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01 Nov 2020, 00:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
59% (02:19) correct
41%
(02:23)
wrong
based on 222
sessions
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Date
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Not Attempted Yet
If p is a positive integer, what is the units digit of Z, if \(Z = 104^{4p + 1} * 277^{p + 1} * 93^{p + 2} * 309^{6p}\)?
(A) 0
(B) 2
(C) 4
(D) 6
(E) 8
(A) 0
(B) 2
(C) 4
(D) 6
(E) 8
01 Nov 2020, 08:29
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Bunuel
Since we need to find the last digit, the number reduces to \(4^{4p + 1} * 7^{p + 1} * 3^{p + 2} * 9^{6p}\)
\(4^{4p + 1}\). 4p + 1 will always be odd, and therefore the last digit of \(4^{4p + 1}\) is 4.
\(9^{6p}\). Since 6p is always even, the last digit of \(9^{6p}\) is 1.
lets take the next 2 terms together \(7^{p + 1} * \space 3^{p + 2}\)
Her we use some numbers to see what we get as the last digit
When p = 1, we get \(7^2 * 3^3\) We just need to consider the last digits of these numbers = 9 * 7. The last digit is 3.
When p = 2, we get \(7^3 * 3^4\) The last digit is 3 * 1 = 3
When p = 3, we get \(7^4 * 3^5 = 7^2 * 7^2 * 3^4 * 3^1\) The last digit is 9 * 9 * 1 * 3 = 3
When p = 4, we get \(7^5 * 3^6 = 7^2 * 7^2 * 7 * 3^4 * 3^2\) The last digit is 9 * 9 * 7 * 1 * 9 = is 3
We see that this is a cycle and the last digit of \(7^{p + 1} * 3^{p + 2}\) is always 3.
Therefore the last digit of \(4^{4p + 1} * 7^{p + 1} * 3^{p + 2} * 9^{6p}\) = 4 * 3 * 1 = 2
Option B
Arun Kumar
01 Nov 2020, 14:03
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Bunuel
\(Z = 104^{4p + 1} * 277^{p + 1} * 93^{p + 2} * 309^{6p} = 104^{4p} * 104 * 277 * 93^2 * (277*93)^p * (309^2)^{3p}\)
\(104^{4p} = (104^2)^{2p} = ($6)^{2p}\) always ends in 6. \((277*93)^p = ($1)^p\) always ends in 1. \((309^2)^{3p} = ($1)^{3p}\) must end with 1, since \(9^2 = 81\). Putting everything together the last digit is:
\(=6 * 104 * 277 * 93^2 * 1 * 1 = (6 * 104)*(277*($9)) => 4*3 => 2\)
Ans: B
