Maths Shortcuts2 - Numbers 1) 2^2n-1 is always divisible by...

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Numbers 1) 2^2n-1 is always divisible by 3 2^2n-1 = (3-1)^2n -1 = 3M +1 -1 = 3M, thus divisible by 3 2) What is the sum of the divisors of 2^5.3^7.5^3.7^2? ANS : (2^6-1)(3^8-1)(5^4-1)(7^3-1)/2.4.6 Funda : if a number 'n' is represented as a^x * b^y * c^z . ... where, {a,b,c,. . } are prime numbers then Quote: (a) the total number of factors is (x+1)(y+1)(z+1) . ... (b) the total number of relatively prime numbers less than the number is n * (1-1/a) * (1-1/b) * (1-1/c). ... (c) the sum of relatively prime numbers less than the number is n/2 * n * (1- 1/a) * (1-1/b) * (1-1/c). ... (d) the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * . ..../(x*y*. ..) 3) what is the highest power of 10 in 203! ANS : express 10 as product of primes; 10 = 2*5 divide 203 with 2 and 5 individually 203/2 = 101 101/2 = 50 50/2 = 25 25/2 = 12 12/2 = 6 6/2 = 3 3/2 = 1 thus power of 2 in 203! is, 101 + 50 + 25 + 12 + 6 + 3 + 1 = 198 divide 203 with 5 203/5 = 40 40/5 = 8 8/5 = 1 thus power of 5 in 203! is, 49 so the power of 10 in 203! factorial is 49 4) x + y + z = 7 and xy + yz + zx = 10, then what is the maximum value of x? ( CAT
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2002 has similar question ) ANS: 49-20 = 29, now if one of the y,z is zero, then the sum of other 2 squares shud be equal to 29, which means, x can take a max value at 5 5) In how many ways can 2310 be expressed as a product of 3 factors? ANS: 2310 = 2*3*5*7*11 When a number can be expressed as a product of n distinct primes, then it can be expressed as a product of 3 numbers in (3^(n+1) + 1)/2 ways 6) In how many ways, 729 can be expressed as a difference of 2 squares? ANS: 729 = a^2 - b^2
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This note was uploaded on 01/27/2010 for the course EE 100 taught by Professor Abc during the Spring '10 term at Punjab Engineering College.

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Maths Shortcuts2 - Numbers 1) 2^2n-1 is always divisible by...

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