If $p = a^{x} . b^{y} . c^{z}$ and $q = 3^{x} . b^{4} {.5}^{z}$ where a,b,c are primes and x,y,z are natural numbers, then which of the following is true if p=q ?

a = b = c

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x = y = z

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a = 3x

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a + c = 2y

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Solution

The correct option is D
a + c = 2y

Since the prime factorization of a number is UNIQUE, we must have a = 3, y = 4, c = 5.
Hence ruling out the options, we find that a+c = 3 + 5 = 8 = 2y is the correct answer.

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If $p = a^{x} . b^{y} . c^{z}$ and $q = 3^{x} . b^{4} {.5}^{z}$ where a,b,c are primes and x,y,z are natural numbers, then which of the following is true if p=q ?

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f (x) = \sqrt{x + \sqrt{0 + \sqrt{x + \sqrt{0 + \sqrt{x + . . . . . .}}}}}

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If p=ax.by.cz and q=3x.b4.5z where a,b,c are primes and x,y,z are natural numbers, then which of the following is true if p=q ?

The correct option is D a + c = 2y Since the prime factorization of a number is UNIQUE, we must have a = 3, y = 4, c = 5. Hence ruling out the options, we find that a+c = 3 + 5 = 8 = 2y is the correct answer.

If $p = a^{x} . b^{y} . c^{z}$ and $q = 3^{x} . b^{4} {.5}^{z}$ where a,b,c are primes and x,y,z are natural numbers, then which of the following is true if p=q ?

The correct option is D
a + c = 2y

Since the prime factorization of a number is UNIQUE, we must have a = 3, y = 4, c = 5.
Hence ruling out the options, we find that a+c = 3 + 5 = 8 = 2y is the correct answer.