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The first four digits of the six-digit initial password for a shopper’s card at a certain grocery store is the customer’s birthd
The first four digits of the six-digit initial password for a shopper’s card at a certain grocery store is the customer’s birthd
admin
2022-10-18
40
问题
The first four digits of the six-digit initial password for a shopper’s card at a certain grocery store is the customer’s birthday in day-month digit form. For example, 15 August corresponds to 1508 and 5 March corresponds to 0503. The 5th digit of the initial password is the units digit of seven times the sum of the first and third digits, and the 6th digit of the initial password is the units digit of three times the sum of the second and fourth digits. What month, and what day of that month, was a customer born whose initial password ends in 16 ?
(1) The customer’s initial password begins with 21, and its fourth digit is 1.
(2) The sum of the first and third digits of the customer’s initial password is 3, and its second digit is 1.
选项
A、Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B、Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C、BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D、EACH statement ALONE is sufficient.
E、Statements (1) and (2) TOGETHER are NOT sufficient.
答案
A
解析
Let d
1
, d
2
, m
1
and m
2
, respectively, represent the first four digits of the customer’s initial password. Then, the entire password has the form d
1
d
2
m
1
m
2
16.
Because the maximum number of days per month is 31 and the number of months in a year is 12, d
1
can be only 0,1,2, or 3 and m
1
can be only 0 or 1. The following summarizes the possible values for d
1
, d
2
, m
1
, and m
2
.
It is given that the fifth digit, which is 1, is the units digit of 7(d
1
+ m
1
). The only relevant multiple of 7 with units digit 1 is (7)(3) = 21, from which it follows that d
1
+ m
1
= 3. Considering the restrictions on the values of the digits, then d
1
= 2 and m
1
= 1 or d
1
= 3 and m
1
= 0. Also, it is given that the sixth digit, which is 6, is the units digit of 3(d
2
+ m
2
). The only relevant multiples of 3 with units digit 6 are (3)(2) = 6 and (3)(12) = 36, from which it follows that d
2
+ m
2
= 2 or d
2
+ m
2
= 12.
Considering the restrictions on the values of the digits, if d
2
+ m
2
= 2, then the only possibilities are d
2
= 0 and m
2
= 2 or d
2
=1 and m
2
= 1 or d
2
= 2 and m
2
= 0. If d
2
+ m
2
= 12, each of d
2
and m
2
is at least 3 because if either of the digits is less than 3, then the sum of the two digits cannot be 12. But if d
1
= 2 and m
1
=1, which is one of the possibilities for d
1
and m
1
above, then m
2
can be only 0,1, or 2; and if d
1
= 3 and m
1
= 0, which is the other possibility above for d
1
and m
1
then d
2
can be only 0 or 1. The table below summarizes the first four digits of the passwords that meet all conditions thus far.
(1) It is given that the customer’s password begins with 21 and the fourth digit is 1. In the table above, only one possible password meets these conditions, so the first four digits of the password are 2111 and the customer’s birthday is the 21st day of November; SUFFICIENT.
(2) It is given that d
1
+ m
1
= 3 and d
2
= 1. In the table above, the possibilities for the first four digits of the customer’s password, where d
2
= 1, are 3101 and 2111, so the customer’s birthday could be the 31st day of January or the 21st day of November; NOT sufficient.
The correct answer is A;
statement 1 alone is sufficient.
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