The remainder obtained when 13-23-33-43- 1003 is divided by 9 is

The correct option is A 1 1^3+2^3+3^3+4^3……100^3= [(100)^2(101)^2 ]/4=25^2×101^2×4 Remainder of a product when it is being divided is going to be the same as ...

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Quantitative Aptitude

Divisibility Rule for Powers of 2 & 5

The remainder...

Question

The remainder obtained when $(1^{3} + 2^{3} + 3^{3} + 4^{3} + \dots \dots 100^{3})$ is divided by 9 is

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Solution

The correct option is B
1

$1^{3} + 2^{3} + 3^{3} + 4^{3} \dots \dots 100^{3} = \frac{[(100)^{2} (101)^{2}]}{4} = 25^{2} \times 101^{2} \times 4$

Remainder of a product when it is being divided is going to be the same as the product of the individual remainders.

$R e m (25^{2} \times 101^{2} \times 4) = R e m (4 \times 4 \times 4) = R e m (64) = 1$

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The remainder obtained when (13+23+33+43+……1003) is divided by 9 is

The correct option is B 1 13+23+33+43……1003=[(100)2(101)2]4=252×1012×4 Remainder of a product when it is being divided is going to be the same as the product of the individual remainders. Rem(252×1012×4)=Rem(4×4×4)=Rem(64)=1

The remainder obtained when $(1^{3} + 2^{3} + 3^{3} + 4^{3} + \dots \dots 100^{3})$ is divided by 9 is