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a 3 digit no. is formed. then find the probability that this 3 digit no. is divisable by 9
if it is given that it is divisable by 11
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For a number to be divisible by both 9 and 11 it must be divisible by 99. There are nine 3-digit numbers divisible by 99. To find the number of 3-digit numbers divisible by 11 subtract the greatest such number from the smallest, divide by 11, and add 1. That gives us (990 - 110)/11 + 1 = 81. Thus the probability that a 3-digit number divisible by 11 is also divisible by 9 is 9/81 = 1/9.
Wrap it in bacon
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hi juantheron
i will try to do this:
let's say we pick a random 3-digit number abc
if it is divisible by 11 then (a+c)-b must be divisible by 11 as well,so a+c=b or a+c=b+11
now if abc is divisible by 9 then a+b+c is divisible by 9 as well.
so: a+b+c=9k
(a+c)+b=9k
b+b=9k or b+11+b=9k
in the case where 2b=9k we see that b is either 0 or 9 which means that a+c is 0 or 9.because a cannot be zero a+c cannot be zero which leaves the case a+c=9 and b=9.so for this case we have 9 numbers.
in the case where 2b=9k-11:
notice that b is in the interval [0,9]
so 2b is in the interval [0,18]
so 2b+11 is in the interval [11,29]
and numbers divisible by 9 from this interval are possibilities for k,and those are 18 and 27.
a) if k=18 then
2b+11=18
2b=7
impossible because b is a natural number (a digit).
b) if k=27
2b+11=27
2b=16
b=8
now:
a+c=b+11
a+c=19
no two digits satisfy this equation.
so by analyzing all the cases we get that a+b=9 and b=9.
So the possibilities for the number abc are:
198,297,396,495,594,693,792,891,990
now we just calculate the number of 3 digit numbers divisible by 11 (which is floor(1000/11))
Last edited by anonimnystefy (2011-12-16 06:28:27)
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
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Hi;
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi bobbym
how did you get that there were 90 3 digit numbers divisible by 11?
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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Hi anonimnystefy;
Sorry all.
You are right!! A slip! I have amended my answer above. It is 9 over 81 not 10 over 90. Forgot about the 3 digits. Thank you!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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yaa thanks to all i have got it
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Yes, lots of answers is what I like!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi bobbym,you're welcome.i made a mistake myself.i counted ten numbers divisible by both 9 and 11,instead of 9.i corrected myself as well.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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