Number Systems

We are given that we have to form six-digit numbers using the digits , with the condition . These numbers are arranged in increasing order, and we need to find the sum of the digits in the 72 number.

Step 1: Total number of possible six-digit numbers. The total number of possible six-digit numbers is given by the number of ways to choose 6 distinct digits from the 9 available digits, which is: Thus, there are 84 possible six-digit numbers.

Step 2: Positioning of the 72 number. The numbers are arranged in increasing order. We need to find the sum of the digits in the 72 number. Let's begin by considering the numbers starting with the smallest digits and counting them.
1. Numbers starting with 1:
The remaining 5 digits are chosen from . The number of such numbers is: So, the first 56 numbers have 1 as the first digit.
2. Numbers starting with 2:
The remaining 5 digits are chosen from . The number of such numbers is: Thus, the next 21 numbers have 2 as the first digit.
So, the 72 number must have 3 as the first digit because the first 56 numbers start with 1, the next 21 numbers start with 2, and the 72 number lies between the 57 and 84 numbers, which start with 3.

Step 3: Finding the remaining digits.
For numbers starting with 3, the remaining 5 digits are chosen from . We need the 72 number, which corresponds to the 72 - 56 = 16 number starting with 3. To find this number, we look at the combinations of the remaining digits:
The possible numbers starting with 3 are formed by selecting 5 digits from , and the 16 number corresponds to the digits , which is the number 35337.

Step 4: Finding the sum of the digits. The sum of the digits in 35337 is:

Given:
is divisible by
is divisible by
is divisible by
is divisible by

Now, we calculate the least common multiple (L.C.M.) of 1746:
The L.C.M. of 1746 is 34.

Therefore, the number is divisible by 34 but not by 14.

Given the equation:

Step 1: Taking the Natural Logarithm

Rewriting the equation:

Let:

The equations become:

Step 2: Rearranging the Equations

From , we get:

Substituting into the second equation:

Factorizing gives:

This implies either:

• or
• , which gives .

Step 3: Case 1 ( )

If , then:

Solving for :

For , the solution is:

Step 4: Case 2 ( )

If , then the solutions are:

Step 5: Summing the Values

For the given conditions, summing :

Final Answer:

We have: Thus, We can simplify this further: Solving for the other terms: From this we can derive , , , and .

The five digits available are , and we need to form five-digit numbers using these digits. The numbers are written in descending order.
Step 1: Total number of possible numbers.
Since the digits can be repeated, the total number of five-digit numbers that can be formed is: Thus, there are 3125 possible five-digit numbers.
Step 2: Numbering in descending order. The numbers are arranged in descending order, and we need to find the serial number of the number 35337.
Step 3: Understanding the descending order. In descending order, the largest number is 77777 and the smallest number is 11111. To find the serial number of 35337, we will first count how many numbers come before 35337 in descending order.
Step 4: Counting numbers greater than 35337. To find how many numbers are greater than 35337, we count the numbers starting with each digit greater than 3:
1. Numbers starting with 7: There are numbers starting with 7.
2. Numbers starting with 5: We need to count the numbers starting with 5 and having digits less than or equal to 35337. These numbers start with 5 and we will consider the second digit.
- For numbers starting with 5 and having 5 as the second digit, there are numbers.
- For numbers starting with 5 and having 3 as the second digit, we need to consider the third digit.
- For numbers starting with 53, there are numbers.
- For numbers starting with 535, there are numbers.
Thus, the numbers greater than 35337 are .
Step 5: Counting numbers less than 35337. To find the serial number of 35337, we count how many numbers are strictly less than 35337. These numbers are formed by starting with 35337, so we count them directly:
Thus, the serial number of 35337 is .

To solve the problem of finding the number of 3-digit numbers that can be formed using the digits 2, 3, 4, 5, and 7 without repetition, and that are not divisible by 3, we start by determining the divisibility rule for 3. A number is divisible by 3 if the sum of its digits is divisible by 3. First, let's calculate the total number of 3-digit combinations:

• Choose the first digit: 5 options (2, 3, 4, 5, 7).
• Choose the second digit: 4 options (remaining 4 digits).
• Choose the third digit: 3 options (remaining 3 digits).

The total number of 3-digit numbers is: 5 × 4 × 3 = 60.

Now, we find numbers divisible by 3. Compute the sum of the available digits: 2 + 3 + 4 + 5 + 7 = 21, which is divisible by 3. Therefore, a combination of digits forming a number that sums to 21 is divisible by 3. We need to identify combinations whose sum is not divisible by 3. We can select digits such that their sum is either equal to (sum when modulo 3 ≠ 0) or (sum when combined results in non-divisibility).

Let's take all permissible sums: 2, 3, 4, 5, 7. Use cases, try (mod 3) results:

• Sum not divisible by 3 could be achieved by avoiding digits: using two or more (mod 3) complementing each other.

Elimination provides:

• Using (3) odd multiples, remaining numbers which could reach a total sum divisible by 3 and how combination is skipped eliminating elements to solve sum remainder checks in construction subsequent permutations cancel.

Consider cases:

Total such numbers: 24 subsequent.

Thus, number of 3-digit numbers not divisible by 3 is: 60 - 24 = 36.

Validated within the given range: 36 ≤ (Total Numbers) ≤ 36.

Let .

Then,

Expanding using the binomial theorem, we get:

for some integer .

Thus, the remainder when is divided by 9 is 1.

The Correct Answer is: 1

We start with the given congruence:

Now, simplify :

Hence,

Therefore, the remainder is:

To determine the largest such that divides , we will use the method of finding the highest power of a prime dividing . The formula is:

For and , the expression becomes:

Calculating each term step by step:

Since all further terms will also be zero, the sum is:

Thus, the largest such that divides is 22.

We are given and the set .

To find , we list the pairs where and .

We compute this for each :

For , the valid pairs are .

For , the valid pairs are .

For , the valid pairs are .

For , the valid pairs are .

For , the valid pairs are .

For , the valid pairs are .

For , the valid pairs are .

For , the valid pairs are .

For , the valid pairs are .

Counting all the valid pairs, we get . Thus, the answer is .

Step 1: Prime factorization of the base.
We write Hence, Step 2: Find powers of 2 and 5 in .
Power of 2 in is Power of 5 in is Step 3: Apply divisibility condition.
For to divide , we must have From the first condition, Step 4: Determine the limiting value.
The limiting condition is .
Step 5: Final conclusion.
Hence, the largest possible value of is 14.

Step 1: Find total outcomes.
Step 2: Count favourable outcomes.
Each element of can be: So, Step 3: Compute probability.
Thus, Step 4: Final answer.