Question

Let N=55^3+17^3-72^3,then N is "not" divisible by A)4 B)13 C)15 D)17

Answer 1

Answer:

Step-by-step explanation:

Formula used:

If a+b+c=0,

then

${a}^3+{b}^3+{c}^3=3abc$

$N={55}^3+{17}^3-{72}^3$

$={55}^3+{17}^3+{(-72)}^3$

since 55+17+(-72)=0,

${55}^3+{17}^3+{(-72)}^3$

=3(55)(17)(-72)

=3×(11×5)(17)(-1)(2×2×2×3×3)

=11×15×17×(-1)×(4×2)×(3×3)

This product contains 4, 15, 17 except 13.

Hence N is not divisible by 13.

Answer 2

we know, if a + b + c = 0 then, a³ + b³ + c³ = 3abc

here , if we assume a = 55, b = 17 and c = -72
then, a + b + c = 55 + 17 - 72 = 0
so, we can apply a³ + b³ + c³ = 3abc

e.g., 55³ + 17³ - 72³ = 3(55)(17)(-72)

now, prime factors of 3(55)(17)(-72) : 2 × 4 × 3 × 3 × 3 × 5 × 11 × 17

here it is clear that N is not divisible by 13. because 13 is not present in prime factors of N.

hence, option (B) is correct.