ash_paglaguy SaysFind the remainder of 200!/100!^2
it will be 0 as 200!/(100!)(100!)
200*199.......101/100*99*.....1
Remainder will be zero as it is numerator is divisible by denominator
it will be 0 as 200!/(100!)(100!)
200*199.......101/100*99*.....1
Remainder will be zero as it is numerator is divisible by denominator
ash_paglaguy SaysFind the remainder of 200!/100!^2
200!=200*199*198*197*196----------------------------------------100 n dats completely divisible by
100..
200*199*198-----100*99*--------1/100!*100!
like dis 12*7/6 ,we can say 12k/6 will always be divisible by 6
this way 99*--------1/100! is also divisible by 100!
so remainder is 0
200!=200*199*198*197*196----------------------------------------100 n dats completely divisible by
100..
200*199*198-----100*99*--------1/100!*100!
like dis 12*7/6 ,we can say 12k/6 will always be divisible by 6
this way 99*--------1/100! is also divisible by 100!
so remainder is 0
answer is 0 but may be u make typo here.
99*_____1 can never be divisible by 100!.it should be 100*99*.....1. Actually it is self visualized. this is 100! itself.
Find the remainder when 100! + 2 is devided by 101 ?
apocalyptic SaysFind the remainder when 100! + 2 is devided by 101 ?
reminder is 1.
100!/101 according to Wilson theorem reminder is -1 ( 101 is prime number )
so -1+2 = 1.
300! is divisible by (24!)^n. what is the max. possible integral value of n?
viveknitw Says300! is divisible by (24!)^n. what is the max. possible integral value of n?
Is the answer 10 ?
Checked the maximum power of the biggest prime
Edit
Sorry its 13 as done by drkunjan below. My methid was same but calculated wron
viveknitw Says300! is divisible by (24!)^n. what is the max. possible integral value of n?
300/23 = 13.
n = 13.
13 is rite answer!
ash_paglaguy SaysFind the remainder of 200!/100!^2
(2n)! is always divisible by (n!)^2
can anybody tell me wat is inverse euler and how to use that to find remainder of 5^26/29
viveknitw Sayscan anybody tell me wat is inverse euler and how to use that to find remainder of 5^26/29
5^26/29
euler of 29=28 ...so 5^26/29 = 5^-2/29
Multiply numerator by 30^2..so remainder = 36/29 = 7Ans
Inverse euler theorem:-
In this you basically multiply the numerator with a term which gives remainder 1 with denominator(so that it wont effect the remainder) and converts your numerator from fraction to an integer
like in this question I have (1/5)^2 in numerator.....Now i need to multiply numerator with a multiple of 5 which leaves remainder 1 with 29
so the cond 5m = 29k+1 => m= 5k+ (4k+1)/5 ...which is true for k=1 and the number is 30
So if we multiply numerator by 30 it wont change the remainder...since we have (1/5)^2 ..multiply numerator by 30^2...and (1/5^2) /29 will change to reaminder of 36/29 = 7
wat is seed concept?
(ex seed of 19- 2)
Find the remainder when 139 + 239 + 339 + 439 + ... + 1239 is divided by 39.
Find the remainder when 139 + 239 + 339 + 439 + ... + 1239 is divided by 39.
139 + 239 + 339 + 439 + ... + 1239 = 39*12 + 7800 which is completely divisible by 39
hence remainder = 0
Find the remainder when 139 + 239 + 339 + 439 + ... + 1239 is divided by 39.
100 + 200 + 300 ....1200 mod 39
100*12*13 mod 39
=0
Find the remainder when 139 + 239 + 339 + 439 + ... + 1239 is divided by 39.
139 + 239 + 339 + 439 + ... + 1239 when divided by 39
100+39+200+39+300+39----------1200+39
100+200+---------------1200+39*12/39
100*156+39*12/39
remainder 0
N is a number such that the ratio of sum of its digit to product of its digits is 3: 40. If N is divisible by 37 and N is the smallest such number, how many factors does N have?
a. 8
b. 16
c. 20
d. 30
a. 8
b. 16
c. 20
d. 30
S is a set of 10 consecutive two-digit integers such that the product of these 10 integers has the highest power of 2 contained in it. How many such sets S are possible?
a. 4
b. 8
c. 10
d. 24
a. 4
b. 8
c. 10
d. 24
S is a set of 10 consecutive two-digit integers such that the product of these 10 integers has the highest power of 2 contained in it. How many such sets S are possible?
a. 4
b. 8
c. 10
d. 24
Good question.
In order to maximize the power of 2 we will consider the set which has highest power of 2.
I.e., we will include multiples of 8
S={64,65,66,67,68,69,70,71,72,73}
S={63,64,65,66,67,68,69,70,71,72}
S={56,57,58,59,60,61,62,63,64,65}
S={55,56,57,58,59,60,61,62,63,64}
Only 4 possible sets.