Let \(a=16^{25}\), then the number is \(\dfrac{a^5-1}{a-1}=1+a+a^2+a^3+a^4\equiv 0(\mod{5})\ (\because a\equiv 1\mod {5})\) This shows that there is nothing sacrosanct about \(16^{25}\) anything that has a remainder \(1\) modulo 5 (in general \(n\)) will do.

Alternatively, you could use Fermat's little theorem in Lines 1 and 2 (Line 2 is logically unnecessary, anyway): \(16^{125}=2^{500}=2^{\phi(625)}\equiv1 (\mod 625)\) .

Haha, nice! I post this in a rush without proofreading it! Obviously the simplest way is to do Samrat's way. By the way, it's Euler's totient function, not Fermat's little theorem.

Which is factorization of two integers, so it's not prime.

Note that with enough patience, one could show that it is divisible by \(101\) by Fermat's little theorem and exponentiation by squaring but it's a little bit tedious.

That is a really nice proof for the earlier unedited version . Thank you for sharing that. What I really like about your proof is that it also proves the edited version :)

If I am not mistaken, any time 6 is multiplied by itself, the resulting number's last digit will always be six. Subtracting 1 will leave 5 as the last digit, and any integer with 5 as its last digit is divisible by 5. Please correct me if I am wrong, as I only come on this site to learn.

It can be easily shown that \( (x^2 + x + 1 - 3^{25}(x+1)) \ne \pm 1 \).

Because \( (x^2 + x + 1 - 3^{25}(x+1)) \) and \( (x^2 + x + 1 + 3^{25}(x+1)) \) are neither \( \ \pm 1 \), then their product is definitely not a prime number.

the answer is 16^50+16^25+1 16 leaves a remainder of 1 when divided by 3. Therefore any power of 16 leaves a remainder 1 when divided by 3. Therefore the expression is divisible by 3 and hence is not prime.

2^1=2, 2^2=4 ,2^3=8, 2^4=16
2^5=32, therefore 2 will come again after every 4 times, now (16)^125=2^500,(dividing 500 by 4 means remainder will zero imply the unit digit of 2^500 will gives 6, now if we subtractions 1 from 6 we will get 5 and similar way id case of denominator we will get the unit digit 5 . now as the numerator is > denominator with unit digit 5 in both the cases . therefore numerator will divisible by denominator which imply the fraction will not be a prime number it will be a composite number .
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## Comments

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TopNewestLet \(a=16^{25}\), then the number is \(\dfrac{a^5-1}{a-1}=1+a+a^2+a^3+a^4\equiv 0(\mod{5})\ (\because a\equiv 1\mod {5})\) This shows that there is nothing sacrosanct about \(16^{25}\) anything that has a remainder \(1\) modulo 5 (in general \(n\)) will do.

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"Sacrosanct" is such a beautiful word. Well done, too.

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We see that \( 16^{125} = (15 + 1)^{125} = 1 + 15 \cdot 125 + 625(\ldots ) \equiv 1 \pmod {5^4} \Rightarrow 16^{125} - 1 \equiv 0 \pmod {5^4} \)

And \(16^{25} = (15 + 1)^{25} = 1 + 25 \cdot 15 + 125 (\ldots ) \equiv 1 \pmod {5^3} \Rightarrow 16^{25} - 1 \equiv 0 \pmod {5^3} \)

But \(16^{25} = (15 + 1)^{25} = 1 + 25 \cdot 15 + 625 (\ldots ) \equiv 376 \pmod {5^4} \Rightarrow 16^{25} - 1 \not \equiv 0 \pmod {5^4} \)

So there's more powers of \(5\) that divides the numerator than the denominator, thus the number is divisible by \(5\).

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Alternatively, you could use Fermat's little theorem in Lines 1 and 2 (Line 2 is logically unnecessary, anyway): \(16^{125}=2^{500}=2^{\phi(625)}\equiv1 (\mod 625)\) .

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Haha, nice! I post this in a rush without proofreading it! Obviously the simplest way is to do Samrat's way. By the way, it's Euler's totient function, not Fermat's little theorem.

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Obviously the number is an integer because letting \(z = 25^5 \), we get

\[ \begin{eqnarray} && \frac {z^{25}-1}{z - 1} \\ &=& z^{24}+z^{23} + \ldots + z + 1 \\ &=& (z^{24} + z^{23} + z^{22} + z^{21} + z^{20} )\\ & \ & + (z^{19} + z^{18} + z^{17} + z^{16} + z^{15} ) \\ & \ & + (z^{14} + z^{13} + z^{12} + z^{11} + z^{10} ) \\ & \ & + (z^{9} + z^{8} + z^{7} + z^{6} + z^{5} ) \\ & \ & + (z^{4} + z^{3} + z^{2} + z^{1} + z^{0} ) \\ &=& (z^4 +z^3+z^2+z + 1)(1 + z^5 + z^{10} + z^{15} + z^{20} ) \\ \end{eqnarray} \]

Which is factorization of two integers, so it's not prime.

Note that with enough patience, one could show that it is divisible by \(101\) by Fermat's little theorem and exponentiation by squaring but it's a little bit tedious.

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Oh ooops. Typo. Fixed!

The question has to be much more interesting than that ...

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I just proved it, essentially the same proof as the earlier problem :). I won't spoil it for others. But I will leave a rather interesting hint:

Think about the square of a sum of the first 3 terms of a geometric progression, with the first term 1 and common ratio \(16^{25}\)

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That is a really nice proof for the earlier unedited version . Thank you for sharing that. What I really like about your proof is that it also proves the edited version :)

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If I am not mistaken, any time 6 is multiplied by itself, the resulting number's last digit will always be six. Subtracting 1 will leave 5 as the last digit, and any integer with 5 as its last digit is divisible by 5. Please correct me if I am wrong, as I only come on this site to learn.

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OH MY GOODNESS THAT SAYS 16. IGNORE THE PREVIOUS COMMENT.

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One could also show that \( \frac {9^{125}-1}{9^{25} - 1} \) is composite. Letting \(x = 9^{25} \)

It will simplify to this too: \(x^4 + x^3 + x^2 + x + 1 \)

Using the algebraic identity

\[ \begin{eqnarray} x^4 + x^3 + x^2 + x + 1 &=& (x^2 + x+ 1)^2 - x(x^2 + 2x + 1) \\ &=& (x^2 + x + 1)^2 - x(x+1)^2 \\ &= &(x^2 + x+ 1)^2 - 9^{25} (x+1)^2 \\ &=& (x^2 + x + 1)^2 - 3^{50} (x+1)^2 \\ &=& (x^2 + x + 1)^2 - (3^{25}(x+1))^2 \\ &=& (x^2 + x + 1 + 3^{25}(x+1)) (x^2 + x + 1 - 3^{25}(x+1)) \\ \end{eqnarray} \]

It can be easily shown that \( (x^2 + x + 1 - 3^{25}(x+1)) \ne \pm 1 \).

Because \( (x^2 + x + 1 - 3^{25}(x+1)) \) and \( (x^2 + x + 1 + 3^{25}(x+1)) \) are neither \( \ \pm 1 \), then their product is definitely not a prime number.

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You wrote : x^2 + 2 x + 1 instead of x^2 + x + 1 at the 4th line

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\( x^4 + x^3 + x^2 + x + 1 = (x^2 + x+ 1)^2 - x(x^2 + 2x + 1) \) is correct.

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the answer is 16^50+16^25+1 16 leaves a remainder of 1 when divided by 3. Therefore any power of 16 leaves a remainder 1 when divided by 3. Therefore the expression is divisible by 3 and hence is not prime.

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Scratch that, it appears to check out.

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16-1/16-1=1

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Note that the exponents are different, so it doesn't evaluate to 1.

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1

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Note that the exponents are different, so it doesn't evaluate to 1.

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answer is 1,048,576

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2^1=2, 2^2=4 ,2^3=8, 2^4=16 2^5=32, therefore 2 will come again after every 4 times, now (16)^125=2^500,(dividing 500 by 4 means remainder will zero imply the unit digit of 2^500 will gives 6, now if we subtractions 1 from 6 we will get 5 and similar way id case of denominator we will get the unit digit 5 . now as the numerator is > denominator with unit digit 5 in both the cases . therefore numerator will divisible by denominator which imply the fraction will not be a prime number it will be a composite number . )

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Haha :) I guess I can prove it

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