Today is the $76^{th}$ day of the year.

$76 = 2^2 \times 19$

$76_{10} = 44_{18}$ which means it is a Brazilian Number, see A125134.

$76$ occurs in two primitive Pythagorean triples.

$(357, 76, 365)$

$(1443, 76, 1445)$

$76 * 76 = 5,776$, since $5,776$ end with the digits $76$ this makes it automorphic, see A003226.

Consider, $76^3 = 76 \times 76^2 = 76 \times 5,776 = 76 \times (5,700 + 76) = (76 \times 57 \times 100) + (76 \times 76)$

The first term is a multiple of $100$ which means that it ends in $00$ whilst the second term is $5,776$ which ends in $76$

Thus we can infer that the sum of these terms and, therefore $76^3$, ends in the digits $76$.

A similar argument applies to all other powers of $76$.

Follow up to THURSDAY, 14 MARCH 2013

In the blog for last Thursday I posed the question "Can anyone find a proof for the hypothesis that $8^n - 1$ is always divisible by $7$?".

I offer the following as an answer.

Assume that $\frac {8^n - 1} {7}$ is an integer for some value of $n$ then

$8^n - 1 = 7k$ for some integer value $k$.

$\therefore 8^n = 7k + 1$

Multiplying both sides by 8 gives

$(8 \times 8^n) = 8 \times (7k + 1)$

Subtract one from both sides gives

$8^{n+1} - 1 = (8 \times 7k) + 8 -1 = (8 \times 7k) + 7$

$\therefore 8^{n+1} - 1 = 7 \times (8k + 1)$

Since $k$ is an integer then $8k + 1$ is an integer which means that $8^{n+1} - 1$ is a multiple of $7$.

However, we can see that for $n = 1$, $8^n - 1$ becomes $8 -1 = 7$ which is clearly divisible by $7$.

Thus, by induction, we have the desired result that $\frac {8^n - 1} {7}$ is always a whole number.

## No comments:

Post a Comment