Question

Carry needs to open a box with a numeric password. The incomplete password written on the box is 427_64_ along with a hint. In how many ways can the box be opened?
(Hint: The password is divisible by 15.)

• A. 6
• B. 9
• C. 15
• D. 8

Answer 1

The correct option is A 6

Let's assume that the incomplete password is 427H64K.
We have to find the values of H and K to get the answer.

This password is divisible by 15.
It means that it is also divisible by 5 and 3.

If it is divisible by 5, then the last digit of the password must be either 0 or 5.

It means that K can take values:
0, 5

Now, the password is also divisible by 3
It means that the sum of the digits of the password must also be divisible by 3.
So, 4 + 2 + 7 + H + 6 + 4 + K = 23 + H + K
is also divisible by 3.

If 23 + H + K is divisible by 3, then the value of 23 + H + K must be a multiple of 3.

Now, let's use the values of K here.
If K = 0,
23 + H is a multiple of 3.
It means that H can take values:
1, 4, 7 (Remember that H and K are one-digit numbers.)
So, there can be 3 possible passwords for the number 427H64K:
4271640, 4274640, 4277640

If K = 5
23 + 5 + H is a multiple of 3.
28 + H is a multiple of 3.
It means that H can take values:
2, 5, 8 (Remember that H and K are one-digit numbers.)
So, there can be 3 possible passwords for the number 427H64K:
4272645, 4275645, 4278645

So, there can be a total of 6 possible passwords.
Hence, the box can be opened in 6 ways.