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20 Sep 2021, 06:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
96% (01:01) correct
4%
(01:09)
wrong
based on 67
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Suppose n is an integer such that \(2 < n^2 <100\). If the units digit of \(n^2\) is 6 and the units digit of \((n —1)^2\) is 5, what is the units digit of \((n +1)^2\)?
(A) 2
(B) 4
(C) 6
(D) 8
(E) 9
(A) 2
(B) 4
(C) 6
(D) 8
(E) 9
20 Sep 2021, 06:59
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Bunuel
Possible set of values for n: {-9, -8, -7......-2 and 9, 8, 7....2}
Units digit of n^2 = 6
So, n can be +-6 or +-4
Units digit of (n-1)^2 = 5
So, n can be 6, -4
Units digit of (n+1)^2 = 7^2 or -3^2
Either way, units digit will be 9
IMO, (E)!
20 Sep 2021, 09:27
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Suppose n is an integer such that 2<n2<1002<n2<100. If the units digit of n2n2 is 6 and the units digit of (n—1)2(n—1)2 is 5, what is the units digit of (n+1)2(n+1)2?
For n^2 unit digit to be 6, n can be either ending in 4 or ending in 6.As (square of all other digits from 0 to 9 , will not end in 6). Now if n is 4 than (n-1)^2 will have unit digit 9 .So n can't be 4. Now n is 6 ,So (n+1)^2 will be 7^2 =49 having unit digit as 9.
Thus option E.
For n^2 unit digit to be 6, n can be either ending in 4 or ending in 6.As (square of all other digits from 0 to 9 , will not end in 6). Now if n is 4 than (n-1)^2 will have unit digit 9 .So n can't be 4. Now n is 6 ,So (n+1)^2 will be 7^2 =49 having unit digit as 9.
Thus option E.
