+0  
 
+1
221
5
avatar+545 

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.

 Jul 8, 2019

Best Answer 

 #4
avatar+19929 
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

 Jul 9, 2019
 #1
avatar+100 
+1

r u in AOPS???

 Jul 8, 2019
 #2
avatar
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

 Jul 8, 2019
 #3
avatar+106539 
+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....!!!!

 

 

cool cool cool

 Jul 8, 2019
edited by CPhill  Jul 9, 2019
 #4
avatar+19929 
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
 #5
avatar+106539 
0

THX, EP.....your answer is the correct one  !!!

 

cool cool cool

CPhill  Jul 9, 2019

33 Online Users

avatar