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# Find the unit's digit in the product (2467)^153 * (341)^72

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Joined: 17 Aug 2010
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Find the unit's digit in the product (2467)^153 * (341)^72  [#permalink]

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Updated on: 24 Mar 2015, 03:54
7
8
00:00

Difficulty:

25% (medium)

Question Stats:

72% (00:55) correct 28% (01:10) wrong based on 560 sessions

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Find the unit's digit in the product (2467)^153 * (341)^72

(A) 0
(B) 1
(C) 2
(D) 7
(E) 9

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Originally posted by zerotoinfinite2006 on 24 Oct 2010, 11:40.
Last edited by Bunuel on 24 Mar 2015, 03:54, edited 2 times in total.
RENAMED THE TOPIC.
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Re: Find the unit's digit in the product (2467)^153 * (341)^72  [#permalink]

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24 Oct 2010, 11:47
2
1
zerotoinfinite2006 wrote:
Find the unit's digit in the product (2467) ^ 153 * (341) 72

(A) 0
(B) 1
(C) 2
(D) 7
(E) 9

I m considering $$(2467) ^ {153} * (341) ^{72}$$

since the unit digits of 341^ any number is 1. We basically need to find the units digit of $$2467 ^{153}$$

Unit digit of $$2467 ^{153}$$ is remainder of $$2467 ^{153}$$ when divided by 10

= 7^153

7^1 =7
7^2 = 49
7^3 = 343
7^4 = ...1
7^5 = ...7

remainder of $$7^{153}$$= remainder of $$7^{152} * 7^1$$ = 7 ( because 152 is multiple of 4 and 7^4 = ....1)

Hence D
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Re: Find the unit's digit in the product (2467)^153 * (341)^72  [#permalink]

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24 Oct 2010, 11:48
It is not clear to me .Can u state the the question again as it is confusing to me because of power and multiplication sign.Thanks.
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Re: Find the unit's digit in the product (2467)^153 * (341)^72  [#permalink]

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24 Oct 2010, 11:51
gurpreetsingh wrote:
zerotoinfinite2006 wrote:
Find the unit's digit in the product (2467) ^ 153 * (341) 72

(A) 0
(B) 1
(C) 2
(D) 7
(E) 9

I m considering $$(2467) ^ {153} * (341) ^{72}$$

since the unit digits of 341^ any number is 1. We basically need to find the units digit of $$2467 ^{153}$$

Unit digit of $$2467 ^{153}$$ is remainder of $$2467 ^{153}$$ when divided by 10

= 7^153

7^1 =7
7^2 = 49
7^3 = 343
7^4 = ...1
7^5 = ...7

remainder of $$7^{153}$$= remainder of $$7^{152} * 7^1$$ = 7 ( because 152 is multiple of 4 and 7^4 = ....1)

Hence D

+1
Thanks.. Your answer is correct. I am little confused with the solution here

7^1 =7
7^2 = 49
7^3 = 343
7^4 = ...1
7^5 = ...7

what does it mean ?
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Re: Find the unit's digit in the product (2467)^153 * (341)^72  [#permalink]

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24 Oct 2010, 12:02
zerotoinfinite2006 wrote:
gurpreetsingh wrote:
zerotoinfinite2006 wrote:
Find the unit's digit in the product (2467) ^ 153 * (341) 72

(A) 0
(B) 1
(C) 2
(D) 7
(E) 9

I m considering $$(2467) ^ {153} * (341) ^{72}$$

since the unit digits of 341^ any number is 1. We basically need to find the units digit of $$2467 ^{153}$$

Unit digit of $$2467 ^{153}$$ is remainder of $$2467 ^{153}$$ when divided by 10

= 7^153

7^1 =7
7^2 = 49
7^3 = 343
7^4 = ...1
7^5 = ...7

remainder of $$7^{153}$$= remainder of $$7^{152} * 7^1$$ = 7 ( because 152 is multiple of 4 and 7^4 = ....1)

Hence D

+1
Thanks.. Your answer is correct. I am little confused with the solution here

7^1 =7
7^2 = 49
7^3 = 343
7^4 = ...1
7^5 = ...7

what does it mean ?

It means the last digit because that only we are needing for this question.
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Re: Find the unit's digit in the product (2467)^153 * (341)^72  [#permalink]

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28 Oct 2010, 21:33
Hence D[/quote]

+1
Thanks.. Your answer is correct. I am little confused with the solution here

7^1 =7
7^2 = 49
7^3 = 343
7^4 = ...1
7^5 = ...7

what does it mean ?[/quote]

Please refer to the section "LAST DIGIT OF A PRODUCT" in the link math-number-theory-88376.html . This is an exhaustive post on Number theory by Bunuel.
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Re: Find the unit's digit in the product (2467)^153 * (341)^72  [#permalink]

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01 Nov 2010, 05:29
1
We only have to look at the first part of a equation, as the second part will always end in 1, having no affect on the first part. If we know that 7 has a cyclicity of 4 (what gurpeet showed), we can simply divided

153/4

which gives us the remainder 1.

7^1 ends in 7 (as does 7^5), so the product also ends in 7.
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Re: Find the unit's digit in the product (2467)^153 * (341)^72  [#permalink]

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24 Mar 2015, 04:22
zerotoinfinite2006 wrote:
Find the unit's digit in the product (2467)^153 * (341)^72

(A) 0
(B) 1
(C) 2
(D) 7
(E) 9

unit digit of 341^72 will be 1.
unti digit of (2467)^153 will be same as that of 7^1 (153 mod 4 = 1)

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Find the unit's digit in the product (2467)^153 * (341)^72  [#permalink]

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15 Oct 2016, 10:00
zerotoinfinite2006 wrote:
Find the unit's digit in the product (2467)^153 * (341)^72

(A) 0
(B) 1
(C) 2
(D) 7
(E) 9

7^x = 7 9 3 1 (repeats)
Where x starts from 1
Hence, units digit of 7^10 = units digit of 7^150 = 9
& units digit of 7^153 = 7

1^72 = 1

Hence ans = 7

It will be good to memorize the foll for exam day:
$$2^x$$ = 2 4 8 6
$$3^x$$= 3 9 7 1
$$4^x$$ = 4 6
$$5^x$$ = 5
$$6^x$$ = 6
$$7^x$$ = 7 9 3 1
$$8^x$$ = 8 4 2 6
$$9^x$$ = 9 1
$$10^x$$ = 0
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Re: Find the unit's digit in the product (2467)^153 * (341)^72  [#permalink]

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15 Oct 2016, 11:31
zerotoinfinite2006 wrote:
Find the unit's digit in the product (2467)^153 * (341)^72

(A) 0
(B) 1
(C) 2
(D) 7
(E) 9

Cyclicity of 7 is 4

So, $$(2467)^{153} = (2467)^{4*38} 2467$$

Unit's digit of (2467)^{4*38} will be 1

Now, Units digit of $$(2467)^{4*38} 2467$$ will be 1 * 7 = 7

$$(341)^{72}$$ will have unit's digit as 1

So, $$(2467)^{153} (341)^{72}$$ will have units digit as $$7*1 = 7$$

Hence answer will be (D) 7
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Re: Find the unit's digit in the product (2467)^153 * (341)^72  [#permalink]

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17 Aug 2017, 17:20
here 7^2=49
7^3=343
7^4=...1(unit digit)
7^152=...1(unit digit)
152 is the multiple of 4 (38*4). The question is 152 is also the multiple of of 2(76*2)..then why 4 is taken?

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Re: Find the unit's digit in the product (2467)^153 * (341)^72  [#permalink]

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29 Sep 2018, 10:55
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