Hi Puys .

i came across this question , Please have a look at it .

In how many ways can the number 105 be written as sum of two or more positive consecutive integers ?

I couldnt really find a concrete method of finding a solution to it ( apart from brute forcing )

Here is what i read as solution which was given in the end .

The number of ways 105 can be written as a sum of two or more consecutive positive integers = number of odd divisors of 105 -1 = 8 - 1=7

What i wanna know is , whether this funda holds goood for all cases ... and moreover how did we reach to this funda ??

Could you please explain how the answer is 7 and what is this brute forcing?

I think 105 could never be written as sum of even no of consecutive positive integers. For eg:- it could not be written as sum of 4 consecutive nos because in that case we will have 2 even and 2 odd nos which gives final sum as even whereas 105 is odd. But, 2 is the exception here:- 105= 52+53.

It could be written as sum of odd no of consecutive positive integer(3,5 and 7).

105= 34+35+36.
105= 19+20+21+22+23.
105= 12+13+14+15+16+17+18.

I cant think of any other way in which 105 could be written as sum of consecutive positive integer.

Puys,

Please solve this:-

N is a natural number between 10 and 1000. P denotes the product of its digits. S denotes the sum of its digits. if 6P + 4S = 4N, how many values can N take?

1)9 2)10 3)11 4)12

D=(10a+b)(10c+d)-(10a+b)(10d+c)
=(10a+b)*9*(c-d)
so D has to b a multiple of 9 and should be written as product of two digit number ab and either9 or a two digit multiple of 9.

options 1 , 2 can be ruled out for they are not multiples of 9
option 4 has a factor 211 wich is a prime number...

so correct answer is option 3)3456

The Value of D could never be 9495. As stated by vaibhav :-

D=(10a+b)(10c+d)-(10a+b)(10d+c)
=(10a+b)*9*(c-d)

Now, lets try to maximize value of D.
Max value (c-d) could take is 9 for c=9 and d=0. Now for maximizing D take a=b=9. Hence,

D= 99*9*9= 8019.

Thats leave 3456 as the only option.

One more:
Akhil was multiplying two 2-digit numbers ab and cd. He accidently interchanged the digits of one of the numbers and the product he obtained differed from the correct product by a value D. Which of the following is a possible value of D?

1. 1234 2. 2345 3. 3456 4. 9495

Puys please post ur approach too.

D=(10a+b)(10c+d)-(10a+b)(10d+c)
=(10a+b)*9*(c-d)
so D has to b a multiple of 9 and should be written as product of two digit number ab and either9 or a two digit multiple of 9.

options 1 , 2 can be ruled out for they are not multiples of 9
option 4 has a factor 211 wich is a prime number...

so correct answer is option 3)3456

Akhil was multiplying two 2-digit numbers ab and cd. He accidently interchanged the digits of one of the numbers and the product

then I think the ans is c i.e. 3456

asghan Says

I think if the diff is d it has to be a single digit no ???, or i have read the question wrong ???

Hey did a mistake in typing the problem.... difference d is not same as the digit in the number cd. Say difference as D and solve...

Thanks.

One more:
Akhil was multiplying two 2-digit numbers ab and cd. He accidently interchanged the digits of one of the numbers and the product he obtained differed from the correct product by a value d. Which of the following is a possible value of d?

1. 1234 2. 2345 3. 3456 4. 9495

Puys please post ur approach too.

I think if the diff is d it has to be a single digit no ???, or i have read the question wrong ???

One more:
Akhil was multiplying two 2-digit numbers ab and cd. He accidently interchanged the digits of one of the numbers and the product he obtained differed from the correct product by a value D. Which of the following is a possible value of D?

1. 1234 2. 2345 3. 3456 4. 9495

Puys please post ur approach too.

30^40 /17
= 900^20/17
=(901-1)^20/17
=(53*17 - 1)^20/17

hence applying binomial theorem........ we get the remainder as 1 :)

this kind of qs can be easily solved by eulers theorm jst visit off. quant thread

30^40 /17
30^(16k+/17
30^8/17
(-4)^8/17
65536/17
gives remainder 1

courtesy to vaibhav2cool

Hi Puys .

i came across this question , Please have a look at it .

In how many ways can the number 105 be written as sum of two or more positive consecutive integers ?

I couldnt really find a concrete method of finding a solution to it ( apart from brute forcing )

Here is what i read as solution which was given in the end .

The number of ways 105 can be written as a sum of two or more consecutive positive integers = number of odd divisors of 105 -1 = 8 - 1=7

What i wanna know is , whether this funda holds goood for all cases ... and moreover how did we reach to this funda ??

Hey here is the explanation:
Once u have an odd divisor say 1 or 3 or 5 say p.. and say the quotient is x then 105 can be written as,
105 = x + x + x+ ... p(odd) nuber of times...

Now consider , 105 = x-2 + x-1 + x + x+1 + x+2 (assume p =5) so we have gotten 5 consecutive integers. So every time u have an odd number as divisor u can find the consecutive integers by by adding and subtracting 1, 2 3 etc... on either side starting from the middle term.

For example:

105 = 21 + 21 + 21 +21 + 21
so 105 = 21-2 + 21-1 + 21 + 21+1 + 21+2 = 19+20+21+22+23.

I hope I am clear.

But 1 is being subtracted .. that is for the number itself as a divisor.

i edited the question .. had made a typing mistake.

Hi Puys .

i came across this question , Please have a look at it .

In how many ways can the number 105 be written as sum of two or more positive integers ?

I couldnt really find a concrete method of finding a solution to it ( apart from brute forcing )

Here is what i read as solution which was given in the end .

The number of ways 105 can be written as a sum of two or more consecutive positive integers = number of odd divisors of 105 -1 = 8 - 1=7

What i wanna know is , whether this funda holds goood for all cases ... and moreover how did we reach to this funda ??

Was the question in how many wasy 105 can be written as sum of two or more positive integers,

or

In how many ways 105 can be written as a sum of 2 consecutive integers ??

Hi Puys .

i came across this question , Please have a look at it .

In how many ways can the number 105 be written as sum of two or more positive consecutive integers ?

I couldnt really find a concrete method of finding a solution to it ( apart from brute forcing )

Here is what i read as solution which was given in the end .

The number of ways 105 can be written as a sum of two or more consecutive positive integers = number of odd divisors of 105 -1 = 8 - 1=7

What i wanna know is , whether this funda holds goood for all cases ... and moreover how did we reach to this funda ??

Your answer is correct but this reasoning needs some correction. There are 25 prime numbers from 1 to 100. Remove 2, 3 and 5 from them. You get 22 such numbers left which are not divisible by either 2 or 3 or 5. 1 is also not divisible by either 2 or 3 or 5. That makes 23 such numbers which are not divisible by either 2 or 3 or 5.

The other three numbers are:
49 = 7*7 (You got this one)
77 = 7*11
91 = 7*13

Making it a total of 26 such numbers not divisible by either 2 or 3 or 5.

Dude my reasoning is perfect .. cos I have used v'diagram to get it and not ne thing else. Yeah i have done a mistake while veryfying my solution, I counted the primes 3 and 5 which is wrong. Any ways thanks for pointing out the other two numbers 77 and 91.

Had I not given the answer from v'diagram as 26, it will be very difficult to count the numbers cos u r not sure there are how many of them.. right? Once we arrive at 26 then search started for the other 3 non prome numbers.

...
Hey I guess my answer is correct .. cos we have 25 prome numbers from 1 to 100. So exclude 2 and include 1 so 25 are there for sure ... can ne one tell me who is this 26th in my answer and that number is not prime???

Hey got the answer... that number is 49. :D

Your answer is correct but this reasoning needs some correction. There are 25 prime numbers from 1 to 100. Remove 2, 3 and 5 from them. You get 22 such numbers left which are not divisible by either 2 or 3 or 5. 1 is also not divisible by either 2 or 3 or 5. That makes 23 such numbers which are not divisible by either 2 or 3 or 5.

The other three numbers are:
49 = 7*7 (You got this one)
77 = 7*11
91 = 7*13

Making it a total of 26 such numbers not divisible by either 2 or 3 or 5.

1)The number of positive integers not greater than 100 which are not divisible by 2,3 or 5 is
2)26 b)18 c)31 d)none

2)the product of all integers from 1 to 100 will have the follwoing number of zeroes at the end
a)20 b)24 c)19 d)22

how do u approach this?explain in detail pls.

For second question we can say that all 5^1 will give on zero hence 20 zeros,
now for every additional power of 5 there will be on extra zero so for 5^2 there will be 4 more zero
hence total zeros will be 24 i.e. option b

1)The number of positive integers not greater than 100 which are not divisible by 2,3 or 5 is
2)26 b)18 c)31 d)none

2)the product of all integers from 1 to 100 will have the follwoing number of zeroes at the end
a)20 b)24 c)19 d)22

how do u approach this?explain in detail pls.

Well first let me get rid of 2nd one. This is same as how many 0's are there in 100!. And by now I remember the answer. It is 24.

2nd one is really worth solving. The answer is 26.
I will solve it by v'diagram. It is easy. But typing is difficult :D.


n(2&3&5) = 3. divisible by 2 and 3 and 5.
n(2&3) = 16. n(3&5)=6. and n(2&5)=10.
Also n(2)= 50, n(3)=33 and n(5)=20.

Now there are overlappings.
so Only(2&3) = 16-3 = 13, Only(2&5) = 10-3=7 and Only(5&3) = 6-3=3.

Similary u calculate Only(2)=50-13-3-7=27
Only(3)=14 and Only(5)=7.
So the answer is 100-(27+7+14 + 7+3+13 + 3) = 26.

Hey don feel it is difficult. On paper u can do all these calculation within a minute.

Hey I guess my answer is correct .. cos we have 25 prome numbers from 1 to 100. So exclude 2 and include 1 so 25 are there for sure ... can ne one tell me who is this 26th in my answer and that number is not prime???

Hey got the answer... that number is 49.

Q. X(1)=1 & X(n+1)=2X(n)+5,where n=1,2....Then X(100)=?
a. 5*2rtp99-6
b. 5*2rtp99+6
c. 6*2rtp99+5
d. 6*2rtp99-5


ans. d

my doubt is how to approach such probs. When ans is in a given format.

U should try to catch the pattern and check it for some simple cases like X(1)=1=6*2^0-5 , X(2)=7=6*2^1-5 ,X(3)=19=6*2^2-5

hi!
for such question I would try to change the value from x(100) to something like x(2) or x(5) etc etc.......and then substitute the corrsponding value of 2^99 to 2^( 2-1) or 2^(5-1) and then calculate the values.....

I really dont see a method for such questions.....

Q. X(1)=1 & X(n+1)=2X(n)+5,where n=1,2....Then X(100)=?
a. 5*2rtp99-6
b. 5*2rtp99+6
c. 6*2rtp99+5
d. 6*2rtp99-5


ans. d

my doubt is how to approach such probs. When ans is in a given format.

1)The number of positive integers not greater than 100 which are not divisible by 2,3 or 5 is
2)26 b)18 c)31 d)none

2)the product of all integers from 1 to 100 will have the follwoing number of zeroes at the end
a)20 b)24 c)19 d)22

how do u approach this?explain in detail pls.

Q. X(1)=1 & X(n+1)=2X(n)+5,where n=1,2....Then X(100)=?
a. 5*2rtp99-6
b. 5*2rtp99+6
c. 6*2rtp99+5
d. 6*2rtp99-5


ans. d

my doubt is how to approach such probs. When ans is in a given format.

K.SHARMA Says

What is the least value of x such that 60! / x will be a odd number?

denotes greatest integer less than or equal to x.
+ + + + = 30 + 15 + 7 + 3 + 1 = 56

The number will be 2^56.

K.SHARMA Says

What is the least value of x such that 60! / x will be a odd number?

Well here your enemy is all the 2's appearing in 60!. Each 2 can make any number an even number. Remove them to get an odd number. And in 60! there are 56 2's so the answer is 2^56.

K.SHARMA Says

What is the least value of x such that 60! / x will be a odd number?

find the max power of 2 in 60!, in this case 56, hence value of x is 2^56

puys pls solve this for me
What is the remainder when ^13333 is divided by 13?
answer is 1..
i did this by taking 6!=720 and dividing it by 13 to get 5 as remainder..
so eqn becomes ^13333
im not getting 1 on solving this..

We can solve it by using fermet's theorem also:-

^13333 = ^

13332 is multiple of 12. Hence remainder would be .

7! is again a multiple of 12. Hence final remainder = 1.

@ravi ..
its not ^13333 .. it is ^13333 .. so ans comes out to be 1

sorry it was my fault..took it as ^13333 instead of ^13333

here we get ^13333
which leads to 7^13333 mod 13..
how do u do this?i mean any shortcut..
if i expand and find remainder each time itl consume time..
pls solve the step 7^13333 mod 13 for me

@ravi ..
its not ^13333 .. it is ^13333 .. so ans comes out to be 1

here we get ^13333
which leads to 7^13333 mod 13..
how do u do this?i mean any shortcut..
if i expand and find remainder each time itl consume time..
pls solve the step 7^13333 mod 13 for me

6!^7!^13333
720^7!^13333%13
5^7!^13333%13
From here 26-1 does not strike me.. so I use euler to solve ir
E(13) =12
so 5^12%13 = 1
7! is a multiple of 12,
so 5^7!^(whatever ) = 1
hence the answer.. hope it was clear..

here we get ^13333
which leads to 7^13333 mod 13..
how do u do this?i mean any shortcut..
if i expand and find remainder each time itl consume time..
pls solve the step 7^13333 mod 13 for me

write 5^2=25=26-1
so remainder becomes ^13333=1

here its (-1) raised to power of (360*7) which will give us (1) raised to power of 13333 finally...

i guess this is wat he has done...

write 5^2=25=26-1
so remainder becomes ^13333=1

here we get ^13333
which leads to 7^13333 mod 13..
how do u do this?i mean any shortcut..
if i expand and find remainder each time itl consume time..
pls solve the step 7^13333 mod 13 for me

puys pls solve this for me
What is the remainder when ^13333 is divided by 13?
answer is 1..
i did this by taking 6!=720 and dividing it by 13 to get 5 as remainder..
so eqn becomes ^13333
im not getting 1 on solving this..

write 5^2=25=26-1
so remainder becomes ^13333=1

puys pls solve this for me
What is the remainder when ^13333 is divided by 13?
answer is 1..
i did this by taking 6!=720 and dividing it by 13 to get 5 as remainder..
so eqn becomes ^13333
im not getting 1 on solving this..

Factorizing the given eqn yields..
(a+b)(b-2)=-31

so possible values of (a,b) are (-34,3),(-34,33),(30,-29),(30,1)

So the answer is four..

Correct if i am wrong.....

Yeah that is correct ..

And now I am an expert Pagal

jha16june Says

Find the number of integral pairs (a,b), which atisfy the equation 2(a+b)=ab+b^2+31.

Factorizing the given eqn yields..
(a+b)(b-2)=-31

so possible values of (a,b) are (-34,3),(-34,33),(30,-29),(30,1)

So the answer is four..

Correct if i am wrong.....

Find the number of integral pairs (a,b), which atisfy the equation 2(a+b)=ab+b^2+31.

siva_nagarajan Says

hi..i guess the answer is 1 coz AB*AB=XYA where your A of AB should match with A of XYA...consider squares 10*10=100(which is wrong and so is 19*19=361).now 11*11=121..where the condition has satisfied.....

Siva,

In above problem AB could be both 11 and 19. But i think the solution lies in problem itself. It says:-

AB^2 = XYA

So, X and B should be different than A. Which is not the case with 11^2 = 121.

I totally agree with Deepak. That method is perfect.

If (AB) x (AB) = XYA, then X stands for :

(1) 3 (2) 1 (3) 6 (4) 9



Hey guys,

How to go abt this?

lemme try this..
a 2-digit no. havin a 3-digit no as its square..so the 2digit no is<32.. in other words A is 1,2 or 3.
but as B^2 has its last digit as A ..therefore A cant be 2 or 3.Hence A=1..and this leads us to B=9..so 19^2=361...
And X=3..

*phew*

hi..i guess the answer is 1 coz AB*AB=XYA where your A of AB should match with A of XYA...consider squares 10*10=100(which is wrong and so is 19*19=361).now 11*11=121..where the condition has satisfied.....

Nice explanation: But there are some problems:

First of all answer is 35 and not 34.
24 primes... 2 is excluded
4*(2, 3, 5, 7, 11, 13, 17, 19, 23) i.e. 9 of them.

Then we have the great 1 = 1-0.
And 4: as 4= 4*1(even*odd so ignore)= 2*2 (even*even) =4-0.
So total = 24+9+1+1 =35.
The why r u assuming a<= b this will give y<=0 in ur explanation.

Finally the idea is to get a unique pair of factor of ne number where the components are both odd or both even. I mean to say 4=2*2 and that is unique. So u will not get any other 4 =a*b where both a nd b are either even or both odd.

You are right. I made a simple mistake. Instead of prime numbers I shouls have said numbers with less than 3 factors which would have included 1 in first case and 4 in second case all by itself. I just missed multiplying 1 by 4 to get 4 also as a possible solution.

Thanks for the insight..

5/2
i dont know how to lengthen this answer...

Hi,

I couldn't completely understand your solutions..
Can you explain with steps for all of them...

Regards,

3). (32^(32^32)) /9 find the remainder..
a.4 b.7 c. 1 d.2
I tried it using remainder theorem it comes out to be 5^(32^32).... but I am unable to get a general pattern of remainders after this step...

4). (32^(32^32))/7 find the reaminder..
a. 4 b.7 c.1 d.2

3> 32^(32^32)mod9=5^(6k+4)(mod9)=>4(mod9) => option (a)

4>32^(32^32)mod7=4^(6k+4)(mod7)=>4(mod7) => option (a)

a. 4 b.7 c.1 d.2

5).Find the remainder- (10^10+10^100+10^1000+.......+10^10000000000 ) /7

a.0 b. 1 c. 2 d.5

same as 10*3^4 (mod7)=5mod7=>option (d)

Hi,

Ya, thanx a lot... hope i'd be able to solve it now..The answer to both 3rd and 4th ques. is 4 .

And what about quest. 5?

Don't really know much about ques 5.....

Anyways, for type of questions of no. 3 and 4, I only remember how to solve when the denominator is prime(which is the case in ques 4)...... I can explain the method and have also got the answer now Committed a small mistake earlier....

32^(32^32) mod 7...... using fermit's theorem.... we first have to calculate
(32^32) mod 6 (This is the top two 32s and we mod it with 7-1)

32 = 2 mod 6
32^2 = 4 mod 6 = (-2) mod 6
32^8 = (-2)^4 = 16 mod 6 = (-2) mod 6
32^32 = (-2) mod 6 = 4 mod 6

So effectively, our expression has been reduced to
32^4 mod 7

Using similar method,
32 = 4 mod 7
32^2 = 2 mod 7
32^4 = 4 mod 7

Now, I'll try to solve the 3rd one as well but I guess it probably needs a different method to deal with... Will get back to you if I get through...

Unfortunately, I've been out of practice since all my exams ended and am not hard-willed enough to go through them again before my college starts.....

Hi,

Ya, thanx a lot... hope i'd be able to solve it now..The answer to both 3rd and 4th ques. is 4 .

And what about quest. 5?

Hi, I have a few doubts regarding the following questions(plz explain the steps as well):

1). Find the last 2 digits of the following - 15*37*63*51*97*17
a.35 b.45 c.55 d.85
What should be the general approach for such type of questions....
2). Find the last 3 digits of the following- 12345*54321
a. 835 b.745 c.845 d.945

3). (32^(32^32)) /9 find the remainder..
a.4 b.7 c. 1 d.2
I tried it using remainder theorem it comes out to be 5^(32^32).... but I am unable to get a general pattern of remainders after this step...

4). (32^(32^32))/7 find the reaminder..
a. 4 b.7 c.1 d.2

5).Find the remainder- (10^10+10^100+10^1000+.......+10^10000000000 ) /7

a.0 b. 1 c. 2 d.5


And can someone plz explain what is this Fermats theorem and euler's number or can someone send the thread where it explained..? I 've seen its usage quite a lot of times in the posts...

I'm pretty sure about the first two answers...... 1st: 35 and 2nd: 745

The approach for finding the last 2/3 digit numbers is to divide the expression by 100/1000(depending on last 2/3 digits which you have to find).

Thus for the 1st ques,

(15*37*63*51*97*17) mod 100

Dividing by 5,
(3*37*63*51*97*17) mod 20

Now, remainder for each number,
3*(-3)*3*(-9)*(-3)*(-3) mod 20

3^7 mod 20 = (3^4)*(3^3) mod 20
(81*27) mod 20 = 7 mod 20

Multiplying by 5 again to make it mod 100
35 mod 100

Similarly, the 2nd question needs to be solved by taking mod 1000. I think you'll be able to do it now

I think answers to 3rd and 4th are 1 and 2 resp. Are they correct?? If they are.... I can explain the method

Hi, I have a few doubts regarding the following questions(plz explain the steps as well):

1). Find the last 2 digits of the following - 15*37*63*51*97*17
a.35 b.45 c.55 d.85
What should be the general approach for such type of questions....
2). Find the last 3 digits of the following- 12345*54321
a. 835 b.745 c.845 d.945

3). (32^(32^32)) /9 find the remainder..
a.4 b.7 c. 1 d.2
I tried it using remainder theorem it comes out to be 5^(32^32).... but I am unable to get a general pattern of remainders after this step...

4). (32^(32^32))/7 find the reaminder..
a. 4 b.7 c.1 d.2

5).Find the remainder- (10^10+10^100+10^1000+.......+10^10000000000 ) /7

a.0 b. 1 c. 2 d.5


And can someone plz explain what is this Fermats theorem and euler's number or can someone send the thread where it explained..? I 've seen its usage quite a lot of times in the posts...

My $0.02 on the prob

Any no 'ab' can be expressed as a diff of 2 perfect squares as
ab = {(a+b)/2}^2 - {(a-b)/2}^2
eg. 24 = 6*4
= 12*2(Note: The no has to written as product of only odd/even pairs)

Using the above we can make following inferences:-
1. All prime nos can be expressed as diff of 2 perfect squares in 1 way.
2. All non multiples of 4 in 0 ways i.e diff of perfect squares cant be 6,10,14.... etc.
3. All prime multiples of 4 in 1 way eg. 8,12,20,28
4. All composite multiples of 4 in more than 1 way.

Coming back to the problem.....there are 24 prime nos less than 100 and 9 prime multiples of 4.Also 1 has to be taken into account.
So answer is 24+9+1=34.
Hope this helps:)

Can you account for the number 4: as 4 = 4-0? So the answer is 35.

My $0.02 on the prob

Any no 'ab' can be expressed as a diff of 2 perfect squares as
ab = {(a+b)/2}^2 - {(a-b)/2}^2
eg. 24 = 6*4
= 12*2(Note: The no has to written as product of only odd/even pairs)

Using the above we can make following inferences:-
1. All prime nos can be expressed as diff of 2 perfect squares in 1 way.
2. All non multiples of 4 in 0 ways i.e diff of perfect squares cant be 6,10,14.... etc.
3. All prime multiples of 4 in 1 way eg. 8,12,20,28
4. All composite multiples of 4 in more than 1 way.

Coming back to the problem.....there are 24 prime nos less than 100 and 9 prime multiples of 4.Also 1 has to be taken into account.
So answer is 24+9+1=35.
Hope this helps:)

It does help but 24 + 9 + 1 =34 not 35 which is the ans

can someone pls explain this?

remainder of 25^102/17?
answer is 4..im not getting 4..
i did it as (17+8 )^102/17
=8^102/17

i seperately calcualted remainders upto 8^4..not getting it still..
pls explain with steps..

We can write 25^102 as (5*5)^102.
Now 5 and 17 are co-primes.So we can apply Fermat's Theorem here.The euler no for 17 is 16 i.e (5^16)/17 gives remainder 1
102=16k+6.
so we just have to calculate remainder (5^6)/17 * (5^6)/17 i.e 2*2=4

can someone pls explain this?

remainder of 25^102/17?
answer is 4..im not getting 4..
i did it as (17+8 )^102/17
=8^102/17

i seperately calcualted remainders upto 8^4..not getting it still..
pls explain with steps..

By euler's theorem 25^16/17 =1 hence (25^16)^6/17 = 1,

Now we are left with 25^6/17 rem 25/17 is 8 hence 8^6/17 = (8^3)^2/17 = 2^2/17 = 4.

can someone pls explain this?

remainder of 25^102/17?
answer is 4..im not getting 4..
i did it as (17+8 )^102/17
=8^102/17

i seperately calcualted remainders upto 8^4..not getting it still..
pls explain with steps..

See,

25 = 8 mod 17
25^2 = 64 mod 17 = (-4) mod 17
25^4 = 16 mod 17 = (-1) mod 17
25^100 = (25^4)^25 = (-1)^25 mod 17 = (-1) mod 17
25^102 = (-1) * (-4) mod 17 = 4 mod 17

How many numbers below 100 can be expressed as a difference of two perfect squares in only one way?
1. 25 2. 26 3.34 4.35

Please do post explanation.

My $0.02 on the prob

Any no 'ab' can be expressed as a diff of 2 perfect squares as
ab = {(a+b)/2}^2 - {(a-b)/2}^2
eg. 24 = 6*4
= 12*2(Note: The no has to written as product of only odd/even pairs)

Using the above we can make following inferences:-
1. All prime nos can be expressed as diff of 2 perfect squares in 1 way.
2. All non multiples of 4 in 0 ways i.e diff of perfect squares cant be 6,10,14.... etc.
3. All prime multiples of 4 in 1 way eg. 8,12,20,28
4. All composite multiples of 4 in more than 1 way.

Coming back to the problem.....there are 24 prime nos less than 100 and 9 prime multiples of 4.Also 1 has to be taken into account.
So answer is 24+9+1=34.
Hope this helps:)

can someone pls explain this?

remainder of 25^102/17?
answer is 4..im not getting 4..
i did it as (17+8 )^102/17
=8^102/17

i seperately calcualted remainders upto 8^4..not getting it still..
pls explain with steps..

Say 1<=k<=100 and (x^2) - (y^2) = k = ab (where a<=b are two factors of k)
Hence,
(x+y)(x-y)=k=ab
We can safely assume..
x + y = a
x - y = b
We get..
x = (a+b)/2
y = (a-b)/2

From above equations for x and y.. we can see that either a and b are both odd or both even. (this is due to the fact that x^2 and y^2 are perfect squares)

We know that for all k, its a possibility that b=1 and a=k. In this case, if a is odd we will have a solution for x and y. If a is even this will yield no solutions.

For odd k, b=1 and a=k will have solutions. Now we have to look for odd numbers k which have no other set of odd factors a and b except for 1 and k. This is same as looking for odd prime numbers. They will have no other combinations of a and b and hence can be represented as a difference of two perfect squares in only one way. Hence we get 24 solutions.
Similarly, for k=1 we will have a=b=1 and this will also yield a solution and as 1 has no other factors this will also be a solution. Hence we get 1 solution.

For even k, b=1 and a=k will give no solutions. Hence all even numbers, which will have a set of factors a and b where both a and b are even will have a solution for x and y. Also the condition is that these even k should have only one set of such factors a and b. For all prime numbers p converted to even by multiplying by 4, will satisfy this criteria as for all such k=4p, possible sets of (b,a) will be (1,4p), (2,2p) and (4,p). Of this only, b=2 and a=2p will give a solution for all odd primes. This will give us 8 solutions (for p = 3,5,7,11,13,17,19,23).
For prime 2, the pairs and (2.2p) and (4,p) will both give solutions but eventually both would be same set of factors i.e. (2,4). This will give us 1 solution.

Hence total number of solutions = 24+1+8+1 = 34

Nice explanation: But there are some problems:

First of all answer is 35 and not 34.
24 primes... 2 is excluded
4*(2, 3, 5, 7, 11, 13, 17, 19, 23) i.e. 9 of them.

Then we have the great 1 = 1-0.
And 4: as 4= 4*1(even*odd so ignore)= 2*2 (even*even) =4-0.
So total = 24+9+1+1 =35.
The why r u assuming a<= b this will give y<=0 in ur explanation.

Finally the idea is to get a unique pair of factor of ne number where the components are both odd or both even. I mean to say 4=2*2 and that is unique. So u will not get any other 4 =a*b where both a nd b are either even or both odd.

How many numbers below 100 can be expressed as a difference of two perfect squares in only one way?
1. 25 2. 26 3.34 4.35

Please do post explanation.

Say 1<=k<=100 and (x^2) - (y^2) = k = ab (where a<=b are two factors of k)
Hence,
(x+y)(x-y)=k=ab
We can safely assume..
x + y = a
x - y = b
We get..
x = (a+b)/2
y = (a-b)/2

From above equations for x and y.. we can see that either a and b are both odd or both even. (this is due to the fact that x^2 and y^2 are perfect squares)

We know that for all k, its a possibility that b=1 and a=k. In this case, if a is odd we will have a solution for x and y. If a is even this will yield no solutions.

For odd k, b=1 and a=k will have solutions. Now we have to look for odd numbers k which have no other set of odd factors a and b except for 1 and k. This is same as looking for odd prime numbers. They will have no other combinations of a and b and hence can be represented as a difference of two perfect squares in only one way. Hence we get 24 solutions.
Similarly, for k=1 we will have a=b=1 and this will also yield a solution and as 1 has no other factors this will also be a solution. Hence we get 1 solution.

For even k, b=1 and a=k will give no solutions. Hence all even numbers, which will have a set of factors a and b where both a and b are even will have a solution for x and y. Also the condition is that these even k should have only one set of such factors a and b. For all prime numbers p converted to even by multiplying by 4, will satisfy this criteria as for all such k=4p, possible sets of (b,a) will be (1,4p), (2,2p) and (4,p). Of this only, b=2 and a=2p will give a solution for all odd primes. This will give us 8 solutions (for p = 3,5,7,11,13,17,19,23).
For prime 2, the pairs and (2.2p) and (4,p) will both give solutions but eventually both would be same set of factors i.e. (2,4). This will give us 1 solution.

Hence total number of solutions = 24+1+8+1 = 34

asghan Says

except 2 all the prime no can be expressed as the diff of 2 perfect square in one way, can't think of a reason for 8, please explain

all the primes except 2 => 24
1 can be expressed as 1-0 =>1
for ne prome p, 4p can be expressed as diff of two squares in only one way
and I am still searching for the reason.
so 8, 12, ... 92, and also 99. so total is 35.

Can u express 2 a prime number .. and then we have 1= 1-0.
then we have 8 = 9-1.
So I have given u enough hint...
Now try to solve it..

except 2 all the prime no can be expressed as the diff of 2 perfect square in one way, can't think of a reason for 8, please explain

I think it is some what related to number of prime numbers below hundred which is 25,
as all the other number can be represented in a minimu of two ways.

lets take an example of 7, x^2-y^2 = 7
so (x+y)(x-y) = 1 * 7, so x+y =7 and x-y = 1
this will give us a value of x = 4 and y =3, and 4^2 - 3^2 = 7.

I think this is the solution.

Can u express 2 a prime number .. and then we have 1= 1-0.
then we have 8 = 9-1.
So I have given u enough hint...
Now try to solve it..

How many numbers below 100 can be expressed as a difference of two perfect squares in only one way?
1. 25 2. 26 3.34 4.35

Please do post explanation.

I think it is some what related to number of prime numbers below hundred which is 25,
as all the other number can be represented in a minimu of two ways.

lets take an example of 7, x^2-y^2 = 7
so (x+y)(x-y) = 1 * 7, so x+y =7 and x-y = 1
this will give us a value of x = 4 and y =3, and 4^2 - 3^2 = 7.

I think this is the solution.