[ankit.3k]

could anyone please give the solution of the problem posted earlier on this thread

find the rem when 17^18 + 13^ 18 is divided by 25??? only answer was given.

also could anyone plz tell me what is diophantine equation as there was a mention of it in one post.


Thanx in advance.

[Stefan R]

Originally Posted by ankit.3k

could anyone please give the solution of the problem posted earlier on this thread

find the rem when 17^18 + 13^ 18 is divided by 25??? only answer was given.

also could anyone plz tell me what is diophantine equation as there was a mention of it in one post.


Thanx in advance.

(15+2)^18 + (15-2)^18 = 2*(15^18 + ....+18C15*15^2*2^16 + 2^18 )

All terms leave remainder 0 except the last. The last term is 2^19

Thus, 2^20 = 1 (mod 25) by Eulers Theorem.

2^19 = 13 (mod 25); Backward tracking...a simple approach,

Diophantine Eq. is not that important.
Remember this;
when ax + by =1 and a,b are co-prime then (x,y) are infinite in integers.

[Stefan R]

Hope Suresh doesn't mind me putting few questions :-)

Let S = 1/1001 + 1/1002 + 1/1003+... + 1/3001, then S lies in

(a) (1,4/3) (b) (4/3,5/3) (c) (5/3,2) (d) (2/3,1)


Let S = (SQroot(100)-SQroot(99)+SQroot(98 )-SQroot(97)+...+SQroot(2)-SQroot(1)); then S lies in

(a) (3,4) (b) (4,5) (c) (5,6) (d) none

[thakarsagar]

Originally Posted by Stefan R

Hope Suresh doesn't mind me putting few questions :-)

Let S = 1/1001 + 1/1002 + 1/1003+... + 1/3001, then S lies in

(a) (1,4/3) (b) (4/3,5/3) (c) (5/3,2) (d) (2/3,1)


Let S = (SQroot(100)-SQroot(99)+SQroot(98 )-SQroot(97)+...+SQroot(2)-SQroot(1)); then S lies in

(a) (3,4) (b) (4,5) (c) (5,6) (d) none

(a) c -> (5/3,2)
solution later....
-sagar

[Stefan R]

Originally Posted by thakarsagar

(a) c -> (5/3,2)
solution later....
-sagar

The answer is wrong. The prob. of getting it right now is 1/3.
Please post the explanation too from next time.

[thakarsagar]

Originally Posted by Stefan R

The answer is wrong. The prob. of getting it right now is 1/3.
Please post the explanation too from next time.

sorry...calculation mistake...
(a) 1 & 4/3

Explanation:
S = 1/1001 + 1/1002 + ... + 1/3001

Let S1 = 1/1001 + 1/1002 +... + 1/2000
Let S2 = 1/2001 + 1/2002 +... + 1/3001

1002>1001
1/1002 < 1/1001
S1 < 1/1001 + 1/1001 + .... + 1/1001
S1 < 1000/1001
S1 < 1 -------------------------->(1) (Ignore the error).

S2 < 1/2001 + 1/2001 + ... + 1/2001
S2 < 1001/2001
S2 < 1/2 -------------------------->(2) (ignore the error)

Using (1) & (2)
S = S1 + S2
< 1 + 1/2
< 3/2 -------------------------->(3)

2000>1999
1/2000<1/1999
S1 > 1/2000 + 1/2000 + ... 1/2000
S1 > 1/2 ------------------------>(4)

S2 > 1/3001 + 1/3001 +... + 1/3001
S2 > 1/3 ------------------------>(5)

Using (4) and (5),
S > 5/6 --------------------------->(6)

Use options (3) and (6) we get the answer as (1,4/3)....
-sagar

[thakarsagar]

Originally Posted by Stefan R

Hope Suresh doesn't mind me putting few questions :-)

Let S = 1/1001 + 1/1002 + 1/1003+... + 1/3001, then S lies in

(a) (1,4/3) (b) (4/3,5/3) (c) (5/3,2) (d) (2/3,1)


Let S = (SQroot(100)-SQroot(99)+SQroot(98 )-SQroot(97)+...+SQroot(2)-SQroot(1)); then S lies in

(a) (3,4) (b) (4,5) (c) (5,6) (d) none

2)

i hope u meant by S = sqrt(100) - sqrt(99) + sqrt(9 - sqrt(97) and so on....
(b) lies between 4 and 5....
-sagar

[thakarsagar]

Originally Posted by thakarsagar

sorry...calculation mistake...
(a) 1 & 4/3

Explanation:
S = 1/1001 + 1/1002 + ... + 1/3001

Let S1 = 1/1001 + 1/1002 +... + 1/2000
Let S2 = 1/2001 + 1/2002 +... + 1/3001

1002>1001
1/1002 < 1/1001
S1 < 1/1001 + 1/1001 + .... + 1/1001
S1 < 1000/1001
S1 < 1 -------------------------->(1) (Ignore the error).

S2 < 1/2001 + 1/2001 + ... + 1/2001
S2 < 1001/2001
S2 < 1/2 -------------------------->(2) (ignore the error)

Using (1) & (2)
S = S1 + S2
< 1 + 1/2
< 3/2 -------------------------->(3)

2000>1999
1/2000<1/1999
S1 > 1/2000 + 1/2000 + ... 1/2000
S1 > 1/2 ------------------------>(4)

S2 > 1/3001 + 1/3001 +... + 1/3001
S2 > 1/3 ------------------------>(5)

Using (4) and (5),
S > 5/6 --------------------------->(6)

Use options (3) and (6) we get the answer as (1,4/3)....
-sagar

guys, any other approach??

[Stefan R]

Originally Posted by thakarsagar

guys, any other approach??

you proved in case of question number 1 that S lies in (5/6,3/2), but how do you deduce from it that the interval is (1,4/3) only. S can be 1.35 or .95 unless proved otherwise.

[ankursanghi]

Originally Posted by Stefan R

you proved in case of question number 1 that S lies in (5/6,3/2), but how do you deduce from it that the interval is (1,4/3) only. S can be 1.35 or .95 unless proved otherwise.

Could any funda of Harmonic Geometric and Arithmetic mean be applied here? I am not able to recall the funda here. Someone please help me recall.. and we could do this question together.