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Bunuel
What is the units digit of \(6^{15} - 7^4 - 9^3\)?

A. 8
B. 7
C. 6
D. 5
E. 4

\(6^{15} - 7^4 - 9^3\)

\(= 6 - 1 - 9\)

\(= 6 - 10\)

\(= -4\)

Thus, Answer must be (C) 6
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"Recall that we can add 10 (or multiples of 10) to a “negative” units digit to make it positive. Therefore, the actual units digit is -4 + 10 = 6"

Can you explain me this rule please?
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Abhishek009
Bunuel
What is the units digit of \(6^{15} - 7^4 - 9^3\)?

A. 8
B. 7
C. 6
D. 5
E. 4

\(6^{15} - 7^4 - 9^3\)

\(= 6 - 1 - 9\)

\(= 6 - 10\)

\(= -4\)

Thus, Answer must be (C) 6


hi guitarist Abhishek009 :) how is it possible that answer is 6 and not -4, can you pls explain the concept

thanks and have a great weekend :)
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Just concentrating on the unit digits..
6-1-9
=6-(1+9)
=6-(0)
=6

Answer: C
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Bunuel
What is the units digit of \(6^{15} - 7^4 - 9^3\)?

A. 8
B. 7
C. 6
D. 5
E. 4

\(6^{15} - 7^4 - 9^3\)

\(= 6 - 1 - 9\)

\(= 6 - 10\)

\(= -4\)

Thus, Answer must be (C) 6


hi guitarist Abhishek009 :) how is it possible that answer is 6 and not -4, can you pls explain the concept

thanks and have a great weekend :)
I can help. So when we talk about numbers it is in the decimal system.
For e.g let us take the example of 20 - 11. So the unit place here would be 9. Or the unit place can be mathematically written as -1. But in the number form you won't write a negative integer.
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ScottTargetTestPrep
"Recall that we can add 10 (or multiples of 10) to a “negative” units digit to make it positive. Therefore, the actual units digit is -4 + 10 = 6"

Can you explain me this rule please?

Response:

Of course, a negative integer cannot be the units digit of any number; however, if a calculation involving units digit turns out to be a negative number, we can use the fact that adding 10 to any number does not change units digits to find an equivalent positive number.

Here’s an alternate explanation:

We found that the units digits of 6^15, 7^4 and 9^3 are 6, 1 and 9, respectively. To find the units digit of 6^15 - 7^4 - 9^3, rather than calculating 6 - 1 - 9, we can instead calculate 16 - 1 - 9 and find 6. Notice that we could also have taken 26 (or any number with a units digit of 6) instead of 16 and we would have found 26 - 1 - 9 = 16, which also has a units digit of 6.
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Bunuel
What is the units digit of \(6^{15} - 7^4 - 9^3\)?

A. 8
B. 7
C. 6
D. 5
E. 4

Asked: What is the units digit of \(6^{15} - 7^4 - 9^3\)?

Unit digit of \(6^{15} = 6\)
Unit digit of \(7^4 = 1\)
Unit digit of \(9^3 = 9\)

Unit digit of \(6^{15} - 7^4 - 9^3 = 16 - 1 - 9 = 6\)

IMO C
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ScottTargetTestPrep Bunuel
Why wasn't comparison of the exponent terms not done here, like it's done for https://gmatclub.com/forum/what-is-the- ... 51159.html and https://gmatclub.com/forum/m26-184441.html
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Bunuel
What is the units digit of \(6^{15} - 7^4 - 9^3\)?

A. 8
B. 7
C. 6
D. 5
E. 4
­
ScottTargetTestPrep Bunuel
Why wasn't comparison of the exponent terms not done here, like it's done for https://gmatclub.com/forum/what-is-the- ... 51159.html and https://gmatclub.com/forum/m26-184441.html

Yes, if \(7^4 + 9^3\) were greater than \(6^{15}\), the answer would have been 4 instead of 6. However, it's easy to see that \(6^{15}\) is a much larger number than \(7^4 + 9^3\):

\(6^{15} > 3^{15} > 3^{14} = 9^7\)
While:

\(7^4 + 9^3 < 9^4 + 9^4 < 2*9^4 < 9^5\)
Therefore, \(6^{15} > 7^4 + 9^3\).

Hope it helps.­
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