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1) Factors• a whole number that divides exactly into another number.
• a whole number that multiplies with another number to make a third number.
Factors Pair• a pair of numbers multiplied together form another number called their product.
Number of factorsA number N can be written as \( N=p^xq^yr^z\), where p,q, and r are prime factors of the number N, and x,y, and z are positive integers.
Number of factors : (including 1 and the number itself)\(=(x+1)(y+1)(z+1)\);
Sum of all the factors\(\left[ \begin{array}{cc|r} \frac{a^{x+1}-1}{a-1} \end{array} \right]\) \(\left[ \begin{array}{cc|r} \frac{b^{y+1}-1}{b-1} \end{array} \right]\) \(\left[ \begin{array}{cc|r} \frac{c^{z+1}-1}{c-1} \end{array} \right]\)
Product all the factors\(N^{\frac{number of factors}{2}}\)
Suppose 120 is our number and the prime factorization is \(2^3*3^1*5^1\)
1) Number of factors: simply take the exponent of each factor and add 1. \((3+1)*(1+1)*(1+1)=16\)
2) Sum of all factors: \(\frac{[(2^0+2^1+2^2+2^3)(3^0+3^1)(5^0+5^1)]}{[(2-1) (3-1)(5-1)]} = 45\)
3) Product of all factors: \(120^{\frac{16}{2}}=120^8\)
1. A number is divisible by 2 if its unit’s digit is even or zero e.g. 128, 146, 34 etc.
2. A number is divisible by 3 if the sum of its digits is divisible by 3 e.g. 102, 192, 99 etc.
3. A number is divisible by 4 when the number formed by last two right hand digits is divisible by ‘4’ e.g. 576, 328, 144 etc.
4. A number is divisible by 5 when its unit’s digit is either five or zero : e.g. 1111535, 3970, 145 etc.
5. A number is divisible by 6 when it’s divisible by 2 and 3 both. e.g. 714, 509796, 1728 etc.
6. A number is divisible by 8 when the number formed by the last three right hand digits is divisible by ‘8’. e.g. 512, 4096, 1304 etc.
7. A number is divisible by 9 when the sum of its digits is divisible by 9 e.g. 1287, 11583, 2304 etc.
8. A number is divisible by 10 when its units digit is zero. e.g. 100, 170, 10590 etc.
9. A number is divisible by 11 when the difference between the sums of digits in the odd and even places is either zero or a multiple of 11. e.g. 17259, 62468252, 12221 etc.
For the number 17259 : Sum of digits in even places = 7 + 5 = 12, Sum of digits in the odd places = 1 + 2 + 9 =12 Hence 12 – 12 = 0.
10. A number is divisible by 12 when it is divisible by 3 & 4 both. e.g. 672, 8064 etc.
11. A number is divisible by 25 when the number formed by the last two Right hand digits is divisible by 25. e.g. 1025, 3475, 55550 etc.
12. A number is divisible by 125, when the number formed by last three right hand digits is divisible by 125. e.g. 2125, 4250, 6375 etc.
NOTE :
1. When any number with even number of digits is added to its reverse, the sum is always divisible by 11. e.g. 2341 + 1432 = 3773 which is divisible by 11.
2. If X is a prime number then for any whole number “a” \((a^X – a)\) is divisible by X e.g. Let X = 3 and a = 5. Then according to our rule \(5^3– 5\) should be divisible by 3.
Now \((5^3– 5) = 120\) which is divisible by 3
2) PrimesA number greater than 1 which has no factor other than 1 and the number itself is a
Prime NumberA number with two distinct factors.
We do have infinite prime numbers and apparently, to our knowledge today, there is NO clear pattern in their unfolding toward the infinite on the number line.
Prime numbers are only positive.
Two (2) is the oNLy positive Even prime numbers
The difference between any two primes greater than 2 is always even.
Verifying the primality (checking whether the number is a prime) of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all integer numbers smaller than \(\sqrt{n}\), thereby checking whether \(n\) is a multiple of \(m\leq{\sqrt{n}}\).
Example: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\), from integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime.
Note that, it is only necessary to try dividing by
prime numbers up to \(\sqrt{n}\), since if
n has any divisors at all (besides 1 and
n), then it must have a prime divisor.
If \(n\) is a positive integer greater than 1, then there is always a prime number \(p\) with \(n < p < 2n\).
3) MultiplesHCF (GCD / GCF) & LCM OF NUMBERSHCF : It is the greatest factor common to two or more given numbers. It is also called GCF OR GCD (greatest common factor or greatest common divisor). e.g. HCF of 10 & 15 = 5, HCF of 55 & 200 = 5, HCF of 64 & 36 = 4
To find the HCF of given numbers, resolve the numbers into their prime factors and then pick the common term(s) from them and multiply them if more than one. This is the required HCF.
LCM : Lowest common multiple of two or more numbers is the smallest number which is exactly divisible by all of them. e.g. LCM of 5, 7, 10 = 70, LCM of 2, 4, 5 = 20, LCM of 11, 10, 3 = 330
To find the LCM resolve all the numbers into their prime factors and then pick all the quantities (prime factors) but not more than once and multiply them. This is the LCM.
NOTE:
1. LCM x HCF = Product of two numbers (valid only for “two”)
2. HCF of fractions = HCF of numerators ÷ LCM of denominators
3. LCM of fractions = LCM of numerators ÷ HCF of denominators
Calculating LCM :
Method 1 :
Factorization MethodRule – After expressing the numbers in terms of prime factors, the LCM is the product of highest powers of all factors.
Q. Find the LCM of 40, 120, 380.
A. 4\(0 = 4 \times 10 = 2 \times 2 \times 2 \times 5 = 2^3 \times 5^1\)
\(120 = 4 \times 30 = 2 \times 2 \times 2 \times 5 \times 3 = 2^3 \times 5^1 \times 3^1\)
\(380 = 2 \times 190 = 2 \times 2 \times 95 = 2 \times 2 \times 5 \times 19 = 2^2 \times 5^1 \times 19^1\)
⇒ Required LCM = \(2^3 \times 5^1 \times 3^1 \times 19^1 = 2280\).
Method 2 :
Division methodQ. Find the LCM of 6, 10, 15, 24, 39
Ans. LCM = 2 x 2 x 3 x 5 x 2 x 13 = 1560.
Calculating HCF FactorizationAfter expressing the numbers in term of the prime factors, the HCF is product of COMMON factors.
Ex. Find HCF of 88, 24, 124
\(88 = 2 \times 44 = 2 \times 2 \times 22 = 2 \times 2 \times 2 \times 11 = 2^3 \times 11^1\)
\(24 = 2 \times 12 = 2 \times 2 \times 6 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3^1\)
\(124 = 2 \times 62 = 2 \times 2^1 \times 31^1 = 2^2 \times 31^1 ⇒ HCF = 2^2\)
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