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

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New post Updated on: 09 Nov 2017, 19:48
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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

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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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

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

Answer B

*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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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
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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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