+1), we can say that (p+1) (q+1)(r+1)………. = 6. This can be done in 2 ways:
Case 1:
If a = \(x^p\) where p = 5, then a will have 6 factors. Depending on the value of x, a can be 1 or 32 or 243 and so on. (a+1) will be 2 or 33 or 244 and so on. The numbers 2 and 33 do not have 6 factors, whereas 244 has 6 factors. So, do we decide that 243 and 244 are our numbers?? Certainly not because depending on the value of x, there can be other values of a and (a+1) that can have 6 factors each.
Case 2:
If a = \(x^p * y^q\), then (p+1) * (q+1) = 6, which can be satisfied by taking p = 2, q = 1 or vice-versa. In this case also, since we don not have an upper limit on a, we can have multiple cases satisfying our condition.
We can safely conclude that statement I alone will be insufficient. Answer option A can be eliminated, the possible answer options at this stage are C or E.

Combining statement I and II, it’s a great thing that we have a limit on 76 because we will now have to deal with a specific set of numbers.
As per case 1, a can be 1 or 32 only since 81 or 243 are greater than 76.
If a=1, (a+1) = 2. None of these numbers have 6 factors.
If a = 32, (a+1) = 33. 33 does not have 6 factors.
We conclude that a cannot be of the form \(x^p\).

Let us look at numbers which can be expressed in the form of \(x^p * y^q\). We can work this out again on a case to case basis.

If x =2, p=2 and y=11, q=1, then a = \(2^2 * 11^1\) = 44 and (a+1) = 45. In this case, a and (a+1) have 6 factors each.

If x = 5, p=2 and y=3, q=1, then a = \(5^2 * 3^1\) = 75, and (a+1) = 76. In this case again, a and (a+1) have 6 factors each.

Clearly, we don’t have a unique value of a. The combination of statements is insufficient, answer option C can be eliminated.
The correct answer option is E.

The number of factors of a composite number does not necessarily depend on the magnitude of the number. As we saw, a smaller number can have the same number of factors as a larger number. This is because the number of factors of a composite number depends on the exponents of the prime factors and not so much on the prime factors themselves.

Hope that helps!
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Took me a hundred years to write out all the possibilities, but I solved it. Here you go GMAT, I answered 1 question correctly before I ran out of time.
Honestly don't know what the Test Prep companies are smoking. I was expecting an elegant official solution, LOL