Question
If 'x' is the lowest positive integer divisible by 10,
18 and 21, then find the second smallest positive integer which is divisible by all the three given numbers?Solution
Lowest positive integer divisible by 10, 18 and 21 = LCM (10, 18, 21) So, LCM of (10, 18, and 21) = 2¹ × 3² × 5¹ × 7¹ = 630 Therefore, the next larger number which is divisible by all the three given numbers = 2 × 630 = 1260
If a number 5678x43267y is completely divisible by 72 then (x+y)?
What is the least number which when divided by 15, 18 and 36 leaves the same remainder 9 in each case and is divisible by 11?
Find the smallest 4-digit number which when divided by 10, 15, 20, 25, 30 leaves remainder 9.
Which of the following numbers is divisible by 11?
On dividing a certain number by 304, we get 43 as the remainder. If the same number is divided by 16, what will be the remainder?
If ‘1x5629’ is a six digit number which is divisible by 9, then which of the following can be the minimum value of ‘x’?
If ‘48b297a54’ is a nine digit number which is divisible by ‘11’, then find the smallest value of (a + b).
A number n when divided by 6, leaves a remainder 3. What will be the remainder when (n2 + 5n + 8) is divided by 6?
A six-digit number 27p5q8 is divisible by 36. What is the greatest possible value for (p×q)?
How many factors of 23 × 33 × 54 ×91× is divisible by 50 but not by 100?