Remainder when 4^37 + 6 ^ 37 is divided by 25....plz give the approach too....
@abhijeetace said: Remainder when 4^37 + 6 ^ 37 is divided by 25....plz give the approach too....
Its 20
(5-1)^37 + (5+1)^37
Using Binomial expansion, Only terms in odd places remains i.e. even power of -1 .. terms in even places cancels out.
every term except, 37C1 5^1 (-1)^36 + 37C1 5^1 (1)^36 , is divisible by 25.
diving this by 25 gives 20.
@abhijeetace said: Remainder when 4^37 + 6 ^ 37 is divided by 25....plz give the approach too....
is the answer 20?
@rahul005
38th term=a+37d=301/25
88th term=a+87d=903/25
sum of first 75 terms= (75/2)*[2a+(75-1)d]=75*(a+37d)=75*301/25=903
sum of last 100 terms= sum of total 175 terms - sum of first 75 terms
=
175/2*[2a+174d]-903=175*(a+87d)-903=175*903/25-903=7*903-903=6*903
hence ratio=903/(6*903)=1/6
38th term=a+37d=301/25
88th term=a+87d=903/25
sum of first 75 terms= (75/2)*[2a+(75-1)d]=75*(a+37d)=75*301/25=903
sum of last 100 terms= sum of total 175 terms - sum of first 75 terms
=
175/2*[2a+174d]-903=175*(a+87d)-903=175*903/25-903=7*903-903=6*903
hence ratio=903/(6*903)=1/6
@JITESHSRI said: @vranjan pls do explain ur ans
á´“(11)=10
And, 51^52 will have its last digit as 1. So, it can be expressed as 10k+1.
So, (50^51^52)=50^(10k+1)
Now, Applying Euler's theorem, remainder will be same as 50/11 which is 6.
@vranjan
@vranjan said: @kkwillbell A) 59Rajesh found a decimal number which when represented in bases 2,3,4,5 and 6 ends in 1,2,3,4 and 5 respectivelyB) 15.Are these the Ans?
A) What is the smallest positive integer satisfying this property ?
B) How many such three-digit numbers are possible ?
How u got these answers?
How u got these answers?
@JITESHSRI said: @vranjan Rajesh found a decimal number which when represented in bases 2,3,4,5 and 6 ends in 1,2,3,4 and 5 respectivelyA) What is the smallest positive integer satisfying this property ?B) How many such three-digit numbers are possible ?How u got these answers?
Last digit in every base will be same as the remainders mentioned as in the question.
So the problem is same as when a no. N is divided by 2,3,4,5,6 leaves remainders as 1,2,3,4,5 respectively.
So, N=(LCM of 2,3,4,5,6)-1 [1=2-1=3-2=4-3=5-4=6-5]
N=59, the general form of N will be 60k+59, and the no. of 3 digit nos. satisfying this form will be 15.
@prati086 said:pleaseeeeeeee solv ths n also telll me the trick behind it.........1.Find maximum value of n such that 157! is perfectly divisible by 10^n2.Find the o of consecutive 0'sat the end in 77! x 42! frnds pls giv me the expained solution (if poss with a tricck to solve such ques)
1> no of zeros at the end funda-- 38 zeroes at the end so 10^38
2>again the same funda- 16 zeroes at the end of 77! and 9 at the end of 42! so add them 25;
do u want the concept to find the no of zeroes too?
@prati086
yes,
last two digits- yaad nai hai funde, lets see if some body else can help u out, ya fir will check it out once i go home and post it..
@rahul005 said:The 38th term and the 88th term of an arithmetic progression are 301/25 and 903/25 respectively.If the total number of terms in the progression is 175, what is the ratio of the sum of the first 75 terms to the sum of the last 100 terms of the progression?ans..1/6(pls show detailed sol.)
Sum of the first 75 terms = 75 * (38th term) = 75 * (301/25) = 903
Sum of the first 175 terms = 175 * (88th term) = 175 * (903/25) = 6321
Sum of last 100 terms = 6321 - 903 = 5418
Thus the required ratio = 903/5418 = 1/6
find the last 2 digits of
65*29*37*63*71*87
with approach
@ipagal said:never mind...got it
You have to multiply only the last 2 digits taking two nos. a ta atime !
Regards,
Never Back Down
Never Back Down
@belldcat17
to find out the remainder when 79^79 is divide by 100!, you need to find out the last 2 digits of 79^79.
the last 2 digits being 19, the answer will be 19.
@prati086 said: @firsty ok firsty i will wait fr ur rply ,bt pls dnt frgt to rply
as promised-- sorry for the delay
@prati086 said: @firsty thnk u so much fr ths help bt there is also a method based on remaindeer theorem concept cn u explain tht one to me
what is the concept??
last two digits= remainder when divided by 100; that is the concept..
@prati086 said: @firsty 62^586 here how to apply Remainder Theorem ,Pls if u tell it to me my doubt will clear
62=(2*31)^586=>((2^10)^58)*(2^6)*(31^586)
((2^10)^58) last 2 digs=76
2^6=64
(31^586)=last 2 digs=81
multiplying 76* any no ends in same no;
effectively- 64*81=84= last two digs:
alll the concepts are there in the article