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

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Joined: 14 Sep 2017
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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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25% (medium)

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69% (00:41) correct 31% (00:40) wrong based on 118 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
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Joined: 25 Feb 2013
Posts: 1216
Location: India
GPA: 3.82
What is the remainder when 47^51 is divided by 10?  [#permalink]

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Updated on: 09 Nov 2017, 19:48
2
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

Originally posted by niks18 on 09 Nov 2017, 09:47.
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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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

We can use units digit cyclicity to find remainders when the divisor is 2, 5, or 10*
$$47^{51}$$ has the same units digit as $$7^{51}$$

(1) Units digit cyclicity of 7 (multiply only the last digit by 7 to get the next power):
$$7^1 = 7$$
$$7^2 =$$__$$9$$
$$7^3=$$ __$$3$$
$$7^4=$$ __$$1$$
------------------------
$$7^5=$$ __$$7$$
$$7^6=$$ __$$9$$
$$7^7=$$ __$$3$$
$$7^8=$$ __$$1$$

Powers of 7 have a cyclicity of 4. (The units digits pattern repeats after 4 terms)

(2) Calculate where $$7^{51}$$ "hits" in a cycle of 4
-- Start with a multiple of 4 that is close to 51
$$7^{48}$$ has the same units digit as $$7^4$$
(48 has exactly 12 cycles of 4. The cycle now starts over.)
$$7^{49}$$ has the same units digit as $$7^1$$
$$7^{50}$$: same units digit as $$7^2$$
$$7^{51}$$: same units digit as $$7^3$$

-- Or divide the exponent by cyclicity of 4
Any remainder, $$r$$, tells us that the units digit of $$7^{51}$$ is the same as that of $$7^{r}$$.
Exponent/cyclicity: $$\frac{51}{4}=12$$ plus $$r=3$$
$$7^{r}=7^{3}$$
$$7^{51}$$ has the same units digit as $$7^3$$

(3) Use the units digit to find the remainder
Above: $$7^3$$ has units digit of 3
$$\frac{3}{10}=0$$ Remainder 3
The units digit of $$7^{51}$$ is also 3

$$7^{51}$$ divided by 10 thus also has a remainder of 3

*Units digit cyclicity works for remainders when the divisor is 10 because any remainder IS the units digit. Whether the number is 3; or 33; or 823,543 = 7^7; or 343 = 7^3, when that number is divided by 10, the remainder is 3.
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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
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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