A natural number N has 991990 factors. Find the minimum number of composite factors of N? (explain plz)
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How many set of two natural numbers are there such that their LCM is 360?
- 54
- 43
- 42
- 53
0 voters
plz help me with the solution of the problem----
Q- A clock has hour and minute hand of same length. then how many times in a day will it be impossible to tell the xact time due to the above mentioned ambiguity?
I found some answers as 22 but wasn't able to figure out.
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The unit digit in the expression (36 ^234 * 33 512 * 39^ 180) - (54 ^ 29 * 25 ^123 *31^ 512) ?
LCM of 2 natural numbers A and B =300. How many different sets of A and B are possible?
LCM of 3 natural numbers = 150. how many different sets of 3 numbers are possible?
if p is a prime number and n is a natural number then HCF (n, n+p) = HCF [n,(n+p-n)].
is this true?
what is the remainder when x is divided by 100, where x is the greatest integer value of [ หลก21+ หลก20]^50 ?
let N be a no,which is the product of all the possible remainder when 6!^5n divided by 7,now when (N+7)! divided by (6!)^p,find the highest value of p,so dat it is completely divisible..
options:
6
2
7
more dan 6
N is a natural number. 2N has 28 factors and 3N has 30 factors. How many factors does 6N have?
find the zeros at the end of (31!)^(31!)
- 49
- none of these
- 7^7
- 7^31!
0 voters
find the zeros at the end of (31!)^(31!)
1) 7^7
2) 7^31!
3) 49
4) none of these Skip
รโโ
find the zeros at the end of (31!)^(31!)
1) 7^7
2) 7^31!
3) 49
4) none of these Skip
รโโ
if we count no's of tens up to 31 there are 7 tens so, the answer will be 7power 31!
I think the answer is 10000.
x^n-y^n = (x-y)[x^n-1 + x^n-2*y +......+y^n-1]
101^100-1 = (101-1)[101^99+101^98+.........+101^3+101^2+101^1+1]
now 101^1 = 101
101^2 = 10201
101
10201
1030301
104060401
10510100501
.
.
.
99terms
Units and tens digits sum up to 100 hundredth digit will be 9....so inside the bracket the digit formed will be of the form XYZ.....900 which is divisible by 100.
Hence 100*100 = 10000
รโโ
what is the highest power of 100 in 101^100 - 1 ?
รโโ
I think the answer is 2 ie 100^2.
x^n-y^n = (x-y)[x^n-1 + x^n-2*y +......+y^n-1]
101^100-1 = (101-1)[101^99+101^98+.........+101^3+101^2+101^1+1]
now 101^1 = 101
101^2 = 10201
101
10201
1030301
104060401
10510100501
.
.
.
99terms
Units and tens digits sum up to 100 hundredth digit will be 9....so inside the bracket the digit formed will be of the form XYZ.....900 which is divisible by 100.
Hence 100*100 = 10000
what's ans.of 31! to power 31!