If x = (16

^{3}+ 17^{3}+ 18^{3}+ 19^{3}), then x divided by 70 leaves a remainder of
(1)0 (2) 1 (3) 69
(4) 35 (5) none of these

Solution follows here:

__Solution:__
x = 16

^{3}+ 17^{3}+ 18^{3}+ 19^{3}
For such type of problems look for a small logic like, 16+19
= 17+18

x = (16

Applying the formula, a^{3}+ 19^{3}) + (17^{3}+ 18^{3})^{3}+b^{3}= (a+b)(a^{2}-ab+b^{2}):
=> x = (16+19)(16

^{2}-16.19+19^{2}) + (17+18) (17^{2}-17.19+19^{2})
=> x = (35)(16

^{2}-16.19+19^{2}) + (35) (17^{2}-17.19+19^{2})
From now onwards, no need of finding the values of these squares
and multiplications. Just go on finding whether the resulting number is odd or
even.

The logic here is:

odd+odd results in even; odd+even results in odd; even+even
results in even; odd*odd results in odd; even*even results in even; odd*even
results in even;

=> x = (35)(even-even+odd) + (35) (odd-odd+odd)

=> x = (35)(even+odd) + (35) (odd) => x = (35)(odd) + (35)
(odd)

=> x = (35)(odd+odd)
=> x = (35)(even) => x = (35)[2*(some
integer)]

=> x = (70)[some integer]
=> ‘x’ is a multiple of 70

=> ‘x’ when divided by 70 leaves a remainder ‘0’.

**Answer (1)**

thanxx sir

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