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  3. 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

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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 d1, d2, m1 and m2, respectively, represent the first four digits of the customer’s initial password. Then, the entire password has the form d1d2m1m216.
Because the maximum number of days per month is 31 and the number of months in a year is 12, d1can be only 0,1,2, or 3 and m1 can be only 0 or 1. The following summarizes the possible values for d1, d2, m1, and m2.

It is given that the fifth digit, which is 1, is the units digit of 7(d1+ m1). The only relevant multiple of 7 with units digit 1 is (7)(3) = 21, from which it follows that d1 + m1 = 3. Considering the restrictions on the values of the digits, then d1= 2 and m1 = 1 or d1 = 3 and m1 = 0. Also, it is given that the sixth digit, which is 6, is the units digit of 3(d2 + m2). The only relevant multiples of 3 with units digit 6 are (3)(2) = 6 and (3)(12) = 36, from which it follows that d2 + m2 = 2 or d2 + m2 = 12.
Considering the restrictions on the values of the digits, if d2 + m2 = 2, then the only possibilities are d2 = 0 and m2 = 2 or d2=1 and m2 = 1 or d2 = 2 and m2 = 0. If d2 + m2 = 12, each of d2 and m2 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 d1 = 2 and m1 =1, which is one of the possibilities for d1 and m1 above, then m2 can be only 0,1, or 2; and if d1 = 3 and m1 = 0, which is the other possibility above for d1 and m1 then d2 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 d1 + m1 = 3 and d2 = 1. In the table above, the possibilities for the first four digits of the customer’s password, where d2 = 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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