IMO A.
1: 10^x=(2^x)*(5^x)
(4^y)(5^z) = 2^2y * 5^z Thus (2^x)*(5^x) = 2^2y * 5^z => x = 2y and z.
If we see x=2y then x is even so sufficient.
2. After solving we get x+5 = 3y+3. Hence x can be odd or even depending on the value of y. so Not Suff.
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ramannjit
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Thanks yes OA is A.shovan85 wrote:IMO A.
1: 10^x=(2^x)*(5^x)
(4^y)(5^z) = 2^2y * 5^z Thus (2^x)*(5^x) = 2^2y * 5^z => x = 2y and z.
If we see x=2y then x is even so sufficient.
2. After solving we get x+5 = 3y+3. Hence x can be odd or even depending on the value of y. so Not Suff.
My little mind is confused with (1), I came up with x=2y so it is even, what about x=z. Help me with this, why we will not consider relation between x and z, please.
Ramannjit
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mj78ind
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x = z is besides the point, what are we going to do with it......since HAS to 2y hence both x and z HAVE to be evenramannjit wrote:Thanks yes OA is A.shovan85 wrote:IMO A.
1: 10^x=(2^x)*(5^x)
(4^y)(5^z) = 2^2y * 5^z Thus (2^x)*(5^x) = 2^2y * 5^z => x = 2y and z.
If we see x=2y then x is even so sufficient.
2. After solving we get x+5 = 3y+3. Hence x can be odd or even depending on the value of y. so Not Suff.
My little mind is confused with (1), I came up with x=2y so it is even, what about x=z. Help me with this, why we will not consider relation between x and z, please.
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