Official Solution:

What is the units digit of \(66^6 - 5^{55}\)?

A. \(0\)
B. \(1\)
C. \(5\)
D. \(6\)
E. \(9\)


A number with the units digit of 6, in any positive integer power, will have 6 as its units digit. So, the units digit of \(66^6\) is 6.

A number with the units digit of 5, in any positive integer power, will have 5 as its units digit. So, the units digit of \(5^{55}\) is 5.

Here, you have to be careful and not assume that the units digit of \(66^6 - 5^{55}\) will be \(6 - 5 = 1\) because you should notice that \(5^{55}\) is MUCH larger number than \(66^6\). Why? \(5^{55}=5^{5*11}=(5^5)^{11}\). Now, \(5^5>66\) and \(11>6\), thus \((5^5)^{11}>66^6\).

Therefore, the units digit of \(66^6 - 5^{55}\) will be \(\{ smaller \ number \ with \ the \ units \ digit \ of \ 6 \} - \{ larger \ number \ with \ the \ units \ digit \ of \ 5\} = 9\). For example, \(16-25=-9\)


Answer: E
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I guess there is a typo in the solution, as it should be something like 16-25, not 15-26, right?

Right. Fixed. Thank you.
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I think I thought of one alternative approach to "see" this question.

Since 5^55 >> 66^6, we have that 66^6 - 5^55 will result in a negative number.

While we know that the units digit of 66^6 will be 6, and the units digit of 5^55 will be 5, we will have (6 - 5). But we know that the "5" number will be larger than the "6" number, then we will need to get a value from the tens digit.

When we "get" the tens digit, we get a 10,

Then we will have 10 - (6 - 5) = 9.

This is like when we have 95 - 46, and we "borrow" 10 from 9 (which then becomes 8), to turn 5 into 15. This 15 we subtract from 6, to have 9 on the units digit.
While on the tens digit we have 8 - 4 = 4, which connecting the numbers we have 49, 95 - 46 = 49.

Great question by the way

I think this is a high-quality question and I agree with explanation.

Are there any more practice questions for such type of questions please? Thank you.

Units digits, exponents, remainders problems to practice: https://gmatclub.com/forum/units-digits ... 75004.html

Hope it helps.
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nice question! I missed the catch here