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chetan2u
Hi nalin,

you are bang on with answer...

A short and sweet method would involve some number properties..

any consecutive integer will be 1 more than the sum of the digit of smaller number.....
\(11=1+1=2...........12=1+2=3............17=1+7=8............18=1+8=9............19=1+9=10=1+0=1..............20=2+0=2................\)
this will continue till you have SUM as 9 and then it will be 10, but 10 will again give us 1...


so our digits will be 1,2,...7,8,9,1,2,3...9,1.. and so on...
so 90 positive integers will have pattern of 9 digits..

and what does these 9 digits consist of
a) EVEN - 2,4,6,8 -FOUR
b) ODD - 1,3,5,7,9 - FIVE

so picking odd in all 90 will be SAME as picking odd in 9 digits..
ans 5/9

I figured there should be a simpler and faster way.
A suggestion - down the line, if the timer statistics justifies it, you may change the difficulty to above 700 level. I have seen 700 level questions that require lesser application.

Regards,
Nalin
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Well.. guess im a little late posting the answer.
Was busy figuring out another way to solve this devilish Q.
okay here it goes..

we can look at the cycle of 1 to 9, we got 5 such odd no.s
Similarly 10 to 18, we got another 5 such odds.
Now, 1 to 90 are 90 numbers.. and 10 such 9 consecutive no. sets can be made ((2 of them as illustrated above)) (..90/9=10)

So, the probability shall be ==>(5*10)/(9*10)==>5/9.

Regards.
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I tried to solve this Q by using the permutations principle. It seems that for any two-digit number to have an ODD sum of digits, its digits must be either both ODD or both EVEN and also the sum must be above more or equal 10.

I tried to solve it somehow along the following lines:
1) Number of applicable one-digit numbers: 5 (1,3,5,7,9)
2) ODD + ODD two-digit numbers: 4 x 5 = 20 (1,3,5,7 & 1,3,5,7,9)
3) ODD + ODD umbers whose digit sum is below 10: 11,13,15,17,31,33,35,51,53,71 or 10 numbers.

However, I never got to the right answer. Is there something flawed with my thinking when it comes to this question? Many thanks for your help! :)
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If # is an operation which results in adding the digits of integer until a single digit is left, what is the probability that a number picked up in first 90 positive integers will have the result of # as an odd digit ?
( example 99 = 9+9 = 18=1+8=9, so #=9)

(a) 4/10
(b) 4/9
(c) 1/2
(d) 6/10
(e) 5/9


self Made

Okay..I really don't think this can be done under the time pressure..but it's a SUPER GOOD question!
I first tried to manually find numbers and find patterns as nalin has done..but no joy..Then I tried to find something related to the digit sum and divisibility by 9..and Bang! Here goes my 2 cents..

The number of possible remainders for 9 is = 9 (0,1,2,3,4,5,6,7,8)
A number whose digit sum is divisible by 9, is itself divisible by 9(Divisibility Rule for 9).
Moreover, both the number and its digit sum leave the same remainder when divided by 9.

Now lets try some numbers..
Remainder 0
9,18,27,36,...81,90

The digit sum is = 9..So we'll count them in. Total numbers of this sort = 10

Remainder 1
10,19,28,37,...82

The digit sum of all these numbers is = 1..So we'll count them in. Total = 9

Remainder 2
11,20,29,38,...83.

Do you see it? all have digit sum = 2..So we'll not count them in.

With similar reasoning..the numbers that leave remainders of 0,1,3,5,7 have to be counted..along with the single digit numbers 1,3,5 and 7. The total comes out to be 50 such numbers.
Probability = 50/90 = 5/9

Answer (E)
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Good question.

First, every multiple of 9 has its digital root (consecutive summation of digits) equal to 9.

So we have arithmetic progression of multiples of 9 to start with: 9, 18, 27 … 90 - total (90-9)/9+1=10

Digital root of every number has a cyclicity of 9. For example let’s start with 1. Digital root of 1 is 1. Next 1+9=10. Digital root of 10 is 1+0=1 still one. Next 10+9=19 => 1+9=10 => 1+0=1 and so on. So we have arithmetic progression 1, 10, 19, … 82. Total (82-1)/9+1=10.

In order for digital root to be odd the numbers in progression should also be odd. We have five odd numbers (1, 3, 5, 7, 9) with 10 elements in progression for each. Total 5*10=50

The sought-for probability will be 50/90 = 5/9

Answer E.
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Hey.. can any body tell me what I have done wrong..
I have used a completely different approach..
Since we are given First 90 positive integers and we have to find those numbers whose digit when added gives an ODD integer.
Now here is my reasoning..

We know that
ODD+ EVEN= ODD
OR
EVEN+ODD= ODD
ALSO
if First digit is 9 then the next digit ought to be 0 (9+0=9 (odd) and we are given first 90 numbers) , hence we get our First number

NOW we need to have an ODD and an EVEN digit in order to get an odd number as the sum

CASE 1: ODD+EVEN
We can fill the first position in 4 ways (there are total 5 odd digits and 9 wont be counted)
and the second place as even here we have 5 ways( 5 even digits)
total numbers=4*5=20 numbers

CASE 2: EVEN+ODD
We can fill the first position in 4 ways(zero cant be filled at the first position)
and the second place as ODD as 5 ways( all Odd integers)
Total numbers=4*5=20 numbers

Required number of numbers are 1+20+20=41 numbers
Total given numbers 90

Probablity will be 41/90

CAN ANYBODY PLEASE HELP?????????????????????????????????????? :( :( :( :( :( :( :( :( :( :cry: :cry: :cry:
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