Numbers Questions

The only even numbers in the list are 2 and 4, but 4 is not a prime. So 2 can be used to illustrate the statement that all primes are not odd.

Clearly 4864 is divisible by 4
So, 9 P 2 must be divisible by 3. So (9+P+2) must be divisible by 3.
So P=1.

Required numbers are 10,15,20,25,...,95
This is an A.P. in which a=10,d=5 and l=95.
Let the number of terms in it be n.Then t=95
So a+(n-1)d=95.
10+(n-1)*5=95,then n=18.
Required sum=n/2(a+l)=18/2(10+95)=945.

Step 1 :2x + 5y

Step 2 :3(2x + 5y) = 6x + 15y

Step 3 :(6x + 15y) - (x +y) = 6x + 15y - x - y (watch that last minus sign!)

This gives 5x + 14y

(2272-875) = 1397, is exactly divisible by N.

Now , 1397 = 11 x 127

The required 3-digit number is 127,the sum of digits is 10.

987 = 3 * 7 * 47.
So, the required number must be divisible by each one of 3, 7, 47
553681 => (Sum of digits = 28, not divisible by 3)
555181 => (Sum of digits = 25, not divisible by 3)
555681 is divisible by each one of 3, 7, 47.

The best approach is to pick a number that satisfies the rules that x and y are greater than 0. So for example you could choose 25 for y.

The value of x before doubling is then 25 - 50/25 = 23

The value after doubling y will be 50 - 50/50 = 49, which is more than double.

(Place value of 7)-(face value of 7)
=7000-7=6993.