If a and b are consecutive positive integers, and ab = 30x ,

Question

If a and b are consecutive positive integers, and  ab = 30x , is x a non-integer?

(1)  a^2  is divisible by 21
(2) 35 is a factor of  b^2

The OA is C but I have a doubt on that: On combining both statements, we get that a is divisible by 3 and 7, b is divisible 5 and 7, and since they are consecutive, ab is divisible by 2 - which leads to ab being divisible by 30. But if they are consecutive integers, how can both a and b be divisible by 7? Doesn't that lead to the conclusion that there is no such a and ab, and hence that x is not an integer?

tihor · Accepted Answer

how about this way:
ab=30x, is x an integer?

ab/30 = x

30 factors into 2,3,5, therefore if x has to be an integer, then the product ab has to consist of 2,3,5 in it.

1) a^2 divisible by 21
therefore a contains at the very least 7,3
not sufficient by itself since we do not know about 2 and 5
2) b^2 divisible by 35
therefore b contains at the very least 7, 5
not sufficient by itself

combine 1 and 2
ab contains 7,7,3,5 at the very least AND ab also contains a 2, since they are consecutive integers
now we know that the product ab contains at least one set of 2,3,5

therefore ab divided by 30 is an integer, or x is an integer....
answer C
reasoning sounds right?

Bunuel · Answer

I agree with 4test1.

This question seems to be wrong.

a^2=21p  -->  a=\sqrt{21p}  --> as  p  is an integer too it should be of a type of  p=21n^2 , for  a  to be an integer. -->  a=\sqrt{21*21n^2}=21n  -->  a  is a multiple of  21 .

Same for  b ,  b=35m .

We got that  a  is a multiple of  21 , and  b  is a multiple of  35 . So they share the same factor  7 .  But consecutive integers are co-prime they don't share any common factor but 1.

Statements contradict.

What is the source of this question?

swatirpr · Answer

Yes/No

1.  a^2/(21)  =  a^2/(7*3) = int  not sufficient.
2.  b^2/(35)  =  b^2/(7*5) = int  not sufficient.

1 & 2
 a^2/(7*3) =  int1   
 b^2/(7*5) =  int2

int1 and int2 some integers

1*2

a^2*b^2/((7*3)(7*5)) = int1* int2

a^2*b^2 = 7*7* 15* int1*int2

given that ab= 30X

so int1*int2 = 15 * 4

therefore X= 7

Answer is NO. Sufficient

Pick C

Hussain15 · Answer

No clue how to apply the prime factorization rule here!!

yangsta8 · Answer

This is a tricky question. I’ll give it a shot. Each statement individually is not sufficient.

(1) a^2 is divisible by 21
This statement tells us that 21 is a factor of a^2. This means that sqrt(21) is a factor of a, or sqrt(3x7) is a factor of a. Does not give us enough info.

(2) 35 is a factor of b^2
This tells us sqrt(35) or sqrt(5x7) is a factor of b. Does not give us enough info.

(1 and 2 together)
Sqrt(3x7) * sqrt(5x7) is a factor of ab. So that means that sqrt(3x5) * 7 is a factor of AB.
The fact that ab and b are consecutive does in fact imply one of them is even meaning the number is also divisible by 2. However, we do not know if it is sufficient as we know the only non-integer factors of ab are 7 and 2.

Hence C is correct.

Edit: I didn’t state this explicitly before but either individual statement does not give us any information to determine non-integer factors. I am factoring it by non-integers, the square roots, to show how I arrive at the final conclusion.

4test1 · Answer

I understand the above reasoning and came to the conclusion C myself on working.

But it struck me that it is given that a and b are consecutive positive integers, and from 1&2 that both a and b are divisible by 7.

Can this be possible - that 2 consecutive positive integers are divisible by 7?

yangsta8 · Answer

The positive integers are not each individually divisible by 7 (and no this is not a possibility). The multiple of the positive integers is divisible by 7, implying only one of either a or b in ab needs to be divisible by 7.

4test1 · Answer

Sorry to harp on this, but 1 says a is divisible by 7, and 2) says b is divisible by 7.

So when we combine them to get to C, the above 1+2 does assume that each is divisible by 7. Which cannot be true for consecutive number a and b.

yangsta8 · Answer

Correct me if I'm wrong but here is my logic:

Neither of the statements say A or B are divisible by 7.
The first one says A^2 is divisible by 21.
 \frac{A^2}{21} = N  where N is some whole number.
 A^2 = 21N 
 A = \sqrt{21N}  since A is positive.
 A = \sqrt{3} * \sqrt{7} * N 
No factor of A is 7. However \sqrt{7}  is a factor of A.

4test1 · Answer

The above won't hold true.
It's given that a is an integer. So A cannot be divisible by  sqrt(7)  - it has to be divisible by 7. Same for b.

So if a,b are consecutive integers, both cannot be divisible by 7. I'll go with C but it doesn't seem possible.

yangsta8 · Answer

Why can’t A be divisible by sqrt(7)? Integers can still be divided by non integers, such as irrational numbers or fractions. 
For example:
If A = 7 = sqrt(7) * sqrt(7) = 7/3 * 3/7

Perhaps we need a maths guru such as Bunuel to clear this up.

AccipiterQ · Answer

The OP has the problem listed incorrectly, it's from Manhattan's advanced Quant book.

The question is as follows:

If a and b are consecutive positive integers, and ab = 30x, is x an integer?

(1)  a^2  is divisible by 25.
(2) 63 is a factor of  b^2 .

If a and b are consecutive positive integers, and ab = 30x ,

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