Fermat numbers

Number Theory Level 3

If $F_n$ denotes the $n$ th Fermat number, then find the value of $n$ :

$F_n=\overline{xyzabcde6}$

for some $\overline {xyzabcde}$ .

Note: Here, $\overline{xyzabcde6}$ is not the product of $\overline{xyzabcde}$ and $6$ . It is in accordance with the base-10 positional system .

Bonus: Prove your answer.

123 437 no value of

n

exists

2^5

500 12

1 solution

Tapas Mazumdar
Oct 17, 2016

The $n^{\text{th}}$ Fermat number is of the form $2^{(2^n)} + 1$ . As we can see that this expression is always odd for positive integral values of $n$ and a number having its unit digit as $6$ is always even, therefore, no such Fermat number exists which is of the required form.

Great! There is also one other feature of Fermat numbers.For $F_x$ ,where $x>1$ ,the number always end in $7$ . Prove it.

Anandmay Patel - 4 years, 8 months ago

Seems like an interesting problem. I'll try it out.

Tapas Mazumdar - 4 years, 8 months ago

Fermat numbers

1 solution

0 pending reports