Quant Quiz By Jp

When a number is divided by 25 and leaves a remainder of 16, it can be written as 25n + 16, where n is an integer.
If we square this number, we get (25n + 16)^2 = 625n^2 + 800n + 256.
When we divide this expression by 25, the remainder will be the constant term, which is 256.
Therefore, the remainder when the square of the number is divided by 25 is 256.

1729 is known as the "Ramanujan Number" because it is the smallest number that can be expressed as the sum of two cubes in two different ways. Specifically, it can be written as 1^3 + 12^3 and also as 9^3 + 10^3. This number is named after the famous Indian mathematician Srinivasa Ramanujan, who discovered this property of 1729 and found it to be a fascinating mathematical curiosity.

The statement suggests that the dividend and the remainder are connected or affected by the same events or actions. Whatever happens to the dividend, whether it increases, decreases, or remains the same, the remainder will also be affected in the same way. This implies that the remainder is directly dependent on the dividend and any changes in the dividend will result in corresponding changes in the remainder.

The statement that all prime numbers are odd is incorrect. While it is true that some prime numbers are odd, there are also prime numbers that are even, such as the number 2. Therefore, the correct answer is false.

The statement is true because all prime numbers (except 2 and 3) can be represented in the form of 6k ± 1, where k is a positive integer. This is known as the "6k ± 1" form. When a number is in this form, it means that it is either 1 more or 1 less than a multiple of 6. This holds true for prime numbers such as 5 (6k - 1), 7 (6k + 1), 11 (6k + 5), 13 (6k - 1), and so on. Therefore, the given statement is correct.

The last digit of a number is determined by its remainder when divided by 10. When we multiply any number by 0, the result will always be 0. Therefore, when we multiply all the given numbers together, the last digit will always be 0.

When a two-digit number of the format 'aa' is squared, the result will have the format 'bbcc'. This means that the first two digits of the square will be the same, and the last two digits will also be the same. The only option that satisfies this condition is 16. Therefore, the sum of the digits is 1 + 6 = 7.

The digit 7 cannot be there at the unit digit's place of the product of these two numbers. This is because when multiplying any number by 7, the unit digit of the product will always be 7. Since the sum of the two 3-digit numbers is 606, it is not possible for the unit digit of their product to be 7.

When a number is divided by 1001, the remainder is obtained by subtracting multiples of 1001 from the number until the result is less than 1001. In this case, since the number P consists of 777 repeated 777 times, we can see that 777 can be divided by 1001 without any remainder. Therefore, the remainder when P is divided by 1001 is 777.

The last two digits of a number are determined by the remainder when the number is divided by 100. Adding the last two digits of 143514587 (87) and 789999982 (82) gives 169. Therefore, the last two digits of the sum are 69. However, since 69 is greater than 52, we need to carry over the tens digit to the left. This results in the last two digits being 52.

The last digit of the sum of 378873 and 873378 is 7.

If n = x + 1, where x is a product of four consecutive positive integers, then n will be odd because adding 1 to any even number will result in an odd number. However, n may not necessarily be prime because it could have factors other than 1 and itself. Additionally, n may or may not be a perfect square since it depends on the values of the four consecutive positive integers that make up x. Therefore, the correct options are 1 and 3 only.

The number of zeros at the end of a factorial number is determined by the number of multiples of 10 in that factorial. Since every multiple of 10 has a factor of 10 and 2, and there are usually more multiples of 2 than 5, the number of zeros is equal to the number of multiples of 5 in the factorial. In this case, the factorial is 100!10!, so we need to find the number of multiples of 5 in both 100! and 10!. There are 24 multiples of 5 in 100! and 2 multiples of 5 in 10!. Therefore, the total number of zeros at the end is 24*10!.

When a two-digit number is reversed, the difference between the original number and the reversed number is always a multiple of 9. In this case, when 13 is reversed to 31, the difference is 18, which is a multiple of 9. Therefore, any two-digit number that increases by 18 when reversed will also be a multiple of 9. The only two-digit numbers that are multiples of 9 are 18, 27, 36, 45, 54, 63, 72, 81, and 90. However, 18 and 90 cannot be reversed to form another two-digit number. So, the other two-digit numbers that exist are 27, 36, 45, 54, 63, and 72, making a total of 6 numbers.

When 531111 is divided by 51, the remainder is 128.