Number System - Questions & Discussions

hi friends plz explain how to find last 3-digits of a number

aryan12o Says
hi friends plz explain how to find last 3-digits of a number


divide the number by 1000..the remainder will be the last 3 digits...

e.g. the last 3 digits of 456789 is 456789%1000 = 789...
aryan12o Says
hi friends plz explain how to find last 3-digits of a number

Hi

As much as I know to find out last 3 digits of a number we divide it by 1000 and the remainder is the answer. But i am not sure if we dont have any other way or not.

If you have any question that may help us to know whether this method is enough or not.

find the last 3-digits of 57^802

aryan12o Says
find the last 3-digits of 57^802


Getting 249 as last three digits..........

As 57^802 mod 1000 = 249

Applied Euler's as 1000 and 57 are coprime so ..........

(1000) = 400 so..........
57^800 mod 1000 = 1 so
57^2 mod 1000 = 249........
aryan12o Says
find the last 3-digits of 57^802


57^2 = 3249.

So remainder of 57^802 is same as R of 249^401 = 249*(249^400)

249^2 = 62001.
Remainder of 249^400 = 1^ 200 = 1.

Remainder of 57^802 = 249 * 1 = 249.

So last three digits of 57^802 is 249.

Hope I am correct.
aryan12o Says
find the last 3-digits of 57^802

E(1000) = 400
57^802 mod 1000 =
57^2 mod 1000 = 249
57^2 = 3249.

So remainder of 57^802 is same as R of 249^401 = 249*(249^400)

249^2 = 62001.
Remainder of 249^400 = 1^ 200 = 1.

Remainder of 57^802 = 249 * 1 = 249.

So last three digits of 57^802 is 249.

Hope I am correct.

yes ans is 249
Hello Frnds , Try solving this question ????
Minimum value that A can take if x2 + Ax + 900 has...

15. Integral roots.

16. Negative integral roots.

17. Positive integral roots.
hello frnds , try solving this question ????
minimum value that a can take if x2 + ax + 900 has...

15. Integral roots.

16. Negative integral roots.

17. Positive integral roots.


15) -901

16) 60

17) -901

Hiii , Frnd ,,,Please provide the solution for this question ??

Hello Frnds , Try solving this question ????
Minimum value that A can take if x2 + Ax + 900 has...
15. Integral roots.
16. Negative integral roots.
17. Positive integral roots.


900 = -2 * -450 (since A shall be minimum for integral roots and +ve roots)
900 = 30 * 30 ( -ve integral)

x^2 - 2x - 450x + 900 = 0
x(x-2) -450(x-2) = 0
(x-450)(x-2) = 0
x= 2, 450

hence minimum value is,
15. -452
16. 60
17. -452

observations
Hello Frnds , Try solving this question ????
Minimum value that A can take if x2 + Ax + 900 has...
15. Integral roots.
16. Negative integral roots.
17. Positive integral roots.


900 = -1 * -900 (since A shall be minimum for integral roots and +ve roots)
900 = 30 * 30 ( -ve integral)

x^2 - 900x - x + 900 = 0
x(x-900) -1(x-2) = 0
(x-900)(x-1) = 0
x= 1, 900

hence minimum value is,
15. -901
16. 60
17. -901

observations welcome
Hello Frnds , Try solving this question ????
Minimum value that A can take if x2 + Ax + 900 has...

15. Integral roots.

16. Negative integral roots.

17. Positive integral roots.



900 is perfect square. so IF we take A = -60.
both roots are negative integrals.

IF we take A = -452. both roots are positive integrals.
X^2 -452X + 900 = (X - 2 ) (X - 450) = 0
s0 X = 2 ,X = 450.

Plz OMIT this.

900 = -2 * -450 (since A shall be minimum for integral roots and +ve roots)
900 = 30 * 30 ( -ve integral)

x^2 - 2x - 450x + 900 = 0
x(x-2) -450(x-2) = 0
(x-450)(x-2) = 0
x= 2, 450

hence minimum value is,
15. -452
16. 60
17. -452

observation


Why are we not taking the combination - 901 ?
For A=-901 (Minimum)
x^2 -x - 900x-900 =0.This will yield positive integral root-->x=900 and x=1
Why are we not taking the combination - 901 ?
For A=-901 (Minimum)
x^2 -x - 900x-900 =0.This will yield positive integral root-->x=900 and x=1


You are correct, i have updated my post. The roots shall be 1, 900.

Didn't get the logic...

achilles95.2 Says
Didn't get the logic...

Which question are you referring to? Quote the post that you're referring to.

Guys I need the approach for a particular prblm....
I dont remember the exact data ....bt it goes smthin like ths
there are 5 students and they have different integral weights
they are weighted in pairs of 2...so we have 10 different combined values...say
C1, C2, C3, C4, C5, C6, C7, C8, C9, C10
how to find out the lightest of the five??

the combined value are arranged in increasing order
maybe like 35,37,.......etc(10 values)