The remainder is 23.
A decimal that has the number formed by the recurring digits for its numerator and a denominator with as many nines as digits in the period.
0! is equal to 1.
6 AM.
4691100843
6.
HCF of fractions = HCF of numerators / LCM of denominators.
Divide 99 by 4 to get a remainder of 3, then calculate 2^3 which equals 8. Thus, the unit digit of 2^99 is 8.
6.
Consider only the unit digit of the number and divide the power by the cyclicity of the unit digit or by 4 to find the remainder.
17.
2772.
When its units place is 0 or divisible by 2.
4.
13.
0.3 = 1/3.
It is denoted as n!.
2³ × 3¹.
The cyclicity will become the power of the unit digit.
If the sum of its digits is divisible by 3.
When the last two digits are 0’s or divisible by 4.
The least number that is exactly divisible by each of the given numbers.
5.
(x - y)²z is even.
(20, 27).
N/(a*b) = a*p*r1 + b*q*r2, where r1 is N/a and r2 is N/b.
The final remainder is 23.
2136/1000.
If the last three digits are 0’s or divisible by 8.
10
Remainder (a - b)/c = remainder(a/c) - remainder(b/c).
10
9
The cyclicity is 2.
7.
315.
Both (a+b) and (a-b) are also divisible by H.
N = LCM of a, b, and c + R.
If the unit digit is 5 or 0.
A number M when divided by N leaves remainder R, and quotient Q can be represented by M = NQ + R.
When n is EVEN.
If the difference between the sum of digits at odd places and the sum at even places is either 0 or 11.
2 pairs.
1
3
3Y(Z - X) is an even integer.
940.
412/100.
It will result in an odd number. Example: 3 × 3 = 9.
If M and N are co-prime and N is a prime number, then remainder (M^N - 1)/N = 1.
5
148.
It must be divisible by both 2 and 3.
If the number of tens added to 5 times the number of units is divisible by 7.
Remainder (a+b)/c = remainder(a/c) + remainder(b/c).
The remainder is 5.
0
The lowest prime number is 2.
Yes
4
3004.
10 min 30 seconds.
H is HCF of (a - p), (b - q), (c - r).
0.37 = 37/99.
It will always result in an even number or zero. Example: 1 + 3 = 4.
Always divisible by a - b.
33
2 is the only even prime number.
6
9729.
252
11111111
47/100.
57/100.
The greatest number that divides two or more numbers exactly.
A zero can be formed by the multiplication of 5 and 2.
If the sum of its digits is divisible by 9.
1
Using Fermat's rule, find remainders for 5^6/7 and 5^10/11, both yielding a remainder of 1.
5, 4, 2
274
HCF * LCM = Product of the two fractions.
2028.
There are 25 prime numbers between 1 to 100.
6
6
It has a remainder of 0.
It will always result in an even number or zero. Example: 2 + 4 = 6.
It will result in an odd number. Example: 4 + 3 = 7.
5
163
1683.
130 (LCM of 6, 7, 9 is 126, then add 4).
The final remainder is 29.
1
The cyclicity is 1.
The remainder is 1.
124.
8
21
It is calculated using the formula: [n/p1] + [n/p2] + [n/p3] + ..., where [n/pi] denotes the quotient when n is divided by p.
There are 24 zeros at the end of 100!.
When n is ODD.
The unit place digit must be 0.
Remainders can also be expressed as negative values; for example, remainder 27/7 = 6 or -1.
18.
48
10080.
7
Remainders (axb)/c = remainder(a/c) x remainder(b/c).
The cyclicity is 4.
The standard form is m × 10^n where m lies between 1 and 10 and n is an integer.
3600
2.
4
