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What is the remainder when 47^51 is divided by 10?

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What is the remainder when 47^51 is divided by 10? [#permalink]

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New post 09 Nov 2017, 09:39
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Question Stats:

56% (00:43) correct 44% (00:39) wrong based on 66 sessions

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What is the remainder when 47^51 is divided by 10?

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9
[Reveal] Spoiler: OA

Kudos [?]: 11 [0], given: 22

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What is the remainder when 47^51 is divided by 10? [#permalink]

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New post 09 Nov 2017, 09:47
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Fedemaravilla wrote:
What is the remainder when 47^51 is divided by 10?

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9


we need to find the unit's digit of the number to find the remainder -

\(47^{51}=47^{48}*47^3 = (47^4)^{12}*47^3\)

Units digit of \(7^4=1\) and \(7^3=3\)

so units digit of \((47^4)^{12}*47^3=1^{12}*3 = 3\)

Hence the remainder will be \(3\)

Option B

Last edited by niks18 on 09 Nov 2017, 19:48, edited 1 time in total.

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What is the remainder when 47^51 is divided by 10? [#permalink]

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New post 09 Nov 2017, 18:48
Fedemaravilla wrote:
What is the remainder when \(47^{51}\) is divided by 10?

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9

\(7^1 = 7\)
\(7^2 =\)__\(9\)
\(7^3=\) __\(3\)
\(7^4=\) __\(1\)
\(7^5=\) __\(7\)
\(7^6=\) __\(9\)
- cyclicity is 4 (starts over again after 4)

You can split the exponents any way you want:

1)\(47^{51} = 47^{1}47^{50}\)
Units digit for \(7^1\)= 7
Units digit for \(7^{50}\):
Divide the exponent by the cyclicity of 4.

\(\frac{50}{4}\) = 12 + remainder 2. The remainder, 2, tells us to use the units digit for \(7^2\).

\(7^{50}\) has the same units digit as \(7^2\): 9

Now we have (_ _ _ 9) * (_7) = _3

Any number divided by 10 will have, as its remainder, the number's last digit.

Remainder 3

Answer B

2) \(47^{51} = 47^347^{48}\)
Units digit of \(47^3 = 7^3 = 3\)
Units digit of \(47^{48}\) (divide by cyclicity of 4):
\(\frac{48}{4} = 12\), no remainder.

When there is no remainder, the units digit is the same as 7's cyclicity; \(7^4\)

\(47^{48}\) has the same units digit as \(7^4\): 1

We have (_ _3) * (_ _ _1) = 3.
A number divided by 10 has, as its remainder, the number's last digit. \(\frac{3}{10}\) = remainder 3.

Answer B

Kudos [?]: 397 [0], given: 640

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Re: What is the remainder when 47^51 is divided by 10? [#permalink]

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New post 09 Nov 2017, 19:51
Fedemaravilla wrote:
What is the remainder when 47^51 is divided by 10?

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9


remember that ALL integers have cyclicity when rraised to power
\(7^1=7......7^2=X9.......7^3=XY3......7^4=XY1\) and so on
so \(7,9,3,1\)

\(47^{51}=47^{(48+3)}=47^{(12*4+3)}\)
so 47^(51) will have same units digit as 7^3, which is 3 as in 7,9,3,1

B
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6097 [0], given: 121

Re: What is the remainder when 47^51 is divided by 10?   [#permalink] 09 Nov 2017, 19:51
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