When a number can be expressed as n(n-1)(n+1) i.e product of three consecutive numbers.....why is it necessarily divisible by 6?
can someone kindly explain?
Ques - N is a natural number formed by writing, in ascending order the first 1002 whole number. What will be the remainder when N is divided by 9?
a)2 b)1 c)4 d)3
Please reply with explanation.
When a number can be expressed as n(n-1)(n+1) i.e product of three consecutive numbers.....why is it necessarily divisible by 6?
can someone kindly explain?
Cos three consecutive natural numbers will always be divisible by 1X2X3 (lowest 3 consecutive natural number) i.e 6.
All other consecutive natural numbers (in pair of three) will be multiples of 1, 2 and 3.
Ques - N is a natural number formed by writing, in ascending order the first 1002 whole number. What will be the remainder when N is divided by 9?
a)2 b)1 c)4 d)3
Please reply with explanation.
number formed 0123...1002
For a number to be divisible by 9, sum of digits should be 9
Sum of digits = (1001*1002)/2=501501=12
Remainder = 12-9=3
d)3
Brain is the most outstanding organ. It works for 24 Hours 365 days right from the time of birth but stops only when we enter the Exam Hall.
I think It should be 1
Use Fermat's little theorem:For any number N not divisible by p, N^(p-1) = 1 mod p where p is a prime
hey cn u explain above example usin fermat theorem...
number formed 0123...1002
For a number to be divisible by 9, sum of digits should be 9
Sum of digits = (1001*1002)/2=501501=12
Remainder = 12-9=3
d)3
Brain is the most outstanding organ. It works for 24 Hours 365 days right from the time of birth but stops only when we enter the Exam Hall.
As per your approach you have calculated the sum of numbers, not sum of digits like you are calculating 10 + 11 as 21 whereas we need to calculate sum of digits i.e 1+0 + 1+1 = 3
What would be the right approach?
Can anyone pls solve :
P Q R | S T
- 30 | - | 12 5
12 4 | 3 - -
Find P:Q:R:S:T?
Anyone ?
number formed 0123...1002
For a number to be divisible by 9, sum of digits should be 9
Sum of digits = (1001*1002)/2=501501=12
Remainder = 12-9=3
d)3
Brain is the most outstanding organ. It works for 24 Hours 365 days right from the time of birth but stops only when we enter the Exam Hall.
hey dude its given natural number is formed by writing first 1002 whole number...so 0 ll nt cm y u hv taken zero so can u plz explain dat
As per your approach you have calculated the sum of numbers, not sum of digits like you are calculating 10 + 11 as 21 whereas we need to calculate sum of digits i.e 1+0 + 1+1 = 3
What would be the right approach?
lets take the example 1234567891011
my approach
sum of digits = 55+11=66=12%9=3
your approch
sum of digits = 55+1+1=57=12%9=3
can try with other numbers in the same series...
Brain is the most outstanding organ. It works for 24 Hours 365 days right from the time of birth but stops only when we enter the Exam Hall
AnupMum Sayshey dude its given natural number is formed by writing first 1002 whole number...so 0 ll nt cm y u hv taken zero so can u plz explain dat
O is a whole number, so we'll consider that as well.
lets take the example 1234567891011
my approach
sum of digits = 55+11=66=12%9=3
your approch
sum of digits = 55+1+1=57=12%9=3
can try with other numbers in the same series...
Brain is the most outstanding organ. It works for 24 Hours 365 days right from the time of birth but stops only when we enter the Exam Hall
You are right!
Can anyone pls solve :
P Q R | S T
- 30 | - | 12 5
12 4 | 3 - -
Find P:Q:R:S:T?
Multiple the first and second relations with 15 and 2 respectively
then ratio of P:Q:R:S:T is 180:60:45:24:10
Ques - N is a natural number formed by writing, in ascending order the first 1002 whole number. What will be the remainder when N is divided by 9?
a)2 b)1 c)4 d)3
Please reply with explanation.
The remainder when any number N divided by 9 is equal to digit sum of N
The digit sum of sum of first 1002 whole numbers is 1001*1002/2 = 1*3 =3
Q1 If n, m are natural numbers and Lcm + HCF = 63. How many ordered pairs of (n,m) exists?
Q2. In how many ways 15! Can be written as product of two coprime numbers?
15!= 2^11*3^4*5^3*11*7*13
factors should not be common in two co-prime no.s . now if i use calculation method and find the pair , it will take time so , please explain the shortcut.
1) My take is 126.
5625 = 625*9.
5126 mod 625 = 126.
126*127/2 mod 9 = 0.
Hence remainder will be of the form - 625k+126 = 9p,ie, 126.
2) My take is 4.
Just subtract sum of alternate digits.
3) My take is 249.
8*1,8*2,...,8*1000 = 8^1000(1*2*...1000) = 8^1000*1000!.
=> 1000/5 = 200/5 = 40/5 = 8/5 = 1. So, 249.
in 2 nd question , how have you find the difference between the sum of alternate digit . it is getting complicated.
The remainder when any number N divided by 9 is equal to digit sum of N
The digit sum of sum of first 1002 whole numbers is 1001*1002/2 = 1*3 =3
how have you find the digit sum ?
If n, m are natural numbers and Lcm + HCF = 63. How many ordered pairs of (n,m) exists?
Ans-(54,9), (60,3),(56,7), (42,21)
is it correct ?
amitshanky Sayshow have you find the digit sum ?
Dude it would be through this:-
n (n+1)/2
If n, m are natural numbers and Lcm + HCF = 63. How many ordered pairs of (n,m) exists?
Ans-(54,9), (60,3),(56,7), (42,21)
is it correct ?
I think it's 6 ordered pairs
Approach:
Let the numbers n be Hx and m be Hy where x and y are co-prime to each other
then LCM = Hxy and HCF = H
so,we can say H*(xy + 1) = 63 = 1*63 , 3*21 ,9*7 ,7*9, 63*1
which means x*y = 62 or 20 or 6 or 8 or 0 and here 0 and 8 are not possible
finding number of ordered pairs of (x,y) is equal to finding number ordered pairs of (n,m)
so,ordered pairs of (x,y) are (31,2)(2,31),(4,5),(5,4),(2,3),(3,2)
Hope it helps;)
I think it's 6 ordered pairs
Approach:
Let the numbers n be Hx and m be Hy where x and y are co-prime to each other
then LCM = Hxy and HCF = H
so,we can say H*(xy + 1) = 63 = 1*63 , 3*21 ,9*7 ,7*9, 63*1
which means x*y = 62 or 20 or 6 or 8 or 0 and here 0 and 8 are not possible
finding number of ordered pairs of (x,y) is equal to finding number ordered pairs of (n,m)
so,ordered pairs of (x,y) are (31,2)(2,31),(4,5),(5,4),(2,3),(3,2)
Hope it helps;)
I think we will include x*y=8
(1,
(8,1)