I created a problem (actually modified it!!). Help me to improve this one....

Let $n$ be a three digit natural number such that $n^{5} - 5$ is divisible by $91$ . Find the least possible value of $n$ .

I contributed this problem to Briiliant but was rejected. So everybody enjoy solving it... and suggest more methods to make this problem even More Thoughtful and Interesting....

For its solution see my reply to Sebastian's comment.

#MathProblem #Math #Opinions

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

do you have a legitimate solution for it?

divyashish choudhary - 8 years, 1 month ago

Could you write a valid solution, please? I found it to be 122 with a computer search.

Sebastian Garrido - 8 years, 1 month ago

Yes you are correct! Its 122. And for a valid solution here it is:

The statement $n^{5} - 5$ is divisible by 91 can be inferred as $n^{5} \equiv 5\pmod{7}$ as well as $\pmod{13}$ . By a direct check, modulo 7, n = 3 is the only value satisfying $n^{5} \equiv 5$ . So $n \equiv 3\pmod{7}$ . Similarly, $n \equiv 5\pmod{13}$ . By the Chinese Remainder Theorem, these two conditions are equivalent to saying that $n \equiv 31\pmod{91}$ . Therefore the least 3 digit possible value of $n = 91 \times 1 + 31 = 122$

Rahul Nahata - 8 years, 1 month ago

You seems to have the belief that $x\equiv a\pmod {pq}\implies x\equiv a\pmod {p}, x\equiv a\pmod{q}$ which is not true.

Abhishek De - 8 years, 1 month ago

@Abhishek De – In this case p and q both are primes to be more specific the numbers p and q need to be co prime) so there is no question about this assumption being correct as 13 and 7 are undoubtedly co prime.

Rahul Nahata - 8 years, 1 month ago

Yes you are correct! Its 122. And for a valid solution here it is:

The statement n5−5 is divisible by 91 can be inferred as n5≡5(mod7) as well as(mod13). By a direct check, modulo 7, n = 3 is the only value satisfying n5≡5. So n≡3(mod7). Similarly, n≡5(mod13). By the Chinese Remainder Theorem, these two conditions are equivalent to saying that n≡31(mod91). Therefore the least 3 digit possible value of n=91×1+31=122

Ashish Gupta - 8 years, 1 month ago

I think the answer is 31. (Gotta say that I cheated--I used C++ programming) You got solution for this??

Sanjay Ambadi - 8 years, 1 month ago

$n$ is a THREE digit number.

Abhishek De - 8 years, 1 month ago

Still your cheat hasn't worked... Answer is 122.

Rahul Nahata - 8 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$