J's answer to Nicholas's Secondary 3 A Maths Singapore question.

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J
J's answer
1022 answers (A Helpful Person)
1st
Quite simple. Notice that the 49 has a power of 15 ? And the binomial expansion has a (1 + x) inside. The idea is to break up the 49 into two parts, such that you can perform binomial expansion on it.
49 = 1 + 48 (note that 48 is divisible by 8)
So 49^15 = (1 + 48)^15
= 1^15 + (15choose1)(1^14)(48) + (15choose2)(1^13)(48²) + ... + (15choose13)(1²)(48^13) + (15choose14)(1)(48^14) + 48^15
Every term in the expansion has a multiple of 48 as a factor, except 1^15. So the sum of all those terms except 1^15 are also divisible by 48 (which in turn implies they are divisible by 8 also).
1^15 = 1 so subtracting 1 from 49^15 takes it away. This would make the remaining expression divisible by 8.
J
J
3 years ago
Oh it's actually (x + 1)ⁿ so I actually did it the other way round. The sum is the same , just flip the terms in reverse order.
Eric Nicholas K
Eric Nicholas K
3 years ago
Basically every of the terms, except 1^15, will have a factor of 48 (since 48^r for r >= 1 is divisible by 48).

We must actually also mention that the other terms in the expansion, including nCr, are also integers.
Nicholas
Nicholas
3 years ago
Ok thx
J
J
3 years ago
There is no need to for the combinations. As for 1^(n-r) , it is self explanatory.

It is understood that for the binomial theorem, the combination (n r) will only be an positive integer since n and r are positive integers and r ≤ n. This is clearly shown in the Pascal's triangle

We do not have to be concerned about the non-integer n and r here (that is under the topic of gamma function)