Splet13. apr. 2024 · The two digit numbers are 10 to 99. So there are 90 two digit numbers. Half of them are even, so 45 two digit numbers are divisible by 2. The lowest two digit number divisible by 3 is 12. And out of 90 numbers, every 3 is divisible by 3. So there are 30 two digit number divisible by 3. So 45 +30 = 75. Answer link SpletSolution Number of possible outcomes = 100 (i) Let E1 be the event of getting a number divisible by 9 and is a perfect square. ∴ Favourable outcomes = {9, 36, 81} Number of favourable outcomes = 3 ∴ P (E1) = 3 100 (ii) Let E2 be the event of getting a prime number greater than 80. ∴ Favourable outcomes = {83, 89, 97}
The number {(101)^100-1} is divisible by - Brainly
SpletAre the numbers in the multiplication table of 101 all prime numbers? No! Only the number 101 is a prime number as 101 is divisible by 1 and 101 only. However, the numbers that appear in the multiplication tables of 101 are divisible by 1, 101 and the number itself. What is 1000 times 101? Answer: 101000 SpletApplying the divisibility test for 3, we get that 1+4+8+1+4+8+1+4+6+8=45, 1+ 4+8+1 +4+8+ 1+4+6 +8 = 45, which is divisible by 3. Hence 1,481,481,468 is divisible by 3. Applying the divisibility test for 4, we get that the last two digits, 68, is divisible by 4. Hence 1,481,481,468 is also divisible by 4. indices notaires-insee
The number 101 ^100 - 1 is divisible by: - Toppr
Splet26. mar. 2024 · Hint: In this question, a 300 digit number is given whose all digits are equal to 1; since it is a 300 digit number, it will be difficult to check whether it is divisible by 37 and 101; hence we will try to break the number into the simplest form and then we will check its divisibility. Complete step-by-step answer: Given that the number is a 300 digit … SpletWhat numbers is 1,001 divisible by? Is 1,001 a prime number? This page will calculate the factors of 1,001 (or any other number you enter). Splet=100 2 (1+ 100 C 2 + 100 C 3 (100)+………+ 100 C 100 100 98) So, 101 100 -1 is divisible by 10 4 Question 3: If n is even positive integer, then the condition that the greatest term in the expansion of (i + x) n may also have the greatest coefficient is a. n / [n + 2] < x < [n + 2] / n b. n / [n + 1] < x < [n + 1] / n locksmith 75050