CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[Aarav]

Originally Posted by Superstar

Hi Aarav,

After folowing all the discussions the entire day, the solution to the problem is still not clear to many of us.

After we get the number 196 in base 10 or 524 in base 6, we need to find out the value of 'x' which is the number of bases which would represent this number in two digits.

Now, what is wrong in dealing in decimals and try to find the number of bases which can show 196 in two digits. We get 182 as the answer or 350 in base 7.

I am not able to understand the solution. I am sure others would also like to finally finish this off today itself.

Please provide the detailed solution if possible by also answering my concern.

Thanks,

Your Fan and Admirer
Superstar

I have deliberately kept the solution short for this as at most one more problem will appear on base systems in QQAD 2008 and wanted people to try hard to understand this well.

Solution:
Let ABC be the largest 3 digit perfect square in base 6 (A, B, C <= 5) => in base 10 ABC will be 36A + 6B + C and this number is a perfect square. Since 5*(36 + 6 + 1) = 215 => 36A + 6B + C = 14^2 = 196 => ABC = 524.

Now, 524 is written as a 2 digit number CD in base b => b*C + D = 524 where C, D < b and C is a positive integer => b ranges from 23 to 524 => x = 502.

The properties (odd/even/prime/perfect square) of a number N in base b is same as that of decimal conversion of N => x is even in base 7 also.

Please note that as the questions asks we are finding the largest 3 digit perfect square in base 6 which is 524 only and x is calculated on 524 here.

Let me know if it's still unclear -> we will take this up from fresh in that case.

[twinkle2]

b ranges from 23 to 524 => x = 502.

how dis range is bng calculatd dis part is not clear.plz explain..
@arav
m findng it little difficult 2 understand alldes concepts. can u suggest me sthng or frm wer shal i brush up my concepts(tricky 1s)
but yes questns r superb

[bankebihari]

Originally Posted by twinkle2

b ranges from 23 to 524 => x = 502.

how dis range is bng calculatd dis part is not clear.plz explain..
@arav
m findng it little difficult 2 understand alldes concepts. can u suggest me sthng or frm wer shal i brush up my concepts(tricky 1s)
but yes questns r superb

According to question:Let x be the number of base systems in which the largest 3 digit perfect square in base 6 can be represented as a 2 digit number....
now the smallest base can be 23 (23*22+1..and obviously largest such base can be 524..now number of such base system will be equal to the number of digits between 23 and 524 including these..hence 502..am i clear??...

[rajeev_hts]

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Quantitative Question # 015
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How many (a, b, c) satisfy log(2ab) = loga*logb, log(bc) = logb*logc, log(2ac) = loga*logc ?

(1) none (2) 1 (3) 2 (4) 4 (5) none of these

-------------------------------------------------------------------

[rajeev_hts]

Ans(2)

a=1/2, b=1, c=1

[pavanpadekal]

Put b=ka and c=pb
log(2ab)=log a*log b
log(2ka^2)=log a*log(ka)
log2+log k+2loga=log a*log k+log a^2
(log a)^2+log a(log k-2)-log k -log 2=0
log a =[-(log k -2)+-sqrt((log k -2)^2+4(log k+log2))]/2

Now for this to give values for log a from 0 to infinity
((log k -2)^2+4(log k+log2)>=0
((log k -2)^2>=-4(log k+log2)
(log k)^2+4-4log k>=-4log k-4log 2
(log k)^2>=-4(1+log 2)
(log k)^2>=-4(1.693)
log k>=sqrt(-6.772)



log(bc) = log b*log c
log(p*b^2)=log b * log p+ (log b)^2
log p+ 2 log b=log b * log p +(log b)^2
(log b)^2 +log b(log p-2)-log p=0
log b = [-(log p -2)+-sqrt[(log p-2)^2+4log p]/2
(log p-2)^2+4log p>=0
(log p) ^2+4>=0
(log p )^2>=-4
log p>=sqrt(-4)

[sabsebadapaagal]

Originally Posted by rajeev_hts

------------------------------------------------------
Quantitative Question # 015
------------------------------------------------------


How many (a, b, c) satisfy log(2ab) = loga*logb, log(bc) = logb*logc, log(2ac) = loga*logc ?

(1) none (2) 1 (3) 2 (4) 4 (5) none of these

-------------------------------------------------------------------

option 1

condition 1 and 3 => b=c

on plugging this in condition2 we get two values of b
b=1, 10^2

on solving these for a .... we get two values for a ie 1 and 10^4...


on verifying these in above eq ... we find that both satisfy the above conditions.... thus option 3

[bankebihari]

Originally Posted by rajeev_hts

------------------------------------------------------
Quantitative Question # 015
------------------------------------------------------


How many (a, b, c) satisfy log(2ab) = loga*logb, log(bc) = logb*logc, log(2ac) = loga*logc ?

(1) none (2) 1 (3) 2 (4) 4 (5) none of these

-------------------------------------------------------------------

log2+loga+logb=loga*logb....(1)
logb+logc=logb*logc............(2)
log2+loga+logc=loga*logc.....(3)
now (1)-(3) will give either loga=1 or(and) log(b/c)=0...now loga=1 doesnt satisfy (1)...so only one is log(b/c)=0..so b=c
putting it in (2)...b=c=100 or logb=0..hence b=c=1
put b=100 in (1)..
hence a=2+log2=200,b=c=100
again put logb=0..it will give a=1/2

so two set of solution is possible...(200,100,100) and (1/2,1,1)..

answer is (3)

[sabsebadapaagal]

Originally Posted by pavanpadekal

Put b=ka and c=pb
log(2ab)=log a*log b
log(2ka^2)=log a*log(ka)
log2+log k+2loga=log a*log k+log a^2
(log a)^2+log a(log k-2)-log k -log 2=0
log a =[-(log k -2)+-sqrt((log k -2)^2+4(log k+log2))]/2

Now for this to give values for log a from 0 to infinity
((log k -2)^2+4(log k+log2)>=0
((log k -2)^2>=-4(log k+log2)
(log k)^2+4-4log k>=-4log k-4log 2
(log k)^2>=-4(1+log 2)
(log k)^2>=-4(1.693)
log k>=sqrt(-6.772)



log(bc) = log b*log c
log(p*b^2)=log b * log p+ (log b)^2
log p+ 2 log b=log b * log p +(log b)^2
(log b)^2 +log b(log p-2)-log p=0
log b = [-(log p -2)+-sqrt[(log p-2)^2+4log p]/2
(log p-2)^2+4log p>=0
(log p) ^2+4>=0
(log p )^2>=-4
log p>=sqrt(-4)

I have highlighted the portion where you have made mistake ....

it should a * loga

[sabsebadapaagal]

Originally Posted by sabsebadapaagal

I have highlighted the portion where you have made mistake ....

it should a * loga

sry I did mistake