Question

Is 1¹¹ + 2¹¹ + 3¹¹ + 4¹¹ + 5¹¹ divisible by 5? Justify.​

Answer 1

$\large\underline{\sf{Solution-}}$

We know,

★ A natural number is divisible by 5 if its unit digit is either 0 or 5.

★ So, in order to prove that 1¹¹ + 2¹¹ + 3¹¹ + 4¹¹ + 5¹¹ divisible by 5.

★ We have to just prove that unit digit of 1¹¹ + 2¹¹ + 3¹¹ + 4¹¹ + 5¹¹ is either 0 or 5.

Now,

Consider,

$\rm :\longmapsto\: {1}^{11}$

$\rm :\longmapsto\:Its \: unit \: digit \: is \: always \: 1$

Consider,

$\rm :\longmapsto\: {2}^{11}$

★ We check the cyclicity of 2 for unit digit.

$\rm :\longmapsto\: {2}^{1} \: ends \: with \: 2$

$\rm :\longmapsto\: {2}^{2} \: ends \: with \: 4$

$\rm :\longmapsto\: {2}^{3} \: ends \: with \: 8$

$\rm :\longmapsto\: {2}^{4} \: ends \: with \: 6$

$\rm :\longmapsto\: {2}^{5} \: ends \: with \: 2$

★ It implies after 4, its start repetition.

So,

$\rm :\longmapsto\: {2}^{11} \: ends \: with \: 8$

Consider

$\rm :\longmapsto\: {3}^{11}$

★ We check the cyclicity for 3

$\rm :\longmapsto\: {3}^{1} \: ends \: with \: 3$

$\rm :\longmapsto\: {3}^{2} \: ends \: with \: 9$

$\rm :\longmapsto\: {3}^{3} \: ends \: with \: 7$

$\rm :\longmapsto\: {3}^{4} \: ends \: with \: 1$

$\rm :\longmapsto\: {3}^{5} \: ends \: with \: 3$

★ It implies after 4, it starts repeatedly.

So,

$\rm :\longmapsto\: {3}^{11} \: ends \: with \: 7$

Consider

$\rm :\longmapsto\: {4}^{11}$

★ Now, we check the cyclicity of 4

$\rm :\longmapsto\: {4}^{1} \: ends \: with \: 4$

$\rm :\longmapsto\: {4}^{2} \: ends \: with \: 6$

$\rm :\longmapsto\: {4}^{3} \: ends \: with \: 4$

★ It implies cyclicity of 4 is 2.

★ It means, after 2 is starts repeating.

So,

$\rm :\longmapsto\: {4}^{11} \: ends \: with \: 4$

Consider,

$\rm :\longmapsto\: {5}^{11}$

$\rm :\longmapsto\:Its \: unit \: digit \: is \: always \: 5$

Hence,

★ The sum of unit digits of 1¹¹ + 2¹¹ + 3¹¹ + 4¹¹ + 5¹¹ is

$\rm \: = \: \: \:1 + 8 + 7 + 4 + 5$

$\rm \: = \: \: \:25$

$\rm :\implies\: {1}^{11} + {2}^{11} + {3}^{11} + {4}^{11} + {5}^{11} \: always \: ends \: with \: 5$

$\bf\implies\:{1}^{11}+{2}^{11}+{3}^{11} +{4}^{11}+{5}^{11} \:is\:divisible\:by\:5$