samrus98 wrote:

yezz wrote:

Thanks samrus for taking the time to re-explain to me, that said and as much as my math skills are very basic ( i dont know wut is Mod

as i dont have a very sofisticated quantitative back ground as óbviously, your , i just try to plug in numbers in the equation given and devide by 5 and it gives me several values ( there is a celecity). how come your way gives one answer and by plugging in numbers in the equation and deviding does not.

Hey yezz, no prob.....its just that I aint tht good at explaining.....

MOD function gives the remainder when a number is divided by another number.

Eg. 11 MOD 7 = 4

Any remainder obtained can also be expressed as a negative number.

Eg. 11 MOD 7 = -3

What this means this 11 is 3 short of 7's next multiple which is 14 in this case.

There are some properties associated with MOD:

Property 1: (a*b*c.....) MOD x = (a MOD x) * (b MOD x) * (c MOD x)....

Eg. 11^5 MOD 7 = (11 MOD 7) * (11 MOD 7) * (11 MOD 7) * (11 MOD 7) * (11 MOD 7)

= (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7)

Property 2: (a MOD x) * (b MOD x) * (c MOD x).... = (a*b*c.....) MOD x

Eg. (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7) = (4^5) MOD 7

NOTE: We started out with (11^5 MOD 7) and have reduced it to (4^5 MOD 7). We could have done this directly, since 11 MOD 7 = 4.

Now (4^5 MOD 7) = (2^10 MOD 7)

Doing some rearranging we get,

= (2* 2^9) MOD 7

= (2 MOD 7) * (8^3 MOD 7)

= 2 * (1^3 MOD 7) .......... (Since 8 MOD 7 = 1)

= 2 * 1

= 2

Basically our goal is to reduce the dividend to such a point where it yields a remainder of either 1 or -1 when divided by the divisor. Initially this might seem daunting and slow, but it becomes really fast by practice!

Property 3: (a

+ b

+ c.....) MOD x = (a MOD x)

+ (b MOD x)

+ (c MOD x)....

Property 4: (a + b*c - d^p) MOD x = (a MOD x) + (b MOD x) * (c MOD x) - (d^p MOD x)

Hope I have been of help!

samrus98, simply superb.

even though orginal prob does not require this method to solve it. The above explained method was an eye-opener.

But, I would like to correct some info in your post, as this post will be reference for learning this property to many ppl.

**Quote:**

NOTE: We started out with (11^5 MOD 7) and have reduced it to (4^5 MOD 7). We could have done this directly, since 11 MOD 7 = 4.

Now (4^5 MOD 7) = (2^10 MOD 7)

Doing some rearranging we get,4

= (2* 2^9) MOD 7

= (2 MOD 7) * (8^3 MOD 7)

= 2 * (1^3 MOD 7) .......... (Since 8 MOD 7 = 1)[/strike]

= 2 * 1

= 2

I have question in the usage of exponents here,

2^9 can be written as 2^(3*3)

(2^3)*(2^3) = 64

or it can be written as,

2^(3^2) = 8^2 = 64

but, you have written as,

2^9 = 2^(3*3) = 8^3 = 512

Whether, writing, 2^9 = 8^3 is correct?

But, it does not have any problem since the base is 1 here in the subsequent steps.

Please, Let us know the steps/properties to consider while simplifying the exponents.

Please correct me if I am wrong.

regards,

hhk

2^9=8^3 is correct.

2^3*2^3 does not equal 2^9. 2^3*2^3=2^(3+3)=2^6 or 2^3*2^3=(2X2)^3=4^3

Please, refer to this link. It has exponent basics and it might be helpful

I am sorry I did not notice that there was a second page and samrus98 has already replied. Anyway, the link might be helpful.