Topic - Number Theory

Solved ? - Yes

Source - ]]>

Topic - Number Theory

Solved ? - Yes

Source - ]]>

Concept : Total number of factors = 1 + (Prime factors) + (Composite Factors) ]]>

Case 1 : (4x^6 + x^3 + 5) mod 7 = 5

Case 2 : (4x^6 + x^3 + 5) mod 7 = 4(1) + 1 + 5 mod 7 = 10 mod 7 = 3

Case 3 : (4x^6 + x^3 + 5) mod 7 = 4(1) - 1 + 5 mod 7 = 8 mod 7 = 1

Hence zero values. ]]>

Factors will be 1,a,b,ab

1 + a + b = 31

a + b = 30

a,b = 7,23 ; 11,19 ; 13,17 => N = 161,209,221

Case 2 : N = a^3

Factors : 1,a,a^2,a^3

1 + a + a^2 = 31

a(a + 1) = 30

a = 5 is the only case

N = 125

Sum = 716

]]>For our question,

C(58,29) mod 29

= C(29,0)^2 + C(29,1)^2 + .. + C(29,29) mod 29

All terms except first and last are divisible by 29.

Remainder = C(29,0)^2 + C(29,29)^2

= 1 + 1 ]]>

Why 105 ?

Step 1 : Find numbers not divisible by 2,3,5 or 7 by simple Euler number's application.

Step 2 : Remove numbers like 103 which being not divisible by 2,3,5 or 7 were removed in euler but do not belong to our range of 1-100 for which answer is to be calculated.

105 Chosen as it is the closest(to100) multiple of LCM of uncancelled denominators 5,7

a + b = 12

As near to each other as possible with they being coprime to each other.

5, 7

Max product = (12 * 5) * (12 * 7) = 60 * 84 = 5040. ]]>

17!(1 + even)

17! * odd

Find power of 2 and subsequently 8 in 17!

Power of 2 : 8 + 4 + 2 + 1 = 15

Power of 8 : [15/3] = 5 ]]>

Power of 3 in 38! = 12 + 4 + 1 = 17

Power of 3 inside bracket = 0

Highest power to divide 58! - 38! = 17 ]]>

26^57 mod 29

26^56 * 26 mod 29

26 ]]>

[(24 - 6)^2 / 48 ] = 7 ]]>

All different : 5c4 * 4! = 120

2 same 2 different : 2c1 * 4c2 * 4!/2! = 144

2 same 2 same : 2c2 * 4!/2!2! = 6

Total : 270 ]]>