if you choose a two digit number at random ,is it more likely that the
number will be divisble by 5 or by 7
ex.
there are 90 possible 2 digit numbers
of those 18 are divisible by 5 and 13 are divisible by 7, the first one is 14 and the last one is 98.
35 and 70 are divisible by both 5 and 7, so in total there are 29 divisible by either 5 or 7
so the probability is 29/90
OOOPS, misread your question
It is actually easier than my first interpretation
there are 18 divisible by 5 and only 13 divisible by 7, so it is more likey to be divisible by 5
To determine whether a randomly chosen two-digit number is more likely to be divisible by 5 or 7, we need to think about the factors of each number.
Let's start with divisibility by 5. A number is divisible by 5 if the units digit is either 0 or 5. Since we are dealing with two-digit numbers, there are a total of 10 possible units digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Out of these, the units digits that are divisible by 5 are 0 and 5. So, out of the 10 possible units digits, 2 of them are divisible by 5.
Now let's consider divisibility by 7. Determining divisibility by 7 can be a bit trickier, but we can notice patterns. The multiples of 7 up to 100 are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, and 98. Counting them, we see that there are 14 two-digit multiples of 7.
Comparing the number of possibilities, we have 2 two-digit numbers divisible by 5 and 14 two-digit numbers divisible by 7. Therefore, it is more likely for a randomly chosen two-digit number to be divisible by 7 than by 5.