A SECRET CODE has 5 digits. Each digit is different. It is a multiple of 4 and 11. If the middle 3 digits are 926, in that order, what is the product of the first and last digits?

The product of the first and last digits is 40.

Step-by-step 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.

hello :

by applying the distributive property :

6 × 4.3 = 6 × ( 4 + 0.3 )

            =   (6 ×   4) + (6 ×   0.3)

            = 24 + 1.8

6 × 4.3 = 25.8