Question

Let S be a set of positive integers such that every element n of S satisfies the conditions 1) 1000<n<1200 and 2) every digit in n is odd .Then, how many elements of S are divisible by 3? illustrate the solution properly. and I would mark as brainliest ​

Answer 1

Answer:

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

A. 1000 < n< 1200

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

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 = 1113)

3 and 1 ( i.e. n = 1131)

= 2 values

CASE 2 - Sum of the odd tens and ones digits is 10

1 and 9 ( i.e. n = 1119)

9 and 1 ( i.e. n = 1191)

3 and 7 ( i.e. n = 1137)

7 and 3 ( i.e. n = 1173)

5 and 5 ( i.e. n = 1155)

= 5 values

CASE 3 - Sum of the odd tens and ones digits is 16

7 and 9 ( i.e. n = 1179)

9 and 7 ( i.e. n = 1197)

= 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.

Step-by-step explanation:

$Hope it helps u ☺️$

$@ chichora SSR ☺️$

Answer 2

Answer:

Let S be a set of positive integers such that every element n of S satisfies the conditions 1) 1000 < n < 1200. 2) Every digit in n is odd Then how many elements of S are divisible by 3?

Answer

Since every digit is an odd digit, the numbers in the set would be:

1111, 1113, 1115, 1117, 1119, 1131, 1133, 1135, 1137, 1139, 1151, 1153, 1155, 1157, 1159, 1171, 1173, 1175, 1177, 1179, 1191, 1193, 1195, 1197 and 1199.

There could only be 25 elements in the set, as shown above and it’s very easy to see that the nine (9) Numbers marked in bold are the ones that are divisible by 3.

The required answer is 9.

Mark it as the brainliest answer.