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Q.  Find the remainder of ‎[7¹²³ + 9¹²³] / 64 ??

[Wonder Boy]

This problem can be easily solved by using binomial theorem.

According to binomial theorem one can expand any power of (x+y) into the sum of the form.

In short form we can write it as

 

so here we will try to expand the given numbers.

9 = 1 + 8

–> 9¹²³ = (1+8)¹²³

= ¹²³C₀ 1¹²³ 8⁰ + ¹²³C₁ 1¹²² 8¹ +………. +¹²³C₁₂₃ 1⁰ 8¹²³

Since leaving first two terms other terms contains power of 8² = 64 so all the other terms are divisible by 64.

Now we have to check the divisibility of only 1st 2 terms i.e. ¹²³C₀ 1¹²³ 8⁰ + ¹²³C₁ 1¹²² 8¹

Again 

7 = -1 + 8

–> 7¹²³ = (-1+8)¹²³

= ¹²³C₀ (-1)¹²³ 8⁰ + ¹²³C₁ (-1)¹²² 8¹ +………. +¹²³C₁₂₃ (-1)⁰ 8¹²³

Since leaving first two terms other terms contains power of 8² = 64 so all the other terms are divisible by 64.

Now we have to check the divisibility of only 1st 2 terms i.e. ¹²³C₀ (-1)¹²³ 8⁰ + ¹²³C₁ (-1)¹²² 8¹

 

So we have

7¹²³ + 9 ¹²³ =  ¹²³C₀ 1¹²³ 8⁰ + ¹²³C₁ 1¹²² 8¹ + 8²x some number  + ¹²³C₀ (-1)¹²³ 8⁰ + ¹²³C₁ (-1)¹²² 8¹ + 8²x some number

= ¹²³C₀ 1¹²³ 8⁰ + ¹²³C₁ 1¹²² 8¹ +¹²³C₀ (-1)¹²³ 8⁰ + ¹²³C₁ (-1)¹²² 8¹ + 8²x some number

= 1 + 123 x 8 + (-1) + 123 x 8  + 8²x some number  = 2x123x8 + 1 + (-1) + 8²x some number

= 16 x 123 +  8²x some number

 

Hence remainder of 7¹²³ + 9 ¹²³ is divided by 64 = remainder of 16×123 /64 = 16x[remainder of 123/4] = 16 x3 = 48 .

 

 

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