Algebra -26 (CAT-2004)

The remainder, when (15²³+23²³) is divided by 19 is:

(1)4                 (2)15               (3)0                 (4)18

Solution follows here:

Solution:

15²³+23²³

Observe that 19 is the arithmetic mean of 15 and 23

=> 15 can be written as (19-4) and 23 can be written as (19+4)

15²³+23²³ = (19-4)²³+(19+4)²³

From Binomial theorem,

(x+y)ⁿ = nC0 xⁿy⁰ + nC1 x^n-1y¹ + nC2 x^n-2y² + ..... + nCn x⁰yⁿ

(x-y)ⁿ = nC0 xⁿy⁰ - nC1 x^n-1y¹ + nC2 x^n-2y² - ..... ± nCn x⁰yⁿ

For odd n,

(x+y)ⁿ + (x-y)ⁿ = 2{ nC0 xⁿy⁰+ nC2 x^n-2y²+.......+ nC(n-1) x¹y^n-1 }

=> (19-4)²³+(19+4)²³ = 2{ 23C0 19²³4¹+ 23C2 19²¹4²+.......+ 23C22 19¹4²²}

= 2*19*{23C0 19²²4¹+ 23C2 19²⁰4²+.......+ 23C22 19⁰4²²}

Observe that this is a multiple of 19

=> when (15²³+23²³) is divided by 19, the remainder is zero

Answer (3)