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(1) \(x\) is a multiple of 10. If \(x=10\) then the answer is NO but if \(x=30\) then the answer is YES. Not sufficient.

(2) \(x^2\) is a multiple of 12. The least perfect square which is a multiple of 12 is 36. Hence, the least value of \(x\) is 6 and in this case the answer is NO, but if for example \(x=12*15\) then the answer is YES. Not sufficient.

Notice that from this statement we can deduce that \(x\) must be a multiple of 3 (else how can this prime appear in \(x^2\)?).

(1)+(2) \(x\) is a multiple of both 10 and 3, hence it's a multiple of 30, so \(x\) must be divisible by 15. Sufficient.

(1) \(x\) is a multiple of 10. If \(x=10\) then the answer is NO but if \(x=30\) then the answer is YES. Not sufficient.

(2) \(x^2\) is a multiple of 12. The least perfect square which is a multiple of 12 is 36. Hence, the least value of \(x\) is 6 and in this case the answer is NO, but if for example \(x=12*15\) then the answer is YES. Not sufficient.

Notice that from this statement we can deduce that \(x\) must be a multiple of 3 (else how can this prime appear in \(x^2\)?).

(1)+(2) \(x\) is a multiple of both 10 and 3, hence it's a multiple of 30, so \(x\) must be divisible by 15. Sufficient.

Answer: C

I couldn't follow the highlighted portion actually.

I seem to find DS Questions with given statements such as below one little difficult/time consuming: 1) x multiple of 55 2) x^3 multiple of 110

Please suggest if there are any familiar types your aware of.

(1) \(x\) is a multiple of 10. If \(x=10\) then the answer is NO but if \(x=30\) then the answer is YES. Not sufficient.

(2) \(x^2\) is a multiple of 12. The least perfect square which is a multiple of 12 is 36. Hence, the least value of \(x\) is 6 and in this case the answer is NO, but if for example \(x=12*15\) then the answer is YES. Not sufficient.

Notice that from this statement we can deduce that \(x\) must be a multiple of 3 (else how can this prime appear in \(x^2\)?).

(1)+(2) \(x\) is a multiple of both 10 and 3, hence it's a multiple of 30, so \(x\) must be divisible by 15. Sufficient.

Answer: C

I couldn't follow the highlighted portion actually.

I seem to find DS Questions with given statements such as below one little difficult/time consuming: 1) x multiple of 55 2) x^3 multiple of 110

Please suggest if there are any familiar types your aware of.

Would be really grateful.

Important thing to know when solving this problem is that exponentiation does not "produce" primes. For example, for integer x if x^(positive integer) is divisible by 3, then x must be divisible by 3. How else would 3 appear in x^(positive integer)?

We are given that \(x^2\), where x is an integer, is a multiple of 12 = 2^2*3.This means that x must be a multiple of both 2 and 3. If they weren't how can x^2 have these primes?

(1) \(x\) is a multiple of 10. If \(x=10\) then the answer is NO but if \(x=30\) then the answer is YES. Not sufficient.

(2) \(x^2\) is a multiple of 12. The least perfect square which is a multiple of 12 is 36. Hence, the least value of \(x\) is 6 and in this case the answer is NO, but if for example \(x=12*15\) then the answer is YES. Not sufficient.

Notice that from this statement we can deduce that \(x\) must be a multiple of 3 (else how can this prime appear in \(x^2\)?).

(1)+(2) \(x\) is a multiple of both 10 and 3, hence it's a multiple of 30, so \(x\) must be divisible by 15. Sufficient.

Answer: C

I couldn't follow the highlighted portion actually.

I seem to find DS Questions with given statements such as below one little difficult/time consuming: 1) x multiple of 55 2) x^3 multiple of 110

Please suggest if there are any familiar types your aware of.

Would be really grateful.

Important thing to know when solving this problem is that exponentiation does not "produce" primes. For example, for integer x if x^(positive integer) is divisible by 3, then x must be divisible by 3. How else would 3 appear in x^(positive integer)?

We are given that \(x^2\), where x is an integer, is a multiple of 12 = 2^2*3.This means that x must be a multiple of both 2 and 3. If they weren't how can x^2 have these primes?

Need great help , I don't understand why (2) can give both yes and no answer , many thanks My understanding: (2)X2 can be 36 or 144 which is x=6 or 12 Both are not divisible by 15 Many thsnks

Need great help , I don't understand why (2) can give both yes and no answer , many thanks My understanding: (2)X2 can be 36 or 144 which is x=6 or 12 Both are not divisible by 15 Many thsnks

Important thing to know when solving this problem is that exponentiation does not "produce" primes. For example, for integer x if x^(positive integer) is divisible by 3, then x must be divisible by 3. How else would 3 appear in x^(positive integer)?

We are given that \(x^2\), where x is an integer, is a multiple of 12 = 2^2*3.This means that x must be a multiple of both 2 and 3. If they weren't how can x^2 have these primes?

I don't think I understand this. Are you saying that if \(x^2\) is divisible by 12 then x is divisible by 12?

So does that mean that if \(x^z\) is divisible by y then x is divisible by y (as long as z is positive)?

Important thing to know when solving this problem is that exponentiation does not "produce" primes. For example, for integer x if x^(positive integer) is divisible by 3, then x must be divisible by 3. How else would 3 appear in x^(positive integer)?

We are given that \(x^2\), where x is an integer, is a multiple of 12 = 2^2*3.This means that x must be a multiple of both 2 and 3. If they weren't how can x^2 have these primes?

So does that mean that if \(x^z\) is divisible by y then x is divisible by y (as long as z is positive)?

...as long as y is prime your statement should be correct.

Important thing to know when solving this problem is that exponentiation does not "produce" primes. For example, for integer x if x^(positive integer) is divisible by 3, then x must be divisible by 3. How else would 3 appear in x^(positive integer)?

We are given that \(x^2\), where x is an integer, is a multiple of 12 = 2^2*3.This means that x must be a multiple of both 2 and 3. If they weren't how can x^2 have these primes?

So does that mean that if \(x^z\) is divisible by y then x is divisible by y (as long as z is positive)?

...as long as y is prime your statement should be correct.

You should read more carefully. As correctly noted by paddy41, the property above is talking about primes.
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how sqrt(180) gives a integer??? we cant take 180 has a sol..

Please re-read the solution. (2) says that x^2 is a multiple of 12, not \(\sqrt{x}\). Next, the solution then considers 180 as one of possible values for x.
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