Type I
(i) `(x^n-a^n)` is divisible by `(x- a),` `AA n in N.`
(ii) `(x^n+a^n)` is divisible by `(x+a),` `AA n in ` Only odd natural numbers.
Type II To show that an expression is divisible by an integer.
Solution Process
(i) If `a, p, n` and `r` are positive integers, then first of all write `a^(pn + r) = a^(pn) . *a ^r =(a^P)^n * a^r`
(ii) If we will show that the given expression is divisible by `c.`
Then , expression `a^p = { 1 + ( a^p - 1) },` if some power of `( a^p - 1)` has `c` as a factor.
or `a^p = { 2 + (a^p- 2)},` if some power of `(a^p- 2)` has `c` as a factor.
or `a^p = { 3 + (a^p- 3)},` if some power of `(a^p- 3)` has `c` as a factor.
.
.
or `a^p = {k + (aP- k)},` if some power of `(a^p- k)` has `c` as a factor.
`text( How to Find Remainder by Using Binomial Theorem )`
If `a, p, n` and `r` are positive integers, then to find the remainder when `a^(pn + r)` is divided by `b`, we adjust power of `a` to `a^(pn + r)` which is very close to `b` , say with difference `1` i.e., `b ± 1.` Also, the remainder is always positive. When number of the type `5n - 2` is divided by `5`, then we have
`(5n-2)/5=(5n-5+3)/5=n-1+3/5`
Hence, the remainder is `3.`
`text( Two Important Results)`
`(i)` `2 <= (1+1/n)^n <3, n>=1 , n in N`
`(ii) ` If `n>6,` then `(n/3)^n < n! <(n/2)^n.`
`text( How to Find Last Digit, Last Two Digits, Last Three Digits, ... and so on)`
If `a, p, n` and r are positive integers, then `a^(pn + r)` is adjust of the form `(10k ± 1)^m,` where k and m are positive integers. For last digit, take 10 common. For last two digits, take 100 common, for last three digits, take 1000 common...., and so on.
`i.e. (10k pm 1)^m = (10k)^m + text()^mC_(1)(10k)^(m-1) (pm1) + text()^mC_(2)(10k)^(m-2) (pm1)^2................
text()^mC_(m-1)(10k) (pm1)^(m-1) + (pm1)^m`
For last digit = `10lambda + (pm1)^m`
For last two digits = `100mu +text()^mC_(m-1)(10k)^(m-1) + (pm1)^m `
Type I
(i) `(x^n-a^n)` is divisible by `(x- a),` `AA n in N.`
(ii) `(x^n+a^n)` is divisible by `(x+a),` `AA n in ` Only odd natural numbers.
Type II To show that an expression is divisible by an integer.
Solution Process
(i) If `a, p, n` and `r` are positive integers, then first of all write `a^(pn + r) = a^(pn) . *a ^r =(a^P)^n * a^r`
(ii) If we will show that the given expression is divisible by `c.`
Then , expression `a^p = { 1 + ( a^p - 1) },` if some power of `( a^p - 1)` has `c` as a factor.
or `a^p = { 2 + (a^p- 2)},` if some power of `(a^p- 2)` has `c` as a factor.
or `a^p = { 3 + (a^p- 3)},` if some power of `(a^p- 3)` has `c` as a factor.
.
.
or `a^p = {k + (aP- k)},` if some power of `(a^p- k)` has `c` as a factor.
`text( How to Find Remainder by Using Binomial Theorem )`
If `a, p, n` and `r` are positive integers, then to find the remainder when `a^(pn + r)` is divided by `b`, we adjust power of `a` to `a^(pn + r)` which is very close to `b` , say with difference `1` i.e., `b ± 1.` Also, the remainder is always positive. When number of the type `5n - 2` is divided by `5`, then we have
`(5n-2)/5=(5n-5+3)/5=n-1+3/5`
Hence, the remainder is `3.`
`text( Two Important Results)`
`(i)` `2 <= (1+1/n)^n <3, n>=1 , n in N`
`(ii) ` If `n>6,` then `(n/3)^n < n! <(n/2)^n.`
`text( How to Find Last Digit, Last Two Digits, Last Three Digits, ... and so on)`
If `a, p, n` and r are positive integers, then `a^(pn + r)` is adjust of the form `(10k ± 1)^m,` where k and m are positive integers. For last digit, take 10 common. For last two digits, take 100 common, for last three digits, take 1000 common...., and so on.
`i.e. (10k pm 1)^m = (10k)^m + text()^mC_(1)(10k)^(m-1) (pm1) + text()^mC_(2)(10k)^(m-2) (pm1)^2................
text()^mC_(m-1)(10k) (pm1)^(m-1) + (pm1)^m`
For last digit = `10lambda + (pm1)^m`
For last two digits = `100mu +text()^mC_(m-1)(10k)^(m-1) + (pm1)^m `