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If n > 0, is (4^n + 1) divisible by 5? (1) n is the integral root of

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If n > 0, is (4^n + 1) divisible by 5? (1) n is the integral root of  [#permalink]

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31 Jul 2018, 06:00
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If n > 0, is ($$4^{n}$$ + 1) divisible by 5?

(1) n is the integral root of equation: $$\frac{n^{2}}{3}$$ + n - 6 = 0
(2) n is an odd integer.

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If n > 0, is (4^n + 1) divisible by 5? (1) n is the integral root of  [#permalink]

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Updated on: 02 Aug 2018, 05:15
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kshitizbansal92 wrote:
If n > 0, is (4^n + 1) divisible by 5?

(1) n is the integral root of equation: $$n^2/3$$ + n - 6 = 0
(2) n is an odd integer.

Statement (1)

Multiply by 3 to get rid of the fraction. $$n^2$$ +3n -18 = 0... Factors to (n+6)(n-3). Roots are 3 and -6. It is given that n>0, so n must = 3.

Plug in n=3...... $$4^3$$ + 1 = 65. Is 65 divisible by 5? YES.

Statement (2)

We know n has to be positive. so our possible values for n are 1,3,5,7,9 etc...

Make n=1 and we get $$4^1$$ +1 =5. Divisible by 5? YES

Make n=3 and we get $$4^3$$ + 1 = 65. Divisible by 5? YES

OR if you are sound in your number theory, you will know that 4 has a cyclicity of 2.
4^1=4
4^2=16
4^3=64
4^4=256

So for all odd integers it will end in units digit 4.
For all even integers it will end in units digit 6.

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Originally posted by MikeScarn on 31 Jul 2018, 06:15.
Last edited by MikeScarn on 02 Aug 2018, 05:15, edited 6 times in total.
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Re: If n > 0, is (4^n + 1) divisible by 5? (1) n is the integral root of  [#permalink]

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31 Jul 2018, 06:58
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For those thinking "What does 'integral root' mean?"

I just googled it. Actually "integral" means integer. So it’s just asking for the integer root.

Which can basically be used as a clue that n is an integer, but we still have to remember that n is positive.

When I first read the question, before looking at the statements, I thought "well if n is a decimal, zero, or a negative number, then the answer will be NO" So "integral" could be used as a quick clue that we will indeed be looking at integers. And n>0 shows us that we will be looking at positive numbers.
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Re: If n > 0, is (4^n + 1) divisible by 5? (1) n is the integral root of  [#permalink]

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16 Sep 2018, 19:55
1)We get the roots of the equation as -6 and 3. We will only consider 3 and after substituting in the equation given in question , the answer comes as yes. Sufficient
2)Any Odd integer the digit at unit's place would be 4 , hence +1 would definitely be divisible by 5.
Re: If n > 0, is (4^n + 1) divisible by 5? (1) n is the integral root of   [#permalink] 16 Sep 2018, 19:55
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