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10^6 represents the 7th digit-- but if \(X^{3}>=10\), then we really have the 8th digit, as we'd be multiplying 10^6 by 10, which would make it the 8th digit.

So if we look at strictly the digit X, and say it's 1, then X^3 would be 1, and we'd have a 1 in the 7th digit since we're multiplying by 10^6 and putting it into the 7th digit. Now the thing we have to watch out for is in the 6th digit spot, 10^5*(3X^{2}Y), and if X=1 then it's 10^5*(3Y), but \(0<=Y<=9\), so the most we could have inside there would be 27, which would make it 10^{5}(27) = 10^{6}(2.7) adding it together we'd get 10^{6}(1) + 10^{6}(2.7 = 10^{6}(3.7), so for sure if X is 1 we'd have 7 or fewer digits. The other digits in the sum are irrelevant, they wouldn't be big enough to affect it.

We can just reuse the 2 examples from (1), since they're still valid when you assume (2) to be true. From the long ass equation from (1), we can see that if M=100, M^3 has 7 digits, but M=220 has 8 digits.

10^6 represents the 7th digit-- but if \(X^{3}>=10\), then we really have the 8th digit, as we'd be multiplying 10^6 by 10, which would make it the 8th digit.

So if we look at strictly the digit X, and say it's 1, then X^3 would be 1, and we'd have a 1 in the 7th digit since we're multiplying by 10^6 and putting it into the 7th digit. Now the thing we have to watch out for is in the 6th digit spot, 10^5*(3X^{2}Y), and if X=1 then it's 10^5*(3Y), but \(0<=Y<=9\), so the most we could have inside there would be 27, which would make it 10^{5}(27) = 10^{6}(2.7) adding it together we'd get 10^{6}(1) + 10^{6}(2.7 = 10^{6}(3.7), so for sure if X is 1 we'd have 7 or fewer digits. The other digits in the sum are irrelevant, they wouldn't be big enough to affect it.

We can just reuse the 2 examples from (1), since they're still valid when you assume (2) to be true. From the long ass equation from (1), we can see that if M=100, M^3 has 7 digits, but M=220 has 8 digits.

INSUFFICIENT

Final Answer, E.

Thats definitely good effort and nice way to solve the question but is little lengthy and time consuming. GMAT questions such as this one should be solved in around 2 minuets.

I like it. Good job and keep it up. _________________

10^6 represents the 7th digit-- but if \(X^{3}>=10\), then we really have the 8th digit, as we'd be multiplying 10^6 by 10, which would make it the 8th digit.

So if we look at strictly the digit X, and say it's 1, then X^3 would be 1, and we'd have a 1 in the 7th digit since we're multiplying by 10^6 and putting it into the 7th digit. Now the thing we have to watch out for is in the 6th digit spot, 10^5*(3X^{2}Y), and if X=1 then it's 10^5*(3Y), but \(0<=Y<=9\), so the most we could have inside there would be 27, which would make it 10^{5}(27) = 10^{6}(2.7) adding it together we'd get 10^{6}(1) + 10^{6}(2.7 = 10^{6}(3.7), so for sure if X is 1 we'd have 7 or fewer digits. The other digits in the sum are irrelevant, they wouldn't be big enough to affect it.

We can just reuse the 2 examples from (1), since they're still valid when you assume (2) to be true. From the long ass equation from (1), we can see that if M=100, M^3 has 7 digits, but M=220 has 8 digits.

INSUFFICIENT

Final Answer, E.

Thats definitely good effort and nice way to solve the question but is little lengthy and time consuming. GMAT questions such as this one should be solved in around 2 minuets.

I like it. Good job and keep it up.

For the basic guys like me, this approach is very preferable! Thanks Hades _________________

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