What is the probability that a three digit number is divis

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ok, total # of 3 digit numbers is: 9x10x10 or 900. we need to have a denominator that is multiple of 900. A is out right away.

now, let's see..105 is a multiple of 7. 994 is a multiple of seven. between, there will be 994/7 - 105/7 +1 multiples, or 128.
128/900 = 64/450 = 32/225

C.

All 3 digit numbers are numbers from 100 to 999 or total 900 numbers [(999-100)+1]

In one set of 100, we have 14 multiples of 7 (14*7=98) Remember we are leaving the 2 left in each 100 for now.
or, in 9 sets of 100 we will have 9*14 =126 multiples of 7

in 9 sets of 100 those 2 left earlier will become 9*2 =18, we get two multiples of 7 from here.
Total multiples = 126 +2 = 128

Probability= 128/900 =64/450=32/225
Answer C

See we need to calculate the number of multiples of 7 which are 3 digit numbers and divide that by 900 which is the total amount of 3 digit numbers..So first we find the first multiple of 7 in a 3 digit world. i.e 105.... 7*15=105 and then find the last which is 994 as 7*142= 994.. So the number of multiples = 142-15+1= 128.. We add one as both 142 and 15 have to be included in counting the multiples.. We can also calculate this as 142-14( the number of multiples which are not 3 digit i.e 2 digit or 1 digit multiples.. ) 128/900 is the answer... i.e 32/225


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What is wrong if I get same answer with different approach.
First 3 digit integer: 100
Last 3 digit integer : 999

So number divisible by 7 = (999-100)/7 = 128.4
Ignore the digits after decimal.
Answer is 128/900= 32/225

In order to find the required probability, we need (No. of 3-digit multiples of 7) / (Total no. of 3 digit numbers).

The no. of multiples of x within a given range (say, between Min and Max, both included) is not always the quotient of (Max - Min)/x. Specifically, if either Max or Min is perfectly divisible by x (or, for that matter, both are perfectly divisible by x), then the quotient of (Max - Min) / x will give a result which is 1 less than the required number of multiples.

In the above eg, if we were asked to find the probability of a 3 digit number being divisible by 2 or 3 (instead of 7, as asked in this question) then the quotient of (999-100)/2 or (999-100)/3 would give the wrong answer.

Largest 3 digit number divisible by 2 = 998
Smallest 3 digit number divisible by 2 = 100
No. of 3 digit multiples of 2 = [(998-100)/2]+1 = 450
Quotient (999-100)/2 = 449

Largest 3 digit number divisible by 3 = 999
Smallest 3 digit number divisible by 3 = 102
No. of 3 digit multiples of 3 = [(999-102)/3]+1 = 300
Quotient (999-100)/3 = 299

Hope this clarifies.

Probability= favourable outcome/total event
Total event= number between 100-999 inclusive, =900
Favourable outcome three difit num that divisible by 7, 105,112,119------ 994. Total 128. So
P=128/900 =32/225

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Smallest 3 digit number = 100
Largest 3 digit number = 999

999 - 100 = 899 + 1 = 900

Smallest 3 digit number divisible by 7: 105
Largest 3 digit number divisible by 7 = 994

994 - 105 = 889

889 / 7 = 127 + 1 = 128

128/900 = 32 / 225

Answer is C.

Someone please tag this as div/mult/fact please

Ok, there are (999-100) + 1 =900 3 digit numbers. First, lets divide 7 into 999 we get 142 with some remainder, so we know 7*142 is the last three digit number divisible by 7. Now lets find the first endpoint 7*12=84, 7*14=98, so we know first 3 digit divisible by 7 is 7(15), so the total number of three digit numbers divisble by 7 is 142-15+1=128, therefore, the probability is 128/900=32/225 OA is C

All 3 digit numbers from 100 to 999 or total numbers 999-99 = 900

First multiple divisble is 105 and last one is 994 which has 128 digits a+(n-1)d = l, 105+n-1*7 = 994
therefore probability being 128/900 = 32 / 225
Hence IMO C

1st 3 digit multiple of 7 ——- 105 = (7) * (15)

105 is the 15th positive multiple of 7


999 when divided by 7 ———> yields a remainder of 5

Thus the last 3 digit number divisible by 7 will be -5 from 999——-> 994

994 = (7) * (142)

994 is the 142nd positive multiple of 7


NUM = (# of favorable outcomes) =

number of 3 digit integers divisible by 7 =

(142nd Factor of 7) - (15th Factor of 7) + 1 =

(142) - (15) + 1 = 128 Numbers


And there are 9 * 10 * 10 = 900 total 3 digit integers = DEN


Answer: (128/900) = 32/225

C

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