Let S be a set of positive integers such that every element n of S

Question

Let S be a set of positive integers such that every element n of S satisfies the conditions

A.  1000 \leq n \leq1200 
B. every digit in n is odd

Then how many elements of S are divisible by 3?

A. 8
B. 9
C. 10
D. 11
E. 12

Ansh13 · Answer

From conditions A and B, we know that n is a 4 digit integer where all the digits are odd. 
For a number to be divisible by 3, the sum of digits has to be divisible by 3. Since all the digits are odd, the sum of 4 odd digits will always be even. Thus, for n to be divisible by 3 the sum of digits should be 6, 12, 18, .... and so on.
The highest sum of digits possible, given the range 1000 to 1200, is 20 (i.e. if n = 1199 -> 1+1+9+9 = 20).
Now that we know the highest sum of digits possible, we can easily infer that the highest sum of digits  divisible by 3 is 18.

To summarize it all -  we need to find all the numbers in the range 1000 to 1200 where all the digits are odd and add up to get a sum of 6, 12 or 18.

Also, note that we can only incorporate the digit '1' at the hundred's place because that is the only odd number between 0 and 2.
Thus, the numbers that we are looking for look like -  11_ _ . 
We need to look for odd numbers to fill up the tens and ones place so that the sum of all the digits is 6, 12 or 18.
OR IN OTHER WORDS
 The sum of the odd numbers that are to be filled in the tens and ones place should be 4 (6-2), 10 (12-2) or 16 (18-2).

CASE 1 - Sum of the odd tens and ones digits is 4 
1 and 3 ( i.e. n = 11 13 )
3 and 1 ( i.e. n = 11 31 )
 = 2 values

CASE 2 - Sum of the odd tens and ones digits is 10 
1 and 9 ( i.e. n = 11 19 )
9 and 1 ( i.e. n = 11 91 )
3 and 7 ( i.e. n = 11 37 )
7 and 3 ( i.e. n = 11 73 )
5 and 5 ( i.e. n = 11 55 )
 = 5 values

CASE 3 - Sum of the odd tens and ones digits is 16 
7 and 9 ( i.e. n = 11 79 )
9 and 7 ( i.e. n = 11 97 )
 = 2 values

Total number of values of n = 2 + 5 + 2 = 9
Therefore,  B is correct .

It took me about 5 minutes to figure out the approach and solve this question. 
Does anyone know any shorter approach?

nick1816 · Answer

Thousand and Hundred digits must be equal to 1.So, s3um of tens and unit digit = 1 mod 3.

Now 1=7=1mod3 So if one of these is at tens or unit digit, other must be 3 or 9(0 mod 3)

Total possible elements = 2*2*2! = 8

If 5=2mod3 is at tens or unit digit, other must be 5(2 mod 3, since their sum would give 1 mod 3).

1 such element is possible.

Total possible numbers = 8+1=9

bond001 · Answer

nick1816
your method looks short but I am not able to grasp your method I did it through a long method and took 5 minutes will you help me with your method. what I am not able to understand your mod what are you doing with that

chetan2u · Answer

All digits are odd, so the thousands digit and hundreds digit have to be 1, as  1000 \leq n \leq1200 .
Sum is 1+1+x+y=2+x+y, so x+y has to be even and so will be 4, 10, and so on, => x+y=3k+1
Now tens and ones can also take 5 values each. ---- 1, 3, 5, 7 and 9
1) If tens value is 1, ones value will be 3k+1-1=3k.......3 and 9  => 2 ways 
2) If tens value is 3, ones value will be 3k+1-3=3k-2.......1 and 7  => 2 ways 
3) If tens value is 5, ones value will be 3k+1-5=3k-4=3k-3-1=3(k-1)-1.......6  => 1 way 
4) If tens value is 7, it will behave same way as 1 since 7=1+6........3 and 9  => 2 ways 
5) If tens value is 9, it will behave same way as 3 since 9=3+6........1 and 7  => 2 ways

Total 2+2+1+2+2=9 ways

B

samarpan.g28 · Answer

Let's divide the numbers in 2 ranges. A. from 1000 to 1099 and B. from 1100 to 1200.
0 is considered even so ignore any number between 1000 to 1099. Let's go to B.
We know, rule of div by 3 is sum of digits is div by 3. Also, units digit in each case shall be any one of 1,3,5,7,9 since number in odd.
1. 1+1+x+1, x can be 3 and 9 so 2 values.
2. 1+1+x+3, x can be 1 and 7 so 2 values.
3. 1+1+x+5, x can be 5 so 1 value.
4. 1+1+x+7, x can be 3 and 9 so 2 values.
5. 1+1+x+9, x can be 1 and 7 so 2 values.
Therefore, total 2+2+1+2+2=9 values(B).

bumpbot · Answer

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Let S be a set of positive integers such that every element n of S

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