Prove that for any natural number n, 11^(n + 2) + 12^(2n + 1) is divisible by 133.?

Answer 1

11^(n + 2) + 12^(2n + 1)
= 11^2.11^n + 12.((12^2)^n
= 121.11^n + 12.144^n
= 133.11^n + 12.(144^n - 11^n)
clearly first term is divisible by 133, second term is also divisible by 133, since a^n - b^n is divisible by a - b for every naturan number n. therefore 144^n - 11^n is divisible by 144 - 11 = 133

Answer 2

The rule is if when a is divided by 133 the remainder is R, then (a^n) when divided by 133 remainder will be R^n.

11^(n+2) + 12^(2n+1)
= 11^n × 11^2 + 12^2n × 12^1
Let n= 1 then
11^n × 11^2 + 12^2n × 12^1
Let n be an odd no. Let n = 1
= 11 × 11^2 + 12^2 × 12
= 1131 + 1728
when 1131 when divided by 133 remainder is 1
when 1728 when divided by 133 remainder is 132
Now total remainder = 1 + 132 = 133 which is divisible by 133
Therefore number when n is odd is divisible by 133
Now Let n be even =2
Then th number will be
= 11^2 × 11^2 + 12^4 × 12^1
= 11^3× 11^1 + 12^3 × 12^2
= 1331 × 11 + 1728 × 144
when 1331 divided by 133 remainder is 1 and 11 divided by 133 remainder will be 11
therefore when 11^4 divided by 133 total remainder will be 1 × 11 = 11
When 1728 divided by 133 remainder is 132 and when 144 divided by 133 remainder is 11
Therefore 12^5 when divided by 133 total remainder will be 132 × 11
Now total remainder of 11^4 + 12^4 is additions of above remainders
= 11 + 132 × 11
= 11(1 +132) = 11 × 133 which is divisible by 133
From this we can conclude that the number is divisible by 133 in either case n is odd or even.

Answer 3

This can be verified using math induction.

11^(n+2) + 12^(2n+1)
Let the above statement be S(n).

verify S(1) is divisible by 133 :
11^(1+2) + 12^(2+1) = 11^3 + 12^3 = 3059
3059/133 = 23.
Thus S(1) is divisible by 133.

Let 11^(m+2) + 12^(2m+1) be divisible by 133 for all positive integral values of m. Let this statement be S(m).

11^(m+2) + 12^(2m+1) = 133Q, where Q is the quotient, Q is a natural number.
12^(2m+1) = 133Q - 11^(m+2)

Consider S(m+1) :
11^(m+1+2) + 12^((2(m+1)+1)
= 11*11^(m+2) + 12^2*12^(2m+1)
= 11*11^(m+2) + 144 ( 133Q - 11^(m+2) )
= 11*11^(m+2) + 144*133Q - 144*11^(m+2)
= 144*133Q + 11*11^(m+2) - 144*11^(m+2)
= 144*133Q + 11^(m+2) [ 11 - 144]
= 144*133Q + 11^(m+2) (-133)
= 144*133Q - 11^(m+2)*133
= 133 [ 144Q - 11^(m+2) ]
Q, m being natural numbers.

11^(m+2) + 12^(2m+1) is divisible by 133 for all natural values of m.
Therefore 11^(n+2) + 12^(2n+1) is divisible by 133 for all natural values of n.

hence proved

Answer 4

Is 11^(n+2) + 12^(2n+1) divisible by 133 if n = 1?
yes
11^3 + 12^3 = 3059
3059/133 = 23

Assume that
11^(k+2) + 12^(2k+1) is divisible by 133, is
11^(k+3) + 12^(2(k+1)+1) divisible by 133?

11^(k+3) + 12^(2k+3)
11 * 11^(k+2) + 144 * 12^(2k+1)
11 * 11^(k+2) + (133 +11) * 12^(2k+1)
11 * 11^(k+2) + 11 * 12^(2k+1) + 133 * 12^(2k+1)
11 * (11^(k+2) + 12^(2k+1)) + 133 * 12^(2k+1)
if we divide this by 133 we get a whole number since 11^(k+2) + 12^(2k+1) is divisible by 133 and 133 * 12^(2k+1) is divisible by 133

So since
11^(n+2) + 12^(2n+1) is divisible by 133 if n = 1
and since assuming
11^(n+2) + 12^(2n+1) is divisible by 133
implies that 11^((n+1)+2) + 12^(2(n+1)+1) is divisible by 133
then
11^(n+2) + 12^(2n+1) is divisible by 133 for all natural numbers

Answer 5

= 121 * 11^n + 12 * 12^2n
= 121 * 11^n + 12 * 144^n
= 121 * 11^n + 12 * (133+11)^n

if we expand binomially the last tern all but 11^n divisible by 133

so we are left with
121* ( 11^n) + 12 * 11^n
= 11^n(121+12)
= 11^n * 133

which is divisible by 133

Answer 6

I suspect that this is one that needs to be done by induction.

Let n =1, then we get 11^3 + 12^3 = 1331 + 1728 =3059 = 133*23

Now assuming that the statement is true for n = k, prove that it is true for n = k+1

Answer 7

i ran this through a computer program and came up with the following results. it couldnt get very far because the numbers became way too big...

(1) 3059
(2) 263473
(3) 3.599286E+07
(4) 5.161552E+09
(5) 7.430278E+11
(6) 1.069934E+14
(7) 1.540702E+16
(8) 2.218611E+18
(9) 3.1948E+20
(10) 4.600512E+22
(11) 6.624737E+24
(12) 9.539622E+26
(13) 1.373706E+29
(14) 1.978136E+31
(15) 2.848516E+33
(16) 4.101863E+35
(17) 5.906682E+37

Answer 8

Start by plugging in natural numbers: 1,2,3,4,5,6,7,8,9,...
11^(1+2)+12^(2(1)+1)=3059: 3059/133=23
11^(2+2)+12^(2(2)+1)=263473: 263473/133=1981
continue with others to prove it.