SUNDAY, 24 MARCH 2013

Today is the $83^{rd}$ day of the year.

$83$ is a prime number. In fact $(2 \times 83) + 1 = 167$ is prime which means that $83$ is a Sophie Germain prime, see A005384.
Given that
$20$ is not prime and $(2 \times 20) + 1 = 41$
$41$ is prime and $(2 \times 41) + 1 = 83$
$83$ is prime and $(2 \times 83) + 1 = 167$
$167$ is prime and $(2 \times 167) + 1 = 335$
$335$ is not prime
Then we have a Cunningham Chain of length $3$ i.e. $(41, 83, 167)$, see A059762.

Take every number less than or equal to $83$, write it down in base $25$, something like this:

$1_{10} = 1_{25}$
$2_{10} = 2_{25}$
$3_{10} = 3_{25}$
$4_{10} = 4_{25}$
$5_{10} = 5_{25}$
$6_{10} = 6_{25}$
$7_{10} = 7_{25}$
$8_{10} = 8_{25}$
$9_{10} = 9_{25}$
$10_{10} = a_{25}$
$11_{10} = b_{25}$
$12_{10} = c_{25}$
$13_{10} = d_{25}$
$14_{10} = e_{25}$
$15_{10} = f_{25}$
$16_{10} = g_{25}$
$17_{10} = h_{25}$
$18_{10} = i_{25}$
$19_{10} = j_{25}$
$20_{10} = k_{25}$
$21_{10} = l_{25}$
$22_{10} = m_{25}$
$23_{10} = n_{25}$
$24_{10} = o_{25}$
$25_{10} = 10_{25}$
$26_{10} = 11_{25}$
$27_{10} = 12_{25}$
$28_{10} = 13_{25}$
...
$59_{10} = 29_{25}$
$60_{10} = 2a_{25}$
$61_{10} = 2b_{25}$
$62_{10} = 2c_{25}$
$63_{10} = 2d_{25}$
$64_{10} = 2e_{25}$
$65_{10} = 2f_{25}$
$66_{10} = 2g_{25}$
$67_{10} = 2h_{25}$
$68_{10} = 2i_{25}$
$69_{10} = 2j_{25}$
$70_{10} = 2k_{25}$
$71_{10} = 2l_{25}$
$72_{10} = 2m_{25}$
$73_{10} = 2n_{25}$
$74_{10} = 2o_{25}$
$75_{10} = 30_{25}$
$76_{10} = 31_{25}$
$77_{10} = 32_{25}$
$78_{10} = 33_{25}$
$79_{10} = 34_{25}$
$80_{10} = 35_{25}$
$81_{10} = 36_{25}$
$82_{10} = 37_{25}$
$83_{10} = 38_{25}$

Now, concatenate each of the base $25$ values starting with $1$ and reversing each value before it is concatenated giving:
$1,234,567,89a,bcd,efg,hij,klm,no0,111,213,141,516,171,819,1a1,b1c,1d1,e1f,1g1,h1i,1j1,k1l,1m1,n1o,102,122,232,425,262,728,292,a2b,2c2,d2e,2f2,g2h,2i2,j2k,2l2,m2n,2o2,031,323,334,353,637,383$
Convert this $142$ digit number in base $25$ to decimal and the result is divisible by $8$.
Not in itself remarkable but given that the number $83$ is the lowest number that can generate a multiple of $8$ in this fashion is remarkable, see A029518.


Consider, the sum of the squares of all the numbers up to and including $83$. This is $194,054$ which is divisible by $83$, see A007310.

Square	Total
$1^2 = 1$	$1$
$2^2 = 4$	$5$
$3^2 = 9$	$14$
$4^2 = 16$	$30$
$5^2 = 25$	$55$
$6^2 = 36$	$91$
$7^2 = 49$	$140$
$8^2 = 64$	$204$
$9^2 = 81$	$285$
$10^2 = 100$	$385$
$11^2 = 121$	$506$
$12^2 = 144$	$650$
$13^2 = 169$	$819$
$14^2 = 196$	$1015$
$15^2 = 225$	$1240$
$16^2 = 256$	$1496$
$17^2 = 289$	$1785$
$18^2 = 324$	$2109$
$19^2 = 361$	$2470$
$20^2 = 400$	$2870$
$21^2 = 441$	$3311$
$22^2 = 484$	$3795$
$23^2 = 529$	$4324$
$24^2 = 576$	$4900$
$25^2 = 625$	$5525$
$26^2 = 676$	$6201$
$27^2 = 729$	$6930$
$28^2 = 784$	$7714$
$29^2 = 841$	$8555$
$30^2 = 900$	$9455$
$31^2 = 961$	$10416$
$32^2 = 1024$	$11440$
$33^2 = 1089$	$12529$
$34^2 = 1156$	$13685$
$35^2 = 1225$	$14910$
$36^2 = 1296$	$16206$
$37^2 = 1369$	$17575$
$38^2 = 1444$	$19019$
$39^2 = 1521$	$20540$
$40^2 = 1600$	$22140$
$41^2 = 1681$	$23821$
$42^2 = 1764$	$25585$
$43^2 = 1849$	$27434$
$44^2 = 1936$	$29370$
$45^2 = 2025$	$31395$
$46^2 = 2116$	$33511$
$47^2 = 2209$	$35720$
$48^2 = 2304$	$38024$
$49^2 = 2401$	$40425$
$50^2 = 2500$	$42925$
$51^2 = 2601$	$45526$
$52^2 = 2704$	$48230$
$53^2 = 2809$	$51039$
$54^2 = 2916$	$53955$
$55^2 = 3025$	$56980$
$56^2 = 3136$	$60116$
$57^2 = 3249$	$63365$
$58^2 = 3364$	$66729$
$59^2 = 3481$	$70210$
$60^2 = 3600$	$73810$
$61^2 = 3721$	$77531$
$62^2 = 3844$	$81375$
$63^2 = 3969$	$85344$
$64^2 = 4096$	$89440$
$65^2 = 4225$	$93665$
$66^2 = 4356$	$98021$
$67^2 = 4489$	$102510$
$68^2 = 4624$	$107134$
$69^2 = 4761$	$111895$
$70^2 = 4900$	$116795$
$71^2 = 5041$	$121836$
$72^2 = 5184$	$127020$
$73^2 = 5329$	$132349$
$74^2 = 5476$	$137825$
$75^2 = 5625$	$143450$
$76^2 = 5776$	$149226$
$77^2 = 5929$	$155155$
$78^2 = 6084$	$161239$
$79^2 = 6241$	$167480$
$80^2 = 6400$	$173880$
$81^2 = 6561$	$180441$
$82^2 = 6724$	$187165$
$83^2 = 6889$	$194054$