Question

A two-digit number from 10 to 99, inclusive, is chosen at random. What is the probability that this number is divisible by 5?

Please note: I don't know how to find the numbers divisible by 5 easily without trying it manually! Is there an easier way to solve it?

Answer 1

Well, you simply divide the range by 5...

Answer 2

From 10 to 99, there are (99-10)+1 numbers.

Answer 3

Starting from 10 (which is divisible by 5), you keep going up.

Answer 4

I don't understand. =|

Answer 5

From 1 to 95, there are 95/5 numbers divisible by 5. 95/5=19

Answer 6

well there
10-15-20-25-30-35-40-45-50-55-60-65-70-75-80-85-90-95
so it would be 18/ 88 can be divided by 5; which can be simplified to 9/44

Answer 7

well theres*

Answer 8

Anyways, the number of numbers divisible by 5 from 1 to 95 is 19. But we are taking the number from 10 to 99, so we would have to subtract 1 number, to get 18 numbers from 10 to 99 that are divisible by 5.

Answer 9

*****my mistake it would be 18/89 can be divided by 5***

Answer 10

I don't understand the subtraction part. Why do we need to subtract one with the range 10-99?

Answer 11

Because, for 1-99 there are 19 numbers divisible by 5. But, we started with 10, and not 1. The number 5, which is between 1 and 10, has been counted in that 19. So, since we don't include the 5, we have to subtract it to get 18.

Answer 12

Does that mean I have to find the multiples of five within the range before solving it?

Answer 13

No. You only have to determine the starting point and finish point, and essentially you could do 1000-9999 inclusive. The work would not be much more tedious if you use my method, although for 10-99 you can just list out all the numbers.

Answer 14

I prefer your method. But could you please explain it again from the beginning?

Answer 15

Well, take 99-10=89. Then, divide by 5 to get 17R4. Since we started with a number divisble by 5 (we started with 10), we add 1 to our answer to get 18. Next, find the number of numbers from 10 to 99, including 10 and 99. To do this, take 99-10+1., which is 90.

To see why we have to add one number, consider the number of numbers from 1 to 5, with 2 and 5 included. We have 2,3,4,5, which is 4 numbers, or 5-2+1. We have to add the extra 1 after the 5-2.

Next, take 18 and divide by 90 to get 18/90. Simplify the fraction.

Answer 16

I'll give you two more examples with larger numbers, but I won't explain them out throuorlhy

Answer 17

What's 17R4? =)

Answer 18

A two-digit number from 17 to 9959, inclusive, is chosen at random. What is the probability that this number is divisible by 5?
Number of numbers divisible by 5:
(9959-17)/5=1988R2.
Because 9959 and 17 aren't divisible by 5, we don't add anything to 1988.

Next, we find the total number of numbers from 17 to 9959. Take 9959-17+1, because we are counting the ends. 9959-17+1=9943

Finally, divide both numbers.
Answer:1988/9943

Answer 19

By the way, I meant a number, not a 2 digit nuimber.

Answer 20

A number from 12 to a one million minus 2, inclusive, is chosen at random. What is the probability that this number is divisible by 3?

(999998-12)/3=333328R2.
Both 999998 and 12 are divisible by 3 (the sum of their digits is divisible by 3). Thus, we add two to the number to get 333330.

(999998-12)+1=999,987.

So, we take 333330/999,987 and simplify the fraction.

Answer 21

Does that help?

Answer 22

from the first example i'm still blanked out on this part:
Next, we find the total number of numbers from 17 to 9959. Take 9959-17+1, because we are counting the ends. 9959-17+1=9943

I don't understand it.

Answer 23

Okay.
"Find the total number of numbers from 5 to 10, inclusive"
we have 5 6 7 8 9 10, which is 6 numbers.
In other words, we take 10-5+1=6

Answer 24

Thank you for your time. I finally get it now haha sorry for the trouble!

Answer 25

No problem - it took me a very long time before I figured out these kinds of problems on my own. Thanks for taking the time to read :)

Answer 26

Of course =)