The units digit of a three-digit number is 6. What is the probability that the number is divisible by 6? Express your answer as a common fraction.

xXxTenTacion Jul 8, 2019

#4**0 **

Number of 3 digit integers = 999-100 +1 = 900

900/6 = 150 of them are divisible by 6

9 x 10 = 90 of them end in 6

every third one is divisible by 6 so 30 30/90 = 1/3 of the three digit numbers that end in 6 (given) are divisible by 6

ElectricPavlov Jul 9, 2019

#2**0 **

**There are: 126 156 186 216 246 276 306 336 366 396 426 456 486 516 546 576 606 636 666 696 726 756 786 816 Total = 24**

**You have: 999 - 100 + 1 =900 - 3-digit numbers**

**Probability =24 / 900 = 2 / 75**

Guest Jul 8, 2019

#3**+1 **

To be divisible by 6, the fist two integers must be divisible by 3......so...we can solve this for n

126 + 30 * (n) = 999

30 * (n) = 873 divide both sides by 30

(n) = 29.1

Take the ceiling of 29.1 = 30

So....30 three-digit integers ending in "6" are divisible by 6

And we have 999 -100 + 1 = 900 three-digit integers

So....the probability is 30 / 900 = 1 / 30

See EP's answer below....!!!!

CPhill Jul 8, 2019

#4**0 **

Best Answer

Number of 3 digit integers = 999-100 +1 = 900

900/6 = 150 of them are divisible by 6

9 x 10 = 90 of them end in 6

every third one is divisible by 6 so 30 30/90 = 1/3 of the three digit numbers that end in 6 (given) are divisible by 6

ElectricPavlov Jul 9, 2019