let the -digit be
ABCD16if
(A+C)*7 produce a number with a unit digit of
1, then
A+C = 3,
and the possible pairs of
(A,C) are
(3,0) and
(2,1) because
C can only be
0 or
1 (
CD is the month which range between
01 and
12)
if
(B+D)*3 produce a number with a unit digit of
6, then
B+D can be equal to
2 or
8 , which produce too many possibilities.
from statement (1), the number is
21C116as we deduced, if
A = 2, then
C = 1 , and the month is
11 (November) -->
sufficientfrom statement (2), the number is
A1CD16 , and we already know that
A+C = 3 -->
insufficientif
B = 1, then
D can be
1 or
7, and
C can be
1 or
0 respectively
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