Numbers and Divisibility

[Explanation: To find the least number which is exactly divisible by 20, 25, 30, we have to calculate the LCM of 20,25,30.LCM of 20,25,30=300 But this is a 3-digit numberSo, to get a 5-digit number 300 should be multiplied with 34 300 x 34=10200]

[Explanation: According to the divisibility rule of 2, 1586 is an even number, therefore, divisible by 2. But last 2 digits, i.e., 86 is not divisible by 4 Therefore, 1586 is not divisible by 4.

[Explanation: Sum of digits= 2+7++4= 13+ To make it divisible by 3, the sum should be 15 Therefore, 2 must be added Now the required number is 2724]

n³ - n = n (n² - 1) = n (n + 1) (n - 1) So n³ - n can be expressed as a product of three consecutive numbers, then at least one of the three consecutive numbers should be a multiple of 3. Hence n³ - n can be divisible by 3.

8x 9y 70 = 8x 9y00 + 70 8x 9y00 can be divisible by 4, but 70 cannot be divisible by 4. So 8x 9y 70 cannot be divisible by 4.

24 = 2*3*4, if x²  is divisible by 24, x² must have factors of 2 and 3. 1)  only, x does not have factor of 3, so x² does not have factor of 3, x² cannot be divisible by 24, sufficient. 2)  only, x is odd, so x does not have factor of 2, so x²  does not have factor of 2, x² cannot be divisible by 24, sufficient.

9! - 2* (7!)² = (7!) (8*9 - 2*7!) = - 18*(7!) - 18 cannot be divisible by 5. 7! = 1*2*3*4*5*6*7 can only be divisible by 5^1. Since n is a positive integer, so the greatest possible value of n should be 1.

1) only, a (50) = a (49) + 5 = a (48) + 5 + 5 = a (48) + 2*5 =…= a (1) + 49*5 = 3 + 49*5, so a (50) - 1 = 49*5 + 2, which cannot be divisible by 5, sufficient. 2) only, a (50) = a(49) + 5, if a (49) is not divisible by 5, then a(50) must not be divisible by 5, but a (50) - 1 might be divisible by 5. Because we don't know the remainder of a(50) divided by 5, we don't know if a (50) - 1 can be divisible by 5 or not, insufficient.

n can always be divisible by itself, so n is not an odd number. n is not divisible by any odd number, so n's factors can only be 2 or 2m. Hence, n can only be 2, 2^2 = 4, 2^3 = 8, 2^4 = 1^6 and 2^5 = 32. There are 5 possible values.

15 = 3*5, 21 = 3*7, 36 = 3*12 So the smallest common multiple of 15, 21 and 36 would be 3*5*7*12 = 1260. D is right.

2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32, 2⁶ = 64… We can find out that the units digit of 2ⁿ is the same as the units digit of 2^{(n + 4)}. So, the units digit of 2²⁰ would be the same as the units digit of 2⁴, which is 6. If the units digit of 2^{(20) - x} is 0 or 5, then it can be divisible by 5. So, x should be 1 or 6.

difference of odd number sum  and  even number sum should be divisble by 11.

This is the correct answer. If the sum of the numbers is divisible by 3, then the number is divisible by 3.

This is the correct answer. A number is divisible by 9 when the sum of the digits is divisible by 9. 

This is the correct answer. If a number is divisible by co-prime numbers, then it is also divisible by the product of the two.