How many 4-digit numbers divisible by 5, all of the digits of which are odd, are there?

I guess that we want to create 4 digit numbers that are divisible by 5, only using the odd numbers in the image.

We know that a number is divisible by 5 only if the last digit (the units digit) is a zero or a five, in the image we only have a five, so our 4-digit numbers need to end with a five, so we have a digit fixed in five and the other 3 digits can be other numbers.

We have two different approaches to this:

First, if each odd number can be used only once, we already used the five, so we can use the other 4 numbers.

Then, for the first digit, we have 4 options.

for the second digit, we have 3 options (because we already used one)

for the third digit, we have 2 options (because we already used 2)

then the number of combinations is equal to the product of the number of options for each selection:

C = 4*3*2 = 24 combinations.

The second approach is If the numbers in the image can be repeated (for example, 5555 or 3435 are allowed)

we still have our last digit fixed in 5, and for the first digit we have 5 options, for the second we also have 5 options, and for the third, we also have 5 options, then, with the same reasoning as above, we have:

C = 5*5*5 = 25*5 = 125 combinations.

step-by-step explanation:

yes it represnts a functio, their are not 2 of the same numbers on the both sides

hi there!

question 1 answer: p/q can be terminating when q = 5^m * 2^n

question 2 answer: -1 and -1

question 3 answer: -4

question 4 answer: 5/12

hope this : )