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

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Senior Manager
Joined: 12 Jan 2019
Posts: 263
If (666 ... n digits)2 + (888 .... n digits) = Y, where n > 1, then fi  [#permalink]

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12 Jan 2019, 11:13
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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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Senior Manager
Joined: 12 Jan 2019
Posts: 263
Re: If (666 ... n digits)2 + (888 .... n digits) = Y, where n > 1, then fi  [#permalink]

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12 Jan 2019, 11:21
$$y=(666 ... n digits)^2 + (888 .... n digits)$$

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

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Re: If (666 ... n digits)2 + (888 .... n digits) = Y, where n > 1, then fi   [#permalink] 12 Jan 2019, 11:21
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