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# Of the integers 2, 7, and 49, which, if any, are divisors of 7^49 + 1?

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Joined: 02 Sep 2009
Posts: 58335
Of the integers 2, 7, and 49, which, if any, are divisors of 7^49 + 1?  [#permalink]

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03 Dec 2018, 01:33
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84% (01:03) correct 16% (01:02) wrong based on 92 sessions

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Of the integers 2, 7, and 49, which, if any, are divisors of 7^49 + 1 ?

A. None
B. 2 only
C. 7 only
D. 7 and 49 only
E. 2, 7, and 49

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Re: Of the integers 2, 7, and 49, which, if any, are divisors of 7^49 + 1?  [#permalink]

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03 Dec 2018, 01:46
Unit Digits of 7:
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7
-> Unit digit of 7^48 is 1
-> UD of 7^49 + 1 = 7+1 = 8
Therefore, 7^49 + 1 is divisible by 2

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Re: Of the integers 2, 7, and 49, which, if any, are divisors of 7^49 + 1?  [#permalink]

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03 Dec 2018, 01:48
Bunuel wrote:
Of the integers 2, 7, and 49, which, if any, are divisors of 7^49 + 1 ?

A. None
B. 2 only
C. 7 only
D. 7 and 49 only
E. 2, 7, and 49

As per cyclicity of 7 the units digit of 7^49 would 7 and adding 1 would give 8

No ending with 8 would be divisible by 2 so we can negate option A,C,D , left with B and E.
7 & 49 will be be divisors of 7^49 but not 7^49+1..
IMO B..
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Re: Of the integers 2, 7, and 49, which, if any, are divisors of 7^49 + 1?  [#permalink]

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12 Dec 2018, 05:05
Archit3110 Reactzz

Guys:
The pattern is

7^1 = 7
7^2 = 49
7^3 = ...3
7^4 = ...1
7^5 = ...7
7^6 = ...9
7^7 = ...3
7^8 = ...1

I went all the way to the power of 8 in order to make the pattern 100% clear.

IMO is that, since 7^7 has 3 as its units digit and since 7 is a multiple of 49, the UD of 7^49 must be 3. Thus, the units digit after adding 1 would be 4. The OA remains the same, however it is not 7 but 3 that is added the 1.

Best, gota900
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Re: Of the integers 2, 7, and 49, which, if any, are divisors of 7^49 + 1?  [#permalink]

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25 Jan 2019, 00:30
Unit Digits of 7:
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7
-> Unit digit of 7^48 is 1
-> UD of 7^49 + 1 = 7+1 = 8
Therefore, 7^49 + 1 is divisible by 2

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Re: Of the integers 2, 7, and 49, which, if any, are divisors of 7^49 + 1?   [#permalink] 25 Jan 2019, 00:30
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