How to show that 43^23 + 23^43 is divisible by 66?

Question

I've thought about this but haven't found a proof yet

Answer 1

Let n = 43^23 + 23^43.
We have to show n is divisible by 2,3 and 11.
Since 43^23 and 23^43 are both odd, n is even,
so 2| n.
Let's look at n mod 3.
Since 43 = 1(mod 3) and 23 = -1(mod 3).
n = 1^23 + (-1)^43(mod 3) = 1 -1 = 0(mod 3).
Thus 3|n.
Finally, let's look at n mod 11.
Note that 43 = -1(mod 11) and 23 = 1(mod 11),
so n = -1^23 + 1^43 = -1 + 1 = 0(mod 11).
So 11|n and we are done.

Answer 2

Note that if N = 43^23 + 23^43 is divisible by 2,3 and 11, N is divisible by 66 too.


i) N is divisible by 2?

Note that:
(43^n mod 2 ) = 1 , for all n
(Note that 43 is odd, so 43^n is always odd.)

(23^n mod 2 ) = 1 , for all n
(Note that 23 is odd, so 43^n is always odd.)

Examples (43):

43^2 = 1849
43^4 = 3418801
43^6 = 6321363049

43^1 = 43
43^3 = 79507
43^5 = 147008443

Examples (23):

23^2 = 529
23^4 = 279841
23^6 = 148035889

23^1 = 23
23^3 = 12167
23^5 = 6436343


So we can write:

43^23 = 2*f+ 1;
23^43 = 2*g+1

where f and g are integers.

N/2 = 43^23 / 2 + 23^43 / 2

N/2 = (2*f+ 1)/2 + (2*g+1)/2

N/2 = f + 1/2 + g + 1/2

N/2 = (f+g)+1/2 +1/2

N/11 = (f+g)+1

=>N is divisible by 2.

***
ii)N is divisible by 3?

Note that:
(43^n mod 3 ) = 1 , for all n


(23^n mod 2 ) = 1 , when n is even
.........................= 2 , when n is odd

Examples (43):

43^2 = 1849 = 616*3 + 1
43^4 = 3418801 = 1139600*3 + 1
43^6 = 6321363049 = 2107121016*3 + 1

43^1 = 43 = 14*3 + 1
43^3 = 79507 = 26502*3 + 1
43^5 = 147008443 = 49002814*3 + 1


Examples (23):

23^2 = 529 = 48*11 + 1
23^4 = 279841 = 25440*11 + 1
23^6 = 148035889 = 13457808*11 + 1


23^1 = 23 = 7*3 + 2
23^3 = 12167 = 4055*3 + 2
23^5 = 6436343 = 2145447*3 + 2


So we can write:

43^23 = 3*h+ 1;
23^43 = 3*i+2

where h and i are integers.

N/3 = 43^23 / 3 + 23^43 / 3

N/3 = (3*h+ 1)/3 + (3*i+2)/3

N/3 = h + 1/3 + i + 2/3

N/3 = (h+i)+1/3 +2/3

N/3 = (h+i)+1

=>N is divisible by 3.

***
iii)N is divisible by 11?

Note that:
(43^n mod 11) = 10 , when n is odd
...........................= 1 , when n is even

(23^n mod 11 ) = 1 for all n

Examples (43):

43^2 = 1849 = 168*11 + 1
43^4 = 3418801 = 310800*11 + 1
43^6 = 6321363049 = 574669368*11 + 1

43^1 = 43 = 3*11 + 10
43^3 = 79507 = 7227*11 + 10
43^5 = 147008443 = 13364403*11 + 10


Examples (23):

23^2 = 529 = 48*11 + 1
23^4 = 279841 = 25440*11 + 1
23^6 = 148035889 = 13457808*11 + 1

23^1 = 23 = 2*11 + 1
23^3 = 12167 = 1106*11 + 1
23^5 = 6436343 = 585122*11 + 1


So we can write:

43^23 = 11*j+ 10;
23^43 = 11*k+1

where j and k are integers.

N/11 = 43^23 / 11 + 23^43 / 11

N/11 = (11*j+ 10)/11 + (11*k+1)/11

N/11 = j+10/11 + k+1/11

N/11 = (j+k)+10/11 +1/11

N/11 = (j+k)+1

=>N is divisible by 11.

******

From i, ii and iii. N is divisible by 2 ,3 and 11.
=> N is divisible by 66 too.

Kisses

=**

Answer 3

Although, there is nothing wrong with the proof above, I think that the point of this exercise is to use number theory and NOT to actually calculate the numbers. I don't think that I can calculate 43^23 by hand but the idea is correct in above answer.

We need to show that this number is divisible by 2, 3, and 11.

Case one, in order to show that the number is even, 43 is odd. 23 is odd. Therefore 43^23 is odd and 23^43 is odd. An odd number plus an odd number is even and divisible by 2.

You can do the rest like that too if you just remember their divisibility laws.

Answer 4

You need to prove it is divisible by factors of 66 ( 11x2x3 = 66) So the number should be divisible with 2,3 and 11.

43^23 + 43^23 = 43^23 x2

1-You can see it is divisible by 2.
2- 43^2 = 1849
43^3 = 79507
43^4=3418801 is divisible by 3 after this point you know the number is divisible by 3 also

3- 43^5 = 147008443 is divisible by 11. After this point you know the whole number is divisible by 11.

Now you know the number you have is divisible by 2,3 and 11 and therefore divisible by 66.