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Updated on: 13 Aug 2018, 02:49
Originally posted by EgmatQuantExpert on 28 Feb 2018, 03:00.
Last edited by EgmatQuantExpert on 13 Aug 2018, 02:49, edited 2 times in total.
Last edited by EgmatQuantExpert on 13 Aug 2018, 02:49, edited 2 times in total.
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D
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:
58% (02:12) correct
42%
(02:32)
wrong
based on 586
sessions
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e-GMAT Question:
If \(a\) is the units digit of \(7^{47}\) and b is the rightmost nonzero digit in \((125^{10}× 28^{15})\). What is the value of \(a+b\)?
A) 1
B) 2
C) 5
D) 6
F) 8
This is
Question 6 of The e-GMAT Number Properties Marathon
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28 Feb 2018, 03:15
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EgmatQuantExpert
Question:
If \(a\) is the units digit of \(7^{47}\) and b is the rightmost nonzero digit in \((125^{10}× 28^{15})\). What is the value of \(a+b\)?
A) 1
B) 2
C) 5
D) 6
F) 8
a= unit digit of 7^47
7 has the unit cycle of 4
so 47 mod 4 = 3
and 7^3 = 343
unit digit 3
a=3
b = right most nonzero digit of 125 ^ 10 x 28 ^ 15
125 = 5^3
125^10 = 5^30
28 = 7 x 4
28^15 = 7^15 x 4^15
4 = 2^2
28^15 = 7^15 x 2^30
now \((125^{10}× 28^{15})\) = 5^30 x 7^15 x 2^30
we know 5x2=10 and the the powers are common so multiply the bases
\((125^{10}× 28^{15})\) = 10^30 x 7^15
for the right most non digit integer
we need 7^15 unit digit
we know 7 has cycle of 4
15 mod 4 = 3
so 7^15 will also end in 3 ( 7^3 = 343 unit digit 3 )
b=3
a+b= 3+3
6
(D) imo
General Discussion
28 Feb 2018, 03:33
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EgmatQuantExpert
Question:
If \(a\) is the units digit of \(7^{47}\) and b is the rightmost nonzero digit in \((125^{10}× 28^{15})\). What is the value of \(a+b\)?
A) 1
B) 2
C) 5
D) 6
F) 8
The cyclicity of number 2 and 7 will be needed to solve the question
-n--------1---2---3---4
-2--------2---4---8---6
-7--------7---9---3---1
Since a is the units digit of \(7^{47}\), it will be the units digit of \(7^{4*11 + 3}\) or \(7^3\), which is 3.
Similarly, b is the units digit of \((125^{10}× 28^{15})\)
\((125^{10} * 28^{15})\) = \((5^3)^{10} * (2^2 * 7)^{15}\) = \(5^{30}*2^{30}*7^{4*3 + 3}\) = \(10^{30} * 7^{4*3 + 3}\) as \(a^m * b^ m = (ab)^m\)
This will translate to the units digit of \(7^3\) which is 3.
Therefore, the value of a+b is 6(Option D)
