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v = 10 x k ,
Let k = 325 , v = 3250,
The sum of the digits for both remains same which is = (3 + 2 + 5 + 0) = 10
When m, n are integers between 1 and 10, and m < n, the number of multiples of m will be more than the number of multiples of n. E.g. for 8 , 9, the number of multiples are respectively 12 and 11
The max value of column A is 15 and min is 10,
The max value of column B is 18 and min is 12.
Hence, no relation can be found.
When a = 14, the remainder of a and a^{2} are respectively 0 and 0.
But when a = 16, the remainder of a and a^{2} are respectively 2 and 4.
S = (3, 2, 2, 3). The members of a set T are the squares of numbers in set S.
S = { 3 , 2 , 2 ,3 }, T = {9 , 4 }, n ( S ) = 4, n (T ) =2
If the integers are (70 ,70 ,70, 70, 70) the median is 70, but if the integers are (60, 65, 65, 80, 80), the median is 65.
Since, they are consecutive odd numbers and R < S < T, (S – R) = 2, (T – S) = 2, Adding (S + T) – (R + S) = 4
⇒(S + T) = ( R + S ) + 4,
⇒( S + T )  1 = ( R + S ) + 3 > ( R + S ) + 1
For x = 2, y = 3, z = – 4, x × y × z = 24, x + y + z = 9
But for x = y = z = 1, x × y × z = 1, x + y + z = 3
x is an integer, and the remainder when 2x is divided by 4 is 0
If x = 4, 2x = 8, 8 / 4,
the remainder is = 0.
Also for 4/4, the remainder is = 0.
But if x = 6 ,2x = 12 , 12 / 4,
the remainder is = 0. But for 6 / 4 , the remainder is ≠ 0.
Each of w and x is less than 5 and greater than 2. Each of y and z is less than 2 and greater than
2 < w < 5, 2 < x < 5.
Adding, we get 4 < ( x + w ) < 10, 1 < y < 2, 1 < z < 2,
Adding, we get 2 < ( y + z ) < 4 < ( w + x ) < 10
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