Olympiad Test: Whole Numbers - Class 6 MCQ

97 is a prime number because it has no divisors other than 1 and itself.To determine which number is a prime number among the options given:

- 123: Not a prime number because it is divisible by 3 (3 x 41).
- 97: This is a prime number because it has exactly two factors: 1 and 97.
- 121: Not a prime number as it is divisible by 11 (11 x 11).
- 105: Not a prime number as it is divisible by 3 and 5 (3 x 5 x 7).

Therefore, the prime number among the options is 97.

So, the number of whole numbers between 25 and 75, inclusive, is:

75 − 25 + 1 = 51

There are 51 whole numbers between 25 and 75, inclusive.

To find the sum of the first 10 odd numbers:

- Odd numbers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
- Adding these numbers: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100
- The sum of the first 10 odd numbers is 100.
- Therefore, option B (100) is the correct answer.

To find the smallest number that satisfies the given condition, we can use the concept of the least common multiple (LCM) of 4, 6, and 9.

The LCM of 4, 6, and 9 is the smallest number that is divisible by all three numbers.

The prime factorization of these numbers is:

• 4 = 2^2
• 6 = 2 * 3
• 9 = 3^2

To find the LCM, we take the highest power of each prime factor that appears in any of the factorizations:

LCM(4, 6, 9) = 2^2 * 3^2 = 4 * 9 = 36

So, the smallest number that when divided by 4, 6, and 9 leaves a remainder of 1 is 36 + 1 = 37.

Whole numbers divisible by 3 between 100 and 200 are 102, 105, ..., 198. There are 34 such numbers.

198=102+(n−1)3

96 = (n - 1)3

n - 1 = 32

n = 33.

Since we know that 99999 is the largest 5-digjt no. So when we divide it by 8 we get 7 as remainder.

When 7 is subtracted from 99999 , 99992 is obtained which is divisible by 8 and is the highest 5-digit no. Divisble by 8.

Another method :-

Since, we know that if a number’s last three digits are divisible by 8 then the whole no. Is divisible by 8 ( for more information search divisibility rule of 8 on google ) .

So on dividing 999 by 8 , 7 is obtained as a remainder and we know that ,

Dividend = Divisor *Quotient + remainder

=> Dividend - remainder = divisor * quotient

=> if remainder is subtracted from the dividend then it is divisible by the divisor.

Therefore ,(999–7) is divisible by 8

=> 992 is divisible by 8 .

As I earlier wrote that if a number's last three digits are divisible by 8 then the whole no. Is divisible by 8 .

992 is divisible by 8

Therefore , 99992 is also divisible by 8 and is the highest 5 digit no. Divisible by 8.

2310 = 2 * 3 * 5 * 7 * 11, and 15 is a factor since 2310 is divisible by 15. Among the above options 15 is the smallest factor of 2310.

The smallest 2-digit prime number is 11 and the largest 1-digit prime number is 7. Their product is 11 * 7 = 77.