What is the number of 4 digit numbers that have 4 as a digit but are not divisible by 4?
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Answers
Answer:
We are talking about a 4-digit number.
Let’s just give a letter to each one of the 4 digits of that number: a, b, c and d.
So now the number looks like:
abcd
where each letter a, b, c and d takes the value of a 1-digit integer (1, 2, 3, 4, 5, 6, 7, 8, 9, 0).
Technically a cannot be 0, because then it becomes a 3-digit number.
For a number to be divisible by 4, the number formed by the last 2 digits of that number must be divisible by 4.
This is part of a larger set of rules:
for a number to be divisible by 2, the last digit must be divisible by 2; it must be an even number …
for a number to be divisible by 4, the last 2 digits must be divisible by 4
for a number to be divisible by 8, the last 3 digits must be divisible by 8
do you see the logic ?
Back to our problem:
This means that cd must be divisible by 4.
In the first 100 number, there are 25 divisible by 4:
04, 08, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 00
And we are told that the number 4 cannot be present in this number,
therefore we take away any number containing a 4.
What is left:
08, 12, 16, 20, 28, 32, 36, 52, 56, 60, 68, 72, 76, 80, 88, 92, 96, 00
That is a total of 18 numbers.
So, for the last 2 digits there are only 18 combinations possible.
We can write this as follows:
P(cd) = 18
[The number of possibilities for cd is 18]
For the second digit, b, the only requirement is that it is not equal to 4.
Therefore there are 9 possibilities for b.
P(b) = 9
And for the 1st digit, a, it cannot be equal to 0 or 4.
Therefore there are 8 possibilities for a.
P(a) = 8
Finally the total number
Finally the total number, N, of combinations is:
N = P(a) x P(b) x P(cd)
Then:
N = 8 * 9 * 18
N = 1296
So, your answer is 1296
All the best!
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