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# Number Theory Help

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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

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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
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r u in AOPS???

Jul 8, 2019
#2
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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
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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   Jul 8, 2019
edited by CPhill  Jul 9, 2019
#4
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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
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