What is the remainder when (32540 x 42400 x 3233) + 222 is divided by 13?

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-1

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Solution

The correct option is C -1
32540 using the divisibility test for 13 ( triplets at odd- triplets at even places ) can be written as

540-32= 508. 508 when divided by 13 can be represented as 13k + 1 (i.e. when the number is divided by 13, the remainder is 1)

Similarly

42400= 400-42= 358 = 13k + 7

3233= 233-3=230= 13k + 9

(13k + 1)( 13k + 7)(13k + 9) will give the non-divisible part as (9x7x1) = 63

therefore 63+222= remainder of 285 when divided by 13= -1 or 12

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Similar questions

(32540 \times 42400 \times 3233) + 222

_{32} 33^{34}

When $32^{33}$ is divided by 34, it leaves the remainder

x^{13} + 1

x - 1

2^{2} + 22^{2} + 222^{2} + 2222^{2} + . . . . (222 . . . .49 t w o s)^{2}

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