[mohit1984]

Hi One more doubt:
I tried solving Yesterday question by splitting into 2

This is how i went
Clubbed 98 and 102
Now 98^5+102^5 is divisible by 200
and 98^5+102^5 when divided by 2 gives 0 as rem.
so 98^5+102^5 isdivisible by 400

Now clubbed 99 and 101
99^5+101^5 is div by 200
When 99^5+101^5 is divided by 2 again we get rem as 0
so when
99^5+101^5 is divided by 400 we get answer 0 again


Then why answer is 200.I know i m making some stupid mistake can someone help.

[Basilisk]

Originally Posted by mohit1984

Hi One more doubt:
I tried solving Yesterday question by splitting into 2

This is how i went
Clubbed 98 and 102
Now 98^5+102^5 is divisible by 200
and 98^5+102^5 when divided by 2 gives 0 as rem.
so 98^5+102^5 isdivisible by 400

Now clubbed 99 and 101
99^5+101^5 is div by 200
When 99^5+101^5 is divided by 2 again we get rem as 0
so when
99^5+101^5 is divided by 400 we get answer 0 again


Then why answer is 200.I know i m making some stupid mistake can someone help.

Any number divisible by 200 is divisible by 2, so rechecking by 2 serves no purpose.......

[Greenspan]

Originally Posted by CAT 2007

Thus a^n + b^n + c^n + d^n + ....... is divisible by a+ b + c + d...... when n is odd and a, b, c, d are in AP...........

Anyone would like to refute it????????

Not possible when there is a mathematical proof to back it.

if taken as a-3d, a-d, a +d, a+3d, then exp has to be divisible by 4a. On binomial exp, d^(odd no) terms get canceled, and d^(even no)terms will all be divisible by 4a. and thats because 2(3^n+1) is always divisible by 4. Forget the proof.....

Good find C07 Or did I go wrong somewhere.... Basil or vin needed..

[CAT 2007]

Originally Posted by mohit1984

Hi One more doubt:
I tried solving Yesterday question by splitting into 2

This is how i went
Clubbed 98 and 102
Now 98^5+102^5 is divisible by 200
and 98^5+102^5 when divided by 2 gives 0 as rem.
so 98^5+102^5 isdivisible by 400
.

If you are splitting the divisor into 2 numbers, then they must be coprime. Here 200 and 2 are not. Try using 25 and 16. It would give you an answer. Albeit, in this case the remainder will be zero because both 98 and 102 are multiples of 2.

Originally Posted by mohit1984

Now clubbed 99 and 101
99^5+101^5 is div by 200
When 99^5+101^5 is divided by 2 again we get rem as 0
so when
99^5+101^5 is divided by 400 we get answer 0 again


Then why answer is 200.I know i m making some stupid mistake can someone help.

In this case the remainder will not be 0. It will be 200. Try it.

[Basilisk]

Originally Posted by Greenspan

Not possible when there is a mathematical proof to back it.

if taken as a-3d, a-d, a +d, a+3d, then exp has to be divisible by 4a. On binomial exp, d^(odd no) terms get canceled, and d^(even no)terms will all be divisible by 4a. and thats because 2(3^n+1) is always divisible by 4. Forget the proof.....

Good find C07 Or did I go wrong somewhere.... Basil or vin needed..

I think we need an even number of terms in the AP and of course for n to be odd.......when both are satisfied this works i guess.........

example, we add 100^5 to yesterdays question still the remainder would remain 200 (as 100^5 is divisible by 400)

I stand corrected........this works for odd number of A.P. terms too.......
example wrong as after adding 100^5, we should check divisibility by 500 and not 400

[mohit1984]

I guess I have really Lost it..Thanks Basilisk Bhai for pointing out the mistake ...Mayb i shud go home and sleep..Before that Let me correct my mistake

Club 98 102 We know they are divisible by 200 Now As Basilisk Bhai told me we need to find another number such that LCm of that Number and 200 is 400.This way we can find out what will be remainder when this Number is divided by 400.So LCM of 200 and 16 is 400 so next we need to divide by 16.

98^5+102^5 on division by 16 will give 0 as rem

Now for 99 and 101 when we divide by 16 we get remainder as 8 so we need to find a number which gives rem of 8 with 16 and 0 with 200 and such number is 200 itself..So total rem comes out to be 200....Sorry for earlier Post

[CAT 2007]

Originally Posted by Basilisk

I think we need an even number of terms in the AP and of course for n to be odd.......when both are satisfied this works i guess.........

example, we add 100^5 to yesterdays question still the remainder would remain 200 (as 100^5 is divisible by 400)

No.. In that case, we should check by 500 instead of 400.. I think it should be for AP of any number of terms...

[rjt163]

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[cimbaish_Nitin]

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