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

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

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

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.

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 \(b^2\).

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.

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?

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.

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.

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.

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.