So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.
264 → 11,3,4 (/)
396 → 11,3,4 (/)
462 → 11,3 (X)
792 → 11,3,4 (/)
968 → 11,4 (X)
2178 → 11,3 (X)
5184 → 3,4 (X)
6336 → 11,3,4 (/)
Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.
Required number of number = 4.
Thus, when 2n is divided by 4, the remainder is 2.
The largest 4 digit number is = 9999
After dividing 9999 with 88, we get 55 as remainder
so largest 4 digit number divisible by 88 = 9999-55 = 9944.
LCM of 2, 3, 7 is 42.
=> (700 ? 300)/42 = 9 22/42 => 9 Numbers.
Since 653xy is divisible by 5 as well as 2, so y = 0.
Now, 653x0 must be divisible by 8.
So, 3x0 must be divisible by 8. This happens when x = 2
x + y = (2 + 0) = 2.
Probability of left handed = 4/5
total student = 40
total left handed = 40 x(4/5)
= 32
(kx22)/100 = 340 - 166.64 = 173.36
k = (173.36 x 100)/22
k = 788
Let the three consecutive odd numbers be x, x+2, x+4
Then,
x + x + 2 + x + 4 = 93
=> 3x + 6 = 93
=> 3x = 87
=> x = 29 => 29, 31, 33 are three consecutive odd numbers.
Therefore, the middle number is 31.
Given the angles of a quadrilateral are in the ratio of 2:4:7:5
Let the angles of a quadrilateral are 2x, 4x, 7x, 5x
But we know that sum of the angles = 360 degrees.
=> 2x + 4x + 7x + 5x = 360
=> x = 20
Therfore, the smallest angle of the quadrilateral = 2x = 2x20 = 40 degrees.
One of the angle of the triangle = 2 x 40 = 80 degrees
The other angle is 180 - (40 + 80) = 60 degrees.
Hence the second largest angle of the triangle is 60 degrees.
we can do this by trial and error method.
Putting x = 2,we get 2²(2² - 1) = 12.
checking with other integers, the above equation always gives a value which is a multiple of 12,
So, x²(x² - 1) is always divisible by 12.
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