Question

What is the remainder of 67^24-54^24 when divided by 11​

Answer 1

( 67 )²⁴ - ( 54 ) ²⁴

=( 67 ¹²)² - ( 54 ¹²)²

following the formula:-  ( a²- b²) = ( a+ b ) ( a - b)

=( 67 ¹² + 54 ¹²) ( 67¹² -54¹² )

=( 67 ¹²+ 54¹² ) { ( 67⁶)² - ( 54⁶ ) ²}

=( 67 ¹²+ 54¹² )( 67⁶ +54⁶ ) ( 67⁶ - 54⁶)

=( 67 ¹²+ 54¹² )( 67⁶ +54⁶ ) { ( 67² )³ - (54 ²) ³}

let's consider ( 67² )³ = a³ and  (54 ²) ³= b ³

now following this formula:-  ( a³ - b³ ) = ( a - b ) ( a² + ab + b² )

=( 67 ¹²+ 54¹² )( 67⁶ +54⁶ ) (67² - 54² ) {( 67²)²+ 67² ₓ 54² + ( 54² ) ²}

=( 67 ¹²+ 54¹² )( 67⁶ +54⁶ ) ( 67 + 54 ) ( 67 - 54 ) ( 67⁴ +67² ₓ 54²+ 54⁴ )

=( 67 ¹²+ 54¹² )( 67⁶ +54⁶ ) ( 67⁴ +67² ₓ 54²+ 54⁴ ) (121 ) (13 )

=( 67 ¹²+ 54¹² )( 67⁶ +54⁶ ) ( 67⁴ +67² ₓ 54²+ 54⁴ ) ( 121 ₓ 13 )

Ans :-  Remainder will be 0

121 will be divisible by 11

Answer 2

Remainder of 67²⁴ - 54²⁴ when divided by 11​ is Zero (0).

Step 1:

Use law of indices xᵃᵇ = (xᵃ)ᵇ and write 24 as 12 x 2

67²⁴ - 54²⁴

= 67¹²ˣ² - 54¹²ˣ²

= (67¹²)² - (54¹²)²

Step 2:

Use a² - b² = (a + b)(a - b)

a = 67¹²

b = 54¹²

= (67¹² + 54¹²)(67¹² - 54¹²)

Step 3:

Writing 12 as 6 x 2 and using again a² - b² = (a + b)(a - b)

= (67¹² + 54¹²)(67⁶ +  54⁶)(67⁶ -  54⁶)

Step 4:

Writing 6 as 2 x 3 and using a³ - b³ = (a - b)(a² + b² + ab)

a = 67²  , b = 54²

= (67¹² + 54¹²)(67⁶ +  54⁶) (67²-  54²)(67⁴ +  54⁴ + 67²*54²)

Step 5:

Using again a² - b² = (a -+ b)(a - b) for  (67²-  54²)

= (67¹² + 54¹²)(67⁶ +  54⁶) (67 +  54)(67 - 54)(67⁴ +  54⁴ + 67²*54²)

= (67¹² + 54¹²)(67⁶ +  54⁶) (121)(13)(67⁴ +  54⁴ + 67²*54²)

Step 6:

121 is 11 x 11 hence divisible by 11  so remainder will be zero

=  11 (67¹² + 54¹²)(67⁶ +  54⁶) (11)(13)(67⁴ +  54⁴ + 67²*54²)

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