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

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Intern
Joined: 14 Sep 2017
Posts: 27

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

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

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09 Nov 2017, 09:39
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35% (medium)

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

Director
Joined: 25 Feb 2013
Posts: 619

Kudos [?]: 304 [0], given: 39

Location: India
GPA: 3.82
What is the remainder when 47^51 is divided by 10? [#permalink]

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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.

Kudos [?]: 304 [0], given: 39

VP
Joined: 22 May 2016
Posts: 1108

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

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

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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

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.

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

Math Expert
Joined: 02 Aug 2009
Posts: 5337

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

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

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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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