## Using all digits from 1 to 9 once to express palindromes 6,77,323,36863

Assign a different digit from 1 to 9 to each letter in the expression

$(A-B)^c \; + \; (C-D)^E \; + (F-G)^I \; = \; 6$
$(A-B)^c \; + \; (C-D)^E \; + (F-G)^I \; = \; 77$
$(A-B)^c \; + \; (C-D)^E \; + (F-G)^I \; = \; 323$
$(A-B)^c \; + \; (C-D)^E \; + (F-G)^I \; = \; 36863$

6, 77, 323, 36863 are palindromes which are the product of two consecutive primes

$6 \; = \; 2 \; \times \; 3$
$77 \; = \; 7 \; \times \; 11$
$323 \; = \; 17 \; \times \; 19$
$36863 \; = \; 191 \; \times \; 193$

1115111 and 3740615160473 are the next two palindromes which the product of two consecutive primes. They cannot be expressed in this manner.

$1115111 \; = \; 1051 \; \times \; 1061$
$3740615160473 \; = \; 1934063 \; \times \; 1934071$

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged , . Bookmark the permalink.

### One Response to Using all digits from 1 to 9 once to express palindromes 6,77,323,36863

1. David @InfinitelyManic says:

Is this what you’re looking for?
–some
(9 – 8)^c + (7 – 6)^5 + (3 – 1)^2 = 6, c=7
(9 – 8)^c + (6 – 3)^4 + (2 – 7)^1 = 77, c=6
(2 – 4)^c + (6 – 3)^1 + (5 – 7)^8 = 323, c=6

–all
(8 – 4)^c + (6 – 7)^3 + (9 – 1)^5 = 36863, c=6
(4 – 8)^c + (6 – 7)^3 + (9 – 1)^5 = 36863, c=6
(2 – 3)^c + (9 – 1)^5 + (8 – 4)^6 = 36863, c=9
(2 – 3)^c + (9 – 1)^5 + (4 – 8)^6 = 36863, c=9