6407522209>3600840049, i.e the numerator is greater than the denominator and hence the fraction will be greater than one. sq rt of a fraction greater than one will be greater than one. No option has any such value. Hence answer is d)none of dese
rahul jain15 Says6407522209>3600840049, i.e the numerator is greater than the denominator and hence the fraction will be greater than one. sq rt of a fraction greater than one will be greater than one. No option has any such value. Hence answer is d)none of dese
rahul i think u did a mistake in reading the question. u r rite that square root will be greater than 1 but in que"2-sq rt." is asked.
y even bother till 5... 9 has cyclicity 2 from 2! onwards all are even ...even+1=odd
this means unit digit 9
Find the last two digits of the number (36)^57
1)N = abcab where 4 N has to be in the form 1001* P1*P2
where 1001 = 7*11*13
N = 1001*P1*P2
as 4 so we have set p1,p2 = (3*17) , (3*19) and (3*23)
hence total 3 values of N are possible
option c
but saying that N=1001*p1*p2 would be wrong as n is the product of three prime numbers wheras 1001 isnt
piyushrohella12 Saysbut saying that N=1001*p1*p2 would be wrong as n is the product of three prime numbers wheras 1001 isnt
sorry sorry my bad dint notice that it was the product of 5 not three
girish142 SaysFind the last two digits of the number (36)^57
the last 2 digits are 96
6^114
(2^10)^11 *16*3^114
24^11*16*9*(81)^28
24*16*9*41=96
wats the oa?
girish142 SaysFind the last two digits of the number (36)^57
solved it, is it 56
ok it should be 96. I had done some silly mistake
the last 2 digits are 96
6^114
(2^10)^11 *16*3^114
24^11*16*9*(81)^28
24*16*9*41=96
wats the oa?
i didn't have the solution for it...evn i solved n got 96...shud be ryt
thanks!!!
why did u took power 114
please explain
its because (36)^57 = (6^2)^57 = (6)^(2*57) = 6^114
Use Euclids division lemma to show that the square of any positive integer is either of
the form 3m or 3m + 1 for some integer m.
each of these and show that they can be rewritten in the form 3m or 3m + 1.]
the form 3m or 3m + 1 for some integer m.
each of these and show that they can be rewritten in the form 3m or 3m + 1.]
3^25/50=1
implies 3^7/50=3^4/50+3^3/50=31+27/50=58?50=8.remainder??????
3^25/50=1
implies 3^7/50=3^4/50+3^3/50=31+27/50=58?50=8.remainder??????
Shouldnt it be 43???
chandrakant.k SaysShouldnt it be 43???
evn im getting 43!!
is it right????
girish142 SaysFind the last two digits of the number (36)^57
36^57=6^114
rem 6^114/100= rem 6^114/25 + rem 6^114/4
= 6^4 * (6^5)^22 / 25 + 36^57 / 4
= 6^4 * (7775 + 1)^22 / 25 + rem(0)
= 6^4 * 1^22
= 36*36 /25
= 11*11 / 25
=121/25 = 21
so on dividng no by 25 gives 21 as remainder and by 4 gives 0
= (25k + +21)/4
k=3
so 25*3+21=96
96
Find the reminder (7^26+5^83)/100
Asfakul SaysFind the reminder (7^26+5^83)/100
answer = 74
any power of 5 will give last 2 digits 25
wheras last digits of 7^4 are 01
so remainder = (01*49 + 25)
=74
Hi Puys...
Can you plzz help me out in this ques
find the reminder when 21^875 is divided by 17..............
hi puys...
Can you plzz help me out in this ques
find the reminder when 21^875 is divided by 17..............
21^875/17 = 4^875/17 = 4.16^437/17 = -4/17 = 13