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shelrod007
I have a big doubt here ..

"Zero as far as i know is neither negative nor positive ."

In the question stem we are given M and N are non-negative which means M and N are positive done .

M + N > 2 , would means M > 0 and N > 0 ? So how does 0 come into play here ?
In this case Statement 1 would be sufficient right ?

Statement 2 is sufficient and the answer would be D not B !!

Please help me clarify my doubt !!

0 is neither positive nor negative integer. Hence, a set of non-negative integers consists of 0 and positive integers (so all integers but negative ones).

Does this make sense?
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Bunuel
shelrod007
I have a big doubt here ..

"Zero as far as i know is neither negative nor positive ."

In the question stem we are given M and N are non-negative which means M and N are positive done .

M + N > 2 , would means M > 0 and N > 0 ? So how does 0 come into play here ?
In this case Statement 1 would be sufficient right ?

Statement 2 is sufficient and the answer would be D not B !!

Please help me clarify my doubt !!

0 is neither positive nor negative integer. Hence, a set of non-negative integers consists of 0 and positive integers (so all integers but negative ones).

Does this make sense?

Units digits, exponents, remainders problems directory: new-units-digits-exponents-remainders-problems-168569.html

Hope it helps.
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Official Solution:


If \(m\) and \(n\) are non-negative integers, what is the units digit of \(5^m + 6^n\) ?

Note that \(m\) and \(n\) are non-negative integers, which means each of them can be 0, 1, 2, and so on. Also, observe that the units digit of 5 raised to a positive integer power is always 5, and the units digit of 6 raised to a positive integer power is always 6. So, if we knew that \(m\) and \(n\) are positive integers, instead of non-negative integers, the answer would be 1, because ...5 + ...6 = ...1

(1) \(m + n > 2\)

Both \(m\) and \(n\) can be positive, giving the units digit of the sum as 1. However, one of them can be 0, and another greater than 2, and in this case, the units digit of \(5^m + 6^n\) will be either 6 (if \(n = 0\)) or 7 (if \(m = 0\)). Not sufficient.

(2) \(mn > 0\)

The above implies that \(m\) and \(n\) are either both negative or both positive. However, since it is given that \(m\) and \(n\) are non-negative, they cannot be both negative, and thus both must be positive. Therefore, the units digit of \(5^m + 6^n\) is 1. Sufficient.


Answer: B
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