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.
87)13601(156
87
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490
435
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551
522
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29
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Required number is 29.
Given
=> A
Therefore, A = B
Given equation 5x-17 = -x+7
Add 1x to each side of the equation
5x-17+x = -x+7+x
6x=24
x=4
Taking Log to both sides
we get
n = 8
(Place value of 7)-(face value of 7)
=7000-7=6993.
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
(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.
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.
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.
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