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Bunuel
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Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
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is there any shorter way to solve this, rather than listing all unit digits of n^2 for 1 to 10, followed by unit digits in multiples of 4n for all n from 1 to 10?
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AbbyJ

n^2 + 4n has units digit ending with 7 i.e. it is odd. Since 4n has to even, n^2 and therefore n are odd numbers.

N's unit digit could end in 1,3,5 or 9 (Since the question states that the unit digit of n is not 7).

On trial and error basis, we find that for n = 9 in the units digit, n^2 + 4n returns 7 in the units digit.

Therefore, n +3 (9+3,19+3,29+3...) will return 2 in the units digit. Ans B.
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Thanks Aman! That helps

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Once we rewrite \(n^2 +n\) as \(n(n+4)\), we can see that taking \(n\) as any even one digit, two digit, three digit.... number - the above equation will always give an even number. Whereas, if we were to take \(n\) as any odd one digit, two digit, three digit.... number, the only \(n\) nos. that we can take are the ones that either have 7 or 9 in the units place. Since we can't take 7, we will take only those \(n's\) whose units digit is 9 and adding three to such a number will give a number which has 2 in the units place.
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