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What are the last two digits of the number 7^45

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What are the last two digits of the number 7^45  [#permalink]

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New post 23 Aug 2019, 11:11
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What are the last two digits of the number 7^45

A 07
B 23
C 49
D 43
E 27
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Re: What are the last two digits of the number 7^45  [#permalink]

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New post 23 Aug 2019, 13:41
The remainder when \(7^{45}\) is divided by 100 is the last 2 digits of \(7^{45}\).

\(7^2\)=49= 49 Mod 100
\(7^4\)= 2401= (01) mod 100

\(7^{4k}\)= 01 MOD 100
\(7^{4k+1}\)= (01*7) MOD 100= 07 MOD 100

45= 4*11+1, hence \(7^{45}\)= 07 MOD 100

A

AbdulMalikVT wrote:
What are the last two digits of the number 7^45

A 07
B 23
C 49
D 43
E 27
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Re: What are the last two digits of the number 7^45  [#permalink]

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New post 24 Aug 2019, 02:19
AbdulMalikVT wrote:
What are the last two digits of the number 7^45

A 07
B 23
C 49
D 43
E 27


we can solve using 7 cyclicity ; 7,9,3,1
so for power 7^45 ; 7^4x+1 ; unit digit will be 3 and tens digit will be same as 7^4 ; 2401
IMO A: 07
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Re: What are the last two digits of the number 7^45  [#permalink]

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New post 24 Aug 2019, 02:33
AbdulMalikVT wrote:
What are the last two digits of the number 7^45

A 07
B 23
C 49
D 43
E 27


What are the last two digits of the number 7^45?

= Rem[7^45/100]

7^2 = 49 = 50-1
7^4 = 49^2 = (50-1)^2 = 50^2 - 2*50 +1 = 2401

7^4k = 01mod100
7^44 = 01mod100
7^45 = 7*7^44 = 07mod100

IMO A
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Re: What are the last two digits of the number 7^45  [#permalink]

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New post 24 Aug 2019, 03:18
To find the last two digits we essentially need to find the remainder when divided by 100.

now, 7 follows cyclicity of 4 i.e. after the fourth power the unit digit gets repeated.

in this case we need last two digits.

7^4 = 2401. now last two digits are 01

so 7^44 = 7^4.
the product reduces to
(01)*7 = 07.

Hence, A becomes the answer.

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Re: What are the last two digits of the number 7^45   [#permalink] 24 Aug 2019, 03:18
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