official quant thread for cat08

[implex]

Originally Posted by themystique

got two numbers- 441 and 881 , hence 1332. right?
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everyone is getting it wrong. i have posted a beautiful solution of this in AoPs.
Art of Problem Solving Forum
100a+10b+c=49(a+b+c)
a=(13b+16c)/17=b+c-(4b+c)/17

we need to insure that a=0...9 and
so 4b+c=17
gives (b,c)=(4,1),(3,5),(8,2)
so numbers are 441,882,735

this question does not require guessing as you may get the notion.

[implex]

Originally Posted by naga25french

i included 147 and 294 as i obtained single digit sum out of those number..

is that wrong???

why didn't u take single digit sums of the other 3 numbers??

[Varun Khullar]

Originally Posted by naga25french

anwer should be 2499..

the numbers are 147,294,441,735,882

147 .. 294 don't qualify..

[naga25french]

Originally Posted by implex

why didn't u take single digit sums of the other 3 numbers??

well other numbers are two digit numbers... i dunno whats wrong in taking the single digit sum... anyways good solution ..


for ur kind info, i didnt guess the numbers..jus used property of 9 .. and found

49 * 9 and 49 * 18... then i checked for 3,6, 12 and 15 as well.. this is how i arrived at the result and not mere guessing..

[implex]

Originally Posted by naga25french

well other numbers are two digit numbers... i dunno whats wrong in taking the single digit sum...

what i desire is consistency
either u take single digit sum of all or double digit sum of all!
You can't do both !

[Varun Khullar]

Originally Posted by naga25french

well other numbers are two digit numbers... i dunno whats wrong in taking the single digit sum... anyways good solution ..


for ur kind info, i didnt guess the numbers..jus used property of 9 .. and found

49 * 9 and 49 * 18... then i checked for 3,6, 12 and 15 as well.. this is how i arrived at the result and not mere guessing..

man nice.. all no have to be factor of 49...49 k table.. yaad kar lo.. and its done..

[linksuresh]

@implex -

do u have the solutions for the those 3 attachments uv put up?

probset1 doesnt seem to have the ans key or the soln..
guess probset2 has only ans keys...

wud be good if u post the solutions as a pdf too....

[linksuresh]

@implex -

could you chk the ans/solutions provided to the following questions on Quant Problem set 3 -

prob nos -

169 - feel ans shd be 7
178 - feel ans shd be 8

correct problem 183-184 to -

expr shd read - [(a/b) + (14b/9a)]
else ans shd be -
183 - option e) many vals
184 - e) infinity

193 - guess ans shd be option e) ull have 2 choices - 15*2 or 5*6.. so...

195 - feel ans shd be option b)
since its given a,b,c are Natural nos.. we can find a set satisfying.. and get sum (a+b+c). no need to know whether they're in AP or not...

198 - feel ans shd be e)
since its not mentioned if p & q are ints or real...

199 - feel ans shd be e)
since its not mentioned if sisters' ages are ints or reals...

didn solve all.. but had doubts in the above...
u may post the q and wait for other s to solve 'em.. else look up em urself

thanks bro ! enjoy

[saurabh1310]

hii puys
i found this question on page 51 of this thread...
the answer posted thr is 3 digits (199). my doubt is shouldn't the answer be 2 digits (89) as it is also satisfying the criteria.
plzz correct me if i'm wrong......
ques is:
A function y=f(n) is defined, for all natural numbers, as the sum of the digits of n.
if k is a natural number such that f(f(f(f(k))))=1 and k>f(f(k)).f(f(f(k)))>1 what is the least number of digits that k can have?

[madadi]

Originally Posted by saurabh1310

hii puys
i found this question on page 51 of this thread...
the answer posted thr is 3 digits (199). my doubt is shouldn't the answer be 2 digits (89) as it is also satisfying the criteria.
plzz correct me if i'm wrong......
ques is:
A function y=f(n) is defined, for all natural numbers, as the sum of the digits of n.
if k is a natural number such that f(f(f(f(k))))=1 and k>f(f(k)).f(f(f(k)))>1 what is the least number of digits that k can have?

Hi Saurabh...
Your answer, k=89 is violating the first condition itself...
The ques says, f(f(f(f(k)))) = 1
If you take k as 89, then,
f(f(f(8+9)))= f(f(f(17)))= f(f(1+7))= f(f( 8 ))=
f( 8 ) = 8 ...
This, you can clearly see, is not equal to 1....
Probably, you din't notice it... Anyways, I guess, it's clear now...