Answer:
The product of the first and last digits is 40.
Explanation:
Multiple of 4 rule:
The last two digits must form a number that is divisible by 4.
Multiple of 11 rule:
We find the alternate sum of the digits, and it must be divisible by 11.
A SECRET CODE has 5 digits. The middle 3 digits are 926
So the code is
x926y
Divisible by 4:
This means that the 6y must be divisible by 4, which means that the possible values of y are 0, 4, 8.
Value of x when y = 0:
x9260
x - 9 + 2 - 6 + 0 = x - 13
It will be divisible by 11 when x - 13 = 22, that is, x = 9. However, the digit 9 is already used, which means that y = 0 is not a possible option.
Value of x when y = 4:
x9264
x - 9 + 2 - 6 + 4 = x - 9
It will be divisible by 11 when x - 9 = 0, that is, x = 9, which is not possible, as we saw above. Or
x - 9 = 11
x = 2
The digit 2 is also possible, so y cannot be 4.
Value of x when y = 8:
x9268
x - 9 + 2 - 6 + 8 = x - 5
x - 5 = 0 -> x = 5
So, the number is:
59268
What is the product of the first and last digits?
5*8 = 40
The product of the first and last digits is 40.