Numbers
1) 2^2n1 is always divisible by 3
2^2n1 = (31)^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 : (2^61)(3^81)(5^41)(7^31)/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 *
(11/a) * (11/b) * (11/c)
....
(c) the sum of relatively prime numbers less than the number is n/2 * n * (1
1/a) * (11/b) * (11/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
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
2002 has similar question )
ANS: 4920 = 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?
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 abc
 Prime number

Click to edit the document details