Using Legendre's formula: floor(100/5) + floor(100/25) + floor(100/125) = 20 + 4 + 0 = 24.
By Fermat's Little Theorem, since 13 is prime and gcd(5,13)=1, we have 5^12 ≡ 1 (mod 13). 100 = 12×8 + 4. So 5^100 ≡ 5^4 (mod 13). 5^4 = 625 = 48×13 + 1 ≡ 1 (mod 13).
2^50 = 1,125,899,906,842,624. Sum of digits = 1+1+2+5+8+9+9+9+0+6+8+4+2+6+2+4 = 76. (Note: This requires calculation; the answer provided may vary based on computation.)
We need to find a number that satisfies two remainder conditions simultaneously. This is a Chinese Remainder Theorem problem.
Step 1: Express the first condition
When the number is divided by 7, remainder is 4:
where \(k\) is a non-negative integer.
This means \(n \in \{4, 11, 18, 25, 32, 39, 46, 53, 60, 67, 74, 81, 88, 95, ...\}\)
Step 2: Apply the second condition
The same number divided by 11 gives remainder 6:
where \(m\) is a non-negative integer.
This means \(n \in \{6, 17, 28, 39, 50, 61, 72, 83, 94, ...\}\)
Step 3: Find the common value
We need a number that appears in both lists and lies between 1 and 100.
From Step 1: \(\{4, 11, 18, 25, 32, 39, 46, 53, 60, 67, 74, 81, 88, 95\}\)
From Step 2: \(\{6, 17, 28, 39, 50, 61, 72, 83, 94\}\)
The common element is \(n = 39\).
Step 4: Verify the answer
\[39 \div 7 = 5\text{ remainder }4\] ✓
\[39 \div 11 = 3\text{ remainder }6\] ✓
Answer: The number is 39 (Option B)
A number is divisible by 9 if sum of its digits is divisible by 9. Here sum is 12, which is not divisible by 9. However, any number divisible by 9 is also divisible by 3. But the given condition states sum of digits is 12, and divisible by 9, which is contradictory. Re-reading: if divisible by 9, then sum must be divisible by 9. Since sum is 12 and divisible by 3, the number is divisible by 3.
Perfect squares from 1 to 1000 are 1², 2², 3², ..., n² where n² ≤ 1000. So n ≤ √1000 ≈ 31.62. Therefore n can be 1, 2, 3, ..., 31. Total = 31 perfect squares.
Let the two consecutive even numbers be n and n+2. Then n(n+2) = 528. So n² + 2n - 528 = 0. Using quadratic formula or factoring: (n+24)(n-22) = 0. So n = 22 (taking positive value). The two numbers are 22 and 24. Larger = 24.
Let number = 10a + b. Reversed = 10b + a. Given: (10b + a) - (10a + b) = 45, so 9b - 9a = 45, thus b - a = 5. Also |a - b| = 5 or a - b = 5. From b - a = 5 and a - b could be -5 or 5. Testing: if b - a = 5 and digits sum conditions... Let a = 2, b = 7: number = 27. Reversed = 72. 72 - 27 = 45. ✓
When a number leaves remainder 3 when divided by 5, we can express it as \(n = 5k + 3\) for some integer \(k\). To find the remainder when divided by 15, we need to examine what values this number can take.
Step 1: Express the general form
Any number satisfying the given condition can be written as:
where \(k\) is a non-negative integer.
Step 2: Test specific values
Let's substitute different values of \(k\) and find remainders when divided by 15:
- If \(k = 0\): \(n = 3\), remainder when divided by 15 is 3
- If \(k = 1\): \(n = 8\), remainder when divided by 15 is 8
- If \(k = 2\): \(n = 13\), remainder when divided by 15 is 13
- If \(k = 3\): \(n = 18\), remainder when divided by 15 is 3
Step 3: Analyze the pattern
The possible remainders are \(\{3, 8, 13, 3, 8, 13, \ldots\}\)
We get three different remainders: 3, 8, and 13 (cycling as \(k\) varies).
Step 4: Conclusion
Since the same condition (remainder 3 when divided by 5) can produce different remainders when divided by 15 depending on which specific number we choose, the remainder cannot be uniquely determined from the given information alone.
Answer: The remainder cannot be determined uniquely. (Option D)
To find the smallest number to add to 5800 to make it a perfect square, we need to find the smallest perfect square greater than 5800.
Step 1: Find the approximate square root of 5800
Calculate \(\sqrt{5800}\) to determine which perfect square is nearest:
Since \(\sqrt{5800}\) lies between 76 and 77, the next perfect square will be \(77^2\).
Step 2: Calculate the next perfect square
Step 3: Find the difference
The number we must add is:
Step 4: Verify the answer
Check that \(5800 + 129 = 5929 = 77^2\) ✓
Answer: The smallest number to be added is 129 (Option D)