If a number A has 24 factors and A x B has 105 factors then, least possible number of factors that B can have? a.12 b.9 c.16 d.7 e. none of these
105 = 3*5*7. A = p^5*q^3. => B = p*q*r^2. Hence, B has min 12 factors.
.Which of the following numbers gives the largest remainder when divided by 101? OPTIONS 1) 123456789123 2) 231456789231 3) 312456789312 4) 213456789213 5) All of the above give the same remainder
As 10^4 mod 101 = 1. => 123456789123 mod 101 = 1234+5678+9123 mod 101 = 77. Likewise you can do it for all.
please solve this question? how to find number of factors of 15! ?
15! = 1*2*3*...*15 Now highest power of 2 in 15! = 15/2 = 7/2 = 3/2 = 1 => 11. of 3 = 15/3 = 5/3 = 1 => 6. of 5 = 15/5 = 3 => 3. of 7 = 15/7 = 2 => 2. of 11 = 1. of 13 = 1. => Prime factorization of 15! = 2^11*3^6*5^3*7^2*11*13. Hence, Number of factors = 12*7*4*3*2*2.
Your friend gave you half information. See, when he told you that you would need to break 36 into 9*4, then something that needs to be pondered upon is why not break it into some other factors of 36? There are 2 reasons behind it, first one is that we know the divisibility rules of these numbers (9 and 4). And the second and most important thing to be noted here is that, the factors you need to break this into should be co-prime to each other (Chinese remainder theorem). Now, we have 16 at hand here. 2^4 = 16. So we cannot break it into two co-prime factors. So this is what we do here. We know that 10^4 mod 16 = 0. Hence we break this number in sets of 4 and take remainder individually and then obtain the sum of all the 4 digits number and then find the remainder with 16. This makes it less arduous.
Hope it helps. :)
I think if you divide 6 by 8 then also the remainder will be 6.
The sequence 1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17,. has one odd number followed by the next two even numbers, then the next three odd numbers followed by the next four even numbers and so on. What is the 2003rd term of the sequence? please explain elaborately...
The sequence 1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17,. has one odd number followed by the next two even numbers, then the next three odd numbers followed by the next four even numbers and so on. What is the 2003rd term of the sequence? please explain elaborately...
If a number A has 24 factors and A x B has 105 factors then, least possible number of factors that B can have? a.12 b.9 c.16 d.7 e. none of these in this question how B is p*q*r^2 plz explain the process ??????????
If a number A has 24 factors and A x B has 105 factors then, least possible number of factors that B can have? a.12 b.9 c.16 d.7 e. none of these in this question how B is p*q*r^2 plz explain the process ??????????
take the factors of 105 -- 3*5*7 ---(2+1)(4+1)(6+1) therefor there are total of three prime factor in multiplication of A*B----p^2*q^4*r^6 it is given A has 24 factors which means factors are q^3*r^5 so now as A*B= p^2*q^4*r^6 PUTTING VALUE OF A= q^3*r^5 WE GET THE VALUE OF B=p^2*q^1*r^1 so no of factors of B are (2+1)*(1+1)*(1+)=12
The sequence 1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17,. has one odd number followed by the next two even numbers, then the next three odd numbers followed by the next four even numbers and so on. What is the 2003rd term of the sequence? please explain elaborately...
find last two digits of ....11 ^ (25)^(31)^(41)...(1001)....?:banghead:
My take is 51.
For the last two digits, 11 has a cyclicity of 10. 11^1 = 11, 11^2 = 21, 11^3 = 31, ...., 11^10 = 01, 11^11 = 11,..., 11^25 = 51. => 11^25^31^41...1001 = 51^25^31^41^...1001. Now 51^odd will always end in 51. So, last two digits of this expression will always be 51.
l1. last 2 digits of 2^2^2003 2. N=12121212..300 digits,find the remainder when N divided by 99
1. My take is 56.
2^2^2003 = 2^(10k+8 ) = 76*56 = xx56.
2. My take is 18.
99=9*11. 2*150 - 1*150 = 150 mod 11 = 7. 1*150+2*150 = 450 mod 9 = 0. Hence, Remainder is of the form - 11k+7=9p, ie, 18.
ans given perfect squares are 18. and fourth root are 6
