What an odd number

As all the digits are odd and distinct , so the sum of the digits must be odd which is a multiple of 9. As the odd multiples of 9 are 9 ,27,45..... but we can easily cancel out 27, 25,63.... since the sum of the digits of any 3 digit multiple of 9 cannot exceed 27. (Maximum 999 - 9+9+9=27) so the sum of the digits must be 9. The 3 odd numbers which gives the sum as 9 are 1,3 and 5. the largest number, that can be formed by these digits is 531

let's reiterate the question. 1. all digits must be distinct and odd 2. it is divisible by 9 so the three digits have to add up to a multiple of 9. for example: 9 or 18

the first number we can conclude is 5 because 9 and 7 do not work.

in the first case, if the first number is 9, that means, the other two digits must add up to 18 or 9. it is impossible to find two distinct odd numbers to fulfill this condition. 9 and 9 are the same so they don't work. any two odd numbers cannot add up to 9 because it must be one even and one odd.

in the second case, if the first number is 7, adding three digits up to 9 is already impossible because 9-7 = 2 so there are no ways of finding two distinct numbers in the first place. if we wanted to add it up to 18, the other two digits have to add up to 11 which is not possible either because two odds always add up to an even.

then, we look at 5 which is the next number on the odds list. 9-5 = 4 which is luckily, an even number. 3 and 1 are odds that equal this and if we are looking at the greatest possibility, then, it should be in this order: 5 3 1

If a 3 digit number is divisible by nine, the sum of its digits must be 9, 18, or 27. Find all three digit numbers divisible by nine: 108, 117, 126...981, 990, 999. Find all three digit numbers divisible by nine with three different odd numbers: 135,153, 315, 351, 513, and 531. 531 is the largest three digit number that fits the criteria. Therefore 531 is the solution

$N$ = $\overline{abc}$ [Here a, b, c are digits] will be divisible by nine if $a+b+c=9$

[This is a famous process to check $9$ 's divisibility]

If a,b,c are distinct odd integers, only possible combination is $1+3+5=9$

As $5>3>1$ , the largest $N$ is $531$ . $(Ans.)$

Divisibility by 9: the sum of the digits is divisible by 9. Using three distinct digits, the only two multiples of 9 that can be reached are 9 and 18. However, 3 odd numbers will never sum to 18 (an even number) so the digits must sum to 9. The only three distinct odd digits that sum to 9 are 5, 3, and 1. 531 is the largest combination of these digits

The largest 3 digit number is 999 and the sum of the digits is 27 . And only 999 will give its sum of the digits as 27 . But the digits are not distinct . So , its not the answer . And 18 cannot be the answer as sum of 3 odd numbers is odd . So , the remaining sum is 9 . And the set of distinct odd possible numbers whose sum is 9 is only 1 , 3 and 5 . So , the largest number is 531 .

Using divisibility rules, we have the sum of the digits as 27, 18, or 9. It cannot be 27 because the digits must be odd and distinct and the only way you can get that is by doing 999, which is not distinct. It cannot be 18 because three odd numbers cannot add up to be even. It has to have a sum of 9. The average of the three digits is 3 {9(sum of digits)/3(number of digits)} so to get distinct odd numbers, you add two to 3 to get the first digit, and you subtract two from 3 to get the third digit. We have the digits 531 that sum to 9 and are distinct. That's the highest number. 351 and 315 and 135 and 153 are lower.

There are only three odd numbers that when summed up equals 9 - 1,3 and 5. To form the greatest number using these digits, arrange it from descending order and you get 531.

531

Já que o número tem que ser formado com 3 dígitos e esses dígitos serem números impares (1,3,5,7 ou 9), mais observando também que esse numero tem que ser divisível por nove, observaremos primeiro o critério de divisibilidade por nove em que a soma dos algarismos tem que formar um numero divisível por nove e o maior numero que atente a essas exigências é o 531, pois 5 + 3 +1 = 9 que é divisível por nove.

divide 9 till you get the answers with the clues

If a number is product of 9, it 'll also divisible by 9 and also the sum of its digit 'll be 9, the only digits that results a sum of 9 are 1, 3, 5 so possible 3 digit distinct odd numbers are, 135, 153, 315, 351, 513, 531......among which.....531 is largest....its my approach....:)

Easy program

odd numbers = 1,3,5,7,9 (1+3+5)=9 so 531 or 135 is N so N=531

the largest 3 digit number whose digits are odd number and is divisible by 9 is 999,and the sum of digit is 27. let the sum of digit be S. since all the digits of N is distinct, then S<27, since sum of any 3 distinct number will not be greater that 27. therefore to satisfy the condition given by the question, the only value that can be taken by S is 9 or 18. but the sum of three odd numbers will never be an even number. this work out S=9, and only sum of 1, 3, 5 satisfy this condition. by simply rearrange the digits, we get 531

First we see the divisibility rule of 9 that the sum of the digits in that number should be divisible by 9,which means that the sum of the digits is either 9/18/27 as it cant exceed 27 because highest sum will be taken from 999.now we see that the digits are odd as well as distinct so we should apply-

arithmetic progression as if we see other cases there will certainly be an even in it. a-First term d-Common difference

equation formed- (a-d)+a+(a+d) where d is maximum and d is odd= 9(x) 3(a)= 9(x) a can be 3/6/9 out of which 6 is not valid as it is even. On taking a as 9 we get max value of a as 9 itself because 9+1 will result in two digits and N will turn 4-digit.Last case a=3 we see that highest possible value of d is 2 as 3-2=smallest single digit number 1

THEREFORE NUMBER IS MADE OF THESE DIGITS IN DESCENDING ORDER TO GET HIGHEST VALUE - 3+2,3,3-2=531 Which is required number

Let us first take 9 as the first digit.For a number to be divisible by 9,sum of its digits should be 9.Here,the remaining two digits should add upto 9 and should also be odd.This is impossible.Thus let us take 7 as first digit.The remaining digits should add upto either 2 or 11 and be odd and distinct.This is also impossible.Thus we take 5 as first digit.The remaining digits should add upto 4 or 13 and be odd and distinct.Only 3 and 1 satisfy these conditions.Thus 531 is the largest number.

Using 1,3,5,7&9

531 will be such number.

since da no. are 3 distinct odd no. so first of all leave the series of 200,400,600 and 800. Now the remaing series are starting from 101 ,301,501,701,901. Start from 901 and leave all those 3 digit no. which are not distinct or distinct but hv evn no. in them. since a no. is divisible by 9 if its digits sum is divisible by 9.In case of series of 9 one will never found a 3 digit no. whose sum is divisible by 9 if there is not an even no. in it. For series of 7 there are no 3 distinct digits whose sum is divisble by 9. Next comes the series of 5. In which we have distinct pairs 597,593,591 then 579,573,571 then 539,537, 531. the sum of digits of 531( 5+3+1) from the third set comes out to b 9. so the answer is 9.