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.
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.
(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.
Let the Number be Y.
Then Y = 296 q + 75
= (37 x 8)q +( 37 x 2) + 1
= 37 (8q + 2) + 1
Thus, when the number is divided by 37, the remainder is 1
(Place value of 7)-(face value of 7)
=7000-7=6993.
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.
Let the two consecutive odd integers be (2x + 1) and (2x + 3)
Then,
(2x + 3)2 - (2x + 1)2
= (2x + 3 + 2x + 1) (2x + 3 - 2x - 1)
= (4x + 4)(2)
= 8 (x + 1), which is always divisible by 8
The pattern is -45, -35, -25, -15
The next number = 20-15= 5
Let the required fraction be x. Then, (1 / x )- x = 9/20
1 - x^(2) / x = 9 / 20 => 20 - 20 * x^(2) = 9 * x.
20 * x^(2) + 9 *x - 20 = 0.
=> (4 * x + 5) (5 * x - 4) = 0.
=> x = 4 / 5.
The sum of three consecutive integers can be written as n + (n + 1) + (n + 2) = 3n + 3
If the sum is 24, we need to solve the equation 3n + 3 = 24;
=> 3n = 21;
=> n = 7
The greatest of the three numbers is therefore 7 + 2 = 9
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.