How can I prove that n (n squared plus twenty) is divisible by forty eight?
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7 Answers
Best
Arpit Agrawal
miss Mathematics out of my life · Author has 374 answers and 1.3M answer views · 8y
X(n) = n(n
2
+ 20)
For n = 2, X = 48 which is divisible.
Let us assume X is divisible by 48 for a general number n = m for m>=2
So that
2
m(m
+ 20)
is divisible by 48
Now we shall prove that X is divisible by 48 for m+2 as well
2
X(m + 1) = (m + 2)[(m + 2)
2
= (m + 2)(m
2
= m(m
2
= m(m
2
= m(m
+ 20]
+ 4m + 4 + 20)
2
+ 4m + 4 + 20) +2(m
+ 4m + 4 + 20)
2
+ 20) +m(4m + 4) + 2(m
2
+ 20) + 6m
+ 4m + 4 + 20)
+ 12m + 48
= 48Z + 6m(m + 2) + 48
Now if we prove that
6m(m + 2)
is divisible by 48 we are done.
Since m is even, one of the factors m or m+2 is a multiple of 4 as well. So the overall m(m+2) is a multiple
of 8 and hence 6m(m+2) is a multiple of 48.
Now, we have proved that if we assume X to be divisible by m, X is also divisible by m+2, and we have
shown X to be divisible by 2, so it concludes that X is divisible by 4 and hence 6 and so on.
This is called Proof by Principal of Mathematical induction
1.8K views · View upvotes
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Siddhant Sharan
Math makes us all Noobs. · 8y
Originally Answered: Show that n (n square plus twenty) is divisible by fourtyeight?
I'd like to do this problem by induction.
See below.
To prove
2
n (n + 20)
is divisible by
48
Now n is even . => n = 2x (x is a natural number)
Which makes our expression
2
8x(x
Now we just have to prove that
+ 5)
2
x(x
+ 5)
is a multiple of 6.
Using induction it can be done quite trivially.
First check the base case. Which is true.
Then assume
2
x(x
+ 5)
is = some
6y
Where y is a natural number.
Now we see that for x+1 in the expression,
3
= (x + 1)
3
= (x
+ 5x + x
+ 5) + 6 + 3x(x + 1)
= 6y + 6 + 3 ∗ (divisible by 2)
Which makes the expression divisible by 6. :)
Hence the original is divisibl… (more)
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Sadik
8y
To Prove :- n(n^2 + 20) = 48k.
Now, n can't be odd.
so take n = 2m
2m (4m^2 + 20) = 8m(m2 + 5).
Now, if we can prove that m(m^2 + 5) is a multiple of 6 then we are done.
for m = 1
1(12 + 5) = 6, divisible by 6.
True for m = 1
Now, Suppose it's true for m = k
so k(k^2 + 5) = 6n
Now, we have to prove it for m = k + 1
(k+1)( (k+1)^2 + 5) = (k + 1) ( k^2 + 2k + 6)
= k(k^2 + 5) + 3k^2 + 3k + 6
Open in App
= 6n + 3( k(k + 1) + 2)
= 6n + 3(2p + 2) as k(k + 1) is always divisible by 2.
= 6(n + p + 1)
hence (k + 1)((k + 1)^2 + 5) is divisible by 6. and so it's true for m = k + 1.
So, we ca… (more)
Ripusudan Agrawal
There is no ignorance, there is only knowledge · 8y
Originally Answered: Show that n (n square plus twenty) is divisible by fourtyeight?
Let us say our even number be n = 2m ,
so by putting this value in
2
8m(m
n(n
2
+ 20)
+ 5)
Now this is to be divisible by 48
or we can say
2
m(m
+ 5)
should be divisible by 6
Now this is our main problem
m can be written as
we get
Sign In
m = 6a + x
(where x can take value from 0 to 5)
Putting this in
m(m
2
(6a + x)(36a
2
+ 5)
2
+ x
we get
+ 12ax + 5)
We will consider different value of x one by one
Case 1: x=0
6a(36a
2
+ 5)
divisible by 6
Case 2: x=1
(6a + 1)(36a
(6a + 1)(36a
6(6a + 1)(6a
2
2
2
+ 1 + 12a + 5)
+ 12a + 6)
+ 2a + 1)
Divisible by 6
Case 3: x=2
(6a + 2)(36a
(6a + 2)(36a
2
2
6(3a + 1)(12a
+ 4 + 24a + 5)
+ 24a + 9)
2
+ 8a + 3)
Divisible by 6
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Jiemba Noah
Janitor at Janitorial Work · Author has 891 answers and 345.2K answer views · 3y
n(n
2
?
+ 20) = 48k
n has to be even, otherwise
n(n
2
+ 20)
is odd.
n = 2p
2
2p(4p
2
p(p
?
+ 20) = 48k
?
+ 5) = 6k
If
p
is even then
Is
p
divisible by
2
p
+ 5 = 9q
Therefore
2
2
p(p
2
p(p
3?
+ 5) mod 2 = 0 .
If yes then
2
p(p
If
p
is odd then
+ 5) mod 3 = 0 .
2
p
+ 5
is even and
Otherwise
2
p(p
p = 3q ± 1
+ 5) mod 2 = 0 .
and
± 6q + 6 mod 3 = 0 .
+ 5)
divides
2
and
3.
Thus
n(n
2
2
+ 20) = 8p(p
+ 5)
divides
2 ∗ 3 ∗ 8 = 48
QED
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Vijay Bhaskar
Passionate about number theory · Author has 58 answers and 151.8K answer views · 7y
N = n*(n^2+20) = 2k(4k^2+20) = 8k(k^2+5)
If k is odd, k^2+5 is even.
Hence k(k^2+5) | 2 ---------- 1
If k = 0 mod 3, k(k^2+5) | 3
If k = 1 mod 3, k^2 = 1 mod 3 => k^2 + 5 = 5+1 = 0 mod 3 => k(k^2+5) | 3
Definitely yes
If k = 2 mod 3 = -1 mod 3, k^2 = 1 mod 3 => k^2 + 5 = 5+1 = 0 mod 3 => k(k^2+5) | 3
Hence k(k^2+5) | 3 for all k ------- 2
From 1 and 2, k(k^2+5) | 6
Hence N | 8*6 or N | 48
Swadesh Kumar Nayak
Works at Reliance Jio (company) · 8y
If n is even, it can be expressed as 2u.
Let P= n(n squared + 20)
Substituting 'n' for '2u', we get the expression for P as 8u(u squared + 5).
Let Q= u(u squared + 5)
If 'u' is odd, (u squared + 5) is even and vice versa. Therefore Q is divisible by 2.
Now 'u' can be of three forms 3v, 3v+1 or 3v+2.
Case 1: u = 3v: Then Q is divisible by 3
Case 2: u = 3v+1: Then Q is divisible by 3 (Please substitute and confirm it for yourself)
Case 3: u=3v+2: Again Q is divisible by 3.
Therefore Q is divisible by 3 and 2.
Since P=8Q, P is divisible by 8*3*2=48. Hence Proved!
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