The number of 6 digit numbers of the form "ABCABC", which are divisible by 13, where A, 1
and C are distinct digits, A and C being even digits is
A) 200
B) 250
C) 160
D) 128
Answers
Answer:
The answer is option (D) 128.
Step-by-step explanation:
Consider the number as ABC
Now we need to make ABC from ABCABC.
ABC x 1000 = ABC000
ABC000 + ABC = ABCABC
Which means (ABC x 1000) + ABC = ABCABC
Since ABC . 1001 = ABCABC
Now, 1001 = 13 x 11 x 7
This means all possible values of ABCABC is divisible by 13, 11 and 7.
Now, possible combinations of ABC.
A & C are even digits.
A, B, C are distinct digits.
Possible values of A - 2, 4, 6, 8 - 4
Possible values of C - 0, 2, 4, 6, 8 - 5
Case # 1: C = 0, Combinations A & C - 4. Here possible values of B - 8 (Since B is different from A & C) - 32
Case # 2: C ≠ 0, Combinations A & C - 4 - 4 x 3. Here possible values of B are - 8 (Since B is different from A & C) - 4 x 3 x 8 = 96
Possible combinations - 96 + 32 = 128