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If a is the units digit of 7^47

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If a is the units digit of 7^47  [#permalink]

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New post Updated on: 13 Aug 2018, 01:49
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Difficulty:

  85% (hard)

Question Stats:

46% (02:15) correct 54% (02:23) wrong based on 121 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

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Originally posted by EgmatQuantExpert on 28 Feb 2018, 02:00.
Last edited by EgmatQuantExpert on 13 Aug 2018, 01:49, edited 2 times in total.
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Re: If a is the units digit of 7^47  [#permalink]

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New post 28 Feb 2018, 02:15
1
EgmatQuantExpert wrote:

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
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Re: If a is the units digit of 7^47  [#permalink]

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New post 28 Feb 2018, 02:33
1
EgmatQuantExpert wrote:

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)
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Re: If a is the units digit of 7^47  [#permalink]

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New post 28 Feb 2018, 04:15
EgmatQuantExpert wrote:

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


7^1 ends in 7
7^2 ends in 9
7^3 ends in 3
7^4 ends in 1
then it repeates
so, 7^47 will have units digit same as 7^3 --> 3 = a

125^10 * 28^15 = 5^30 * 2^30 * 7^14 = 7*14 * 10^30 which will have 30 zeros preceded by units digit of 7^14 (same as that of 7^3 = 3 = b)

hence, a+b=6 (option d).
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Re: If a is the units digit of 7^47  [#permalink]

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New post 28 Feb 2018, 11:12

Solution:



We need to the find the sum of \(a\) and \(b\) where, \(a\) is the units digit of \(7^{47}\) and \(b\) is the rightmost nonzero digit in \(125^{10}× 28^{15}\).
Let’s now calculate \(a\)and \(b\).

Units digit of\(7^{47}\)
    \(47= 4*11+3\)
    \(4*11+3= 4k+3\)
    Units digit of \(7^(4k+3)\) is \(3\)
.
Thus, units digit of \(7^{47}\) is \(3\).
    \(a= 3\)
Rightmost nonzero digit in 125^{10}× 28^{15}
    125^{10}× 28^{15}= (5^3)^10× (2^2*7)^15.
    125^{10}× 28^{15}= 5^30 ×2^30 ×7^15
    125^{10}× 28^{15}= 10^30 ×7^15
We know, 10^n only gives n number of zeroes at the end of the number.
Hence, the units digit of 7^15 will be same as the rightmost nonzero digit of 125^{10}× 28^{15}.
    \(15= 4*3+3\)
    \(4*3+3= 4k+3\)
    Units digit of \(7^(4k+3)\) is \(3\).
Thus,
    \(b= 3\)
Hence,
    \(a+b= 3+3\)
    \(a+b=6\)
Answer: Option D
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Re: If a is the units digit of 7^47 &nbs [#permalink] 28 Feb 2018, 11:12
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