Competitive exam Questions and Answers on Numbers.
In this section, we’ll explore three methods: A number is said to be divisible by 7 if it can be divided by 7 without leaving any remainder. Instead of performing long division every time, there is a shortcut rule: 1.…
Let , S = 10a + b If “S” is divisible by 7 then if we substract or add any multiple of “7” to “S ” then resultant number will still be divisible by “7” So we get, S -7a…
From image find value of X = ????
Solved using divisibility rule for ‘ 7 ‘
We know 4^2 = 16 4^4 = 256 So all even powers of 4 will give 6 at units place. So in the expansion of both given expressions we get 6 at units place. Hence we get, 6 + 6…
As we know 30! = 30 x 29 x 28 ………………………..x 3 x 2 x 1 Lets divide the answer into three parts First lets mutiply 1 to 10 1 x 2x 3 x 4 x 5 x 6 x…
We know 10! = 10x9x8………x3x2x1 So we get one trailing zero from 10 itself Smilarly we get another trailing zero from 5×2 = 10 Thus answer is ‘ 2 ‘
As we know 100! = 100 x 99 x 98 ………………………..x 3 x 2 x 1 First lets mutiply 1 to 10 1 x 2x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10…
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