Bill and Clinton take the square of a certain decimal number and express it in base 5 and 6 respectively. Then Bush comes and he takes the two representations and assuming that the expressions are in base 10, adds the numbers. Which of the following cannot be the value of the unit's digit of the sum obtained?
a.
0
b.
2
c.
8
d.
6
e.
3
The product of two numbers '231' and 'ABA' is 'BA4AA' in a certain base system (where base is less than 10), where A and B are distinct digits. What is the base of that system.
The product of two numbers '231' and 'ABA' is 'BA4AA' in a certain base system (where base is less than 10), where A and B are distinct digits. What is the base of that system.a. 5b. 6c. 7d. 8e. 4
1) 6Squares in base 5 end in 0,1 and 4 while squares in base 6 end in 0,1,4 and 3so we cannot get 6 as the sum...2) (2x^2+3x+1)(ax^2+bx+a) = bx^4+ax^3+4x^2+ax+aoptions...6(72+18+1)(36a+6b+a)=(1296b+216a+144+6a+a)91(37a+6b) = (223a+1296b+144)3367a+546b = 223a+1296b+1443144a-750b = 144Solution -> (a,b) = (1,4)put it back in the equation and it satisfies...so answer should be 6....this is a lenghty approach....cld not figure out a better one.
ya mann the approach for the 2nd question was quite lengthy.
1)what whole no below 1000 has the maximum factors ?ans) 840 with 32 factors.2)what whole no below 100 has the maximum factors ?ans) 60, 72, 84, 90 and 96 each have 12 factors.I know how to find the no of factors for a particular no but how to find this ?
1. it has to involve more number of primes...here 2*3*5*7 will be the highest in that way..if we use next prime 11 then it exceeds 1000 now 2*3*5*7 = 210 ..nw to maximise ths value..increase power of 2..
==> 2^3* 3*5*7 =840 total factors = 32 ,
2. 2^n *3^m ==> n = 3 m=2 ; 2^a*3^b*5^c ==> a=2,b=1,c=1 ..similarly you can write for all numbers in this format
The product of two numbers '231' and 'ABA' is 'BA4AA' in a certain base system (where base is less than 10), where A and B are distinct digits. What is the base of that system.a. 5b. 6c. 7d. 8e. 4
2x^2+3x+1)(Ax^2+Bx+A)=Bx^4+Ax^3+4x^2+Ax+A (2A-B)x^3+(2B+2A)x^2+(3A+3B-4)x+(2A+B)=0 There is no better option than hit and trail for these kind of questions. here (A,B) =1,4 and x=6. So base system is 6.
1. Find the number ABCD, such that ABCD*4 =DCBA? 2.Find the average of all 5 digit numbers formed by 1,2,3,4 and 5?
2x^2+3x+1)(Ax^2+Bx+A)=Bx^4+Ax^3+4x^2+Ax+A (2A-B)x^3+(2B+2A)x^2+(3A+3B-4)x+(2A+B)=0 There is no better option than hit and trail for these kind of questions. here (A,B) =1,4 and x=6. So base system is 6.
1. Find the number ABCD, such that ABCD*4 =DCBA? 2.Find the average of all 5 digit numbers formed by 1,2,3,4 and 5?
1) 2178*4 = 8712
also 21978*4 = 87912
i remember this....if you want i can show the working...but i advice everyone remembers such stuff.....
2) avg of all 5 digit numbers formed by digits - 1,2,3,4,& 5
5!(15)(11111)/5! = 166665
pl let me know if this is correct....i feel i have forgotten how to do this 😞
1) 2178*4 = 8712also 21978*4 = 87912 i remember this....if you want i can show the working...but i advice everyone remembers such stuff.....2) avg of all 5 digit numbers formed by digits - 1,2,3,4,& 55!(15)(11111)/5! = 166665 pl let me know if this is correct....i feel i have forgotten how to do this
😛 not every one remembers those who are regular here can !! 2nd one is simple 12345+54321/2 =33333
2x^2+3x+1)(Ax^2+Bx+A)=Bx^4+Ax^3+4x^2+Ax+A (2A-B)x^3+(2B+2A)x^2+(3A+3B-4)x+(2A+B)=0There is no better option than hit and trail for these kind of questions. here (A,B) =1,4 and x=6. So base system is 6.1. Find the number ABCD, such that ABCD*4 =DCBA?2.Find the average of all 5 digit numbers formed by 1,2,3,4 and 5?
2. sum of all numbers /total no.of numbers = 11111*4!* (1+2+3+4+5)/ 120 = 11111*3 =33333
1. ABCD ==> A=1 ,2 ..A can't be 1 bcs any num multiplied by 4 never ends with 1
so, A=2 ==> D = 3 or 8 ..now any four digit multiplied by 4 will always be > 4000 ==> D=8
now the number is 2BC8 *4 = 8CB2 ...here 2BC8 is always now B can be 0,1,2 and 4C+3 = _B . ==> B can never be 0 or 1 .. am nt able to go further..bt hit n trail..using 20C8 and 21C8 ==> gav me 2178
Ten tickets are numbered 1,2,3.........10. six tickets are selected at random one at a time with replacement. prob that largest no. appearing on the selected ticket is 7 is ??
Ten tickets are numbered 1,2,3.........10. six tickets are selected at random one at a time with replacement. prob that largest no. appearing on the selected ticket is 7 is ??
Ten tickets are numbered 1,2,3.........10. six tickets are selected at random one at a time with replacement. prob that largest no. appearing on the selected ticket is 7 is ??
The tickets should not have 8,9,10
Total =10^6
Favourable cases= (All cases less thanor equal to 7) - (cases less than 7 when 7 is not selected)
The probability of a bomb hitting a bridge is 1/2 and two direct hits are needed to destroy it .The least number of bombs required so that the probability of the bridge being destroyed is greater than 0.9 is????/
Ten tickets are numbered 1,2,3.........10. six tickets are selected at random one at a time with replacement. prob that largest no. appearing on the selected ticket is 7 is ??
