when numbers are written in base b we have 12x25 333 The value of b is a 6 b 8

ans is 7
12=(2*7^0)+(1*7^1)=9
25=(5*7^0)+(2*7^1)=19
333=(3*7^0)+(3*7^1)+(3*7^2)=171
ie. 9*19=171
12 years agoHelpfull: Yes(15) No(6)
when a no. is written in base b
(1*b^1+2*b^0)*(2*b^1+5*b^0)=(3*b^2+3*b^1+3*b^0)
(b+2)*(2b+5)=(3b^2+3b+3)
solve it further a quadratic eq. is formed

b^2-6b-7
after solving it can be written as
(b-7)(b+1)
therefore b=7
b=7
12 years agoHelpfull: Yes(15) No(4)
what the meaning base in this questions
8 years agoHelpfull: Yes(9) No(1)
Let the base = b
So, (b+2)(2b+5) = (b+2)(2b+5)=3b2+3b+3
2b2+9b+10=3b2+3b+3
b2−6b−7=0
Solving we get b = 7 or -1
So b = 7
8 years agoHelpfull: Yes(2) No(1)
12*25=5*2=10
but given 3
therefore 10-3 =7 is the base
8 years agoHelpfull: Yes(2) No(1)
5*2=3,carry 1
5+5=3,carry 1
therefore
12*25=333
12 years agoHelpfull: Yes(1) No(25)
it will be b=7 b=-1 so -1 discarded my friend.
9 years agoHelpfull: Yes(0) No(0)
Let the base = b
So, (b+2)(2b+5) = (b+2)(2b+5)=3b2+3b+3(b+2)(2b+5)=3b2+3b+3
2b2+9b+10=3b2+3b+32b2+9b+10=3b2+3b+3
b2−6b−7=0b2−6b−7=0
Solving we get b = 7 or -1
So b = 7
8 years agoHelpfull: Yes(0) No(3)
a
when a no. is written in base b
(1*b^1+2*b^0)*(2*b^1+5*b^0)=(3*b^2+3*b^1+3*b^0)
(b+2)*(2b+5)=(3b^2+3b+3)
solve it further a quadratic eq. is formed

b^2-6b-7
after solving it can be written as
(b-7)(b-1)
therefore b=7
6 years agoHelpfull: Yes(0) No(0)

when numbers are written in base b, we have 12x25=333. The value of b is
a)6 b)8 c)7 d)none of these