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If (666 ... n digits)2 + (888 .... n digits) = Y, where n > 1, then fi

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Joined: 12 Jan 2019
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If (666 ... n digits)2 + (888 .... n digits) = Y, where n > 1, then fi  [#permalink]

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New post 12 Jan 2019, 10:13
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If \((666 ... n digits)^2 + (888 .... n digits) = Y\), where n > 1, then find the hundred's digit in Y.
a. 2
b. 4
c. 3
d. 0
e. None of these

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Re: If (666 ... n digits)2 + (888 .... n digits) = Y, where n > 1, then fi  [#permalink]

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New post 12 Jan 2019, 10:21
\(y=(666 ... n digits)^2 + (888 .... n digits)\)

= 36(111 ... n digits)^2 + (888 .... n digits)
= 36( ... 4321) + (888 .... n digits)
=(.... 556) + (888 .... n digits)
= ( ...444 )

thus, answer is 4.

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Re: If (666 ... n digits)2 + (888 .... n digits) = Y, where n > 1, then fi &nbs [#permalink] 12 Jan 2019, 10:21
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If (666 ... n digits)2 + (888 .... n digits) = Y, where n > 1, then fi

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