Official Quant thread for CAT 2013

MBA CAT 2024
13783
@bs0409 said:
Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number should satisfy the following requirements:a. The number should be divisible by 9.b. If the most right digit is removed, the remaining number should be divisible by 8.c. If then again the most right digit is removed, the remaining number should be divisible by 7.d. etc. until the last remaining number of one digit which should be divisible by 1.
i have a very long approach for this one....

let the number be abcdefghi
e has to be 5
so N becomes abcd5fghi as of now
we can also deduce that b,d,f and h are even while a,c,e,g and i are odd
also a+b+c = 3p form
and a+b+c+d+5+f = 3k form (since it is div by 6)
=> 3p+d+5+f = 3k
=> d+5+f = 3k
so (d,f) can be (2,8) or (6,4)

as of now our number is abc258ghi or abc654ghi

h is even and g is odd (for a number to be multiple of 8, we chk the last 2 digits)
gh can be 16,32,72 and 96

so we have 4 possible numbers now - abc25816i or abc25896i or abc65432i or abc65472i

now we can say that b can be either 4 or 8 only

and a+b+c = 3k
possibilities are - 147,183,189,381,387,741,789,981 and 987

apply all the possibilities and divide the number with 7 (because I don't know the div rule for that, if any)

the number comes out to be 381654729...


Phew...what a scorcher....exam mein aayega toh kya badhiya normalize karega score ko agar sahi kar diya toh
13784
@bs0409 said:
Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number should satisfy the following requirements:a. The number should be divisible by 9.b. If the most right digit is removed, the remaining number should be divisible by 8.c. If then again the most right digit is removed, the remaining number should be divisible by 7.d. etc. until the last remaining number of one digit which should be divisible by 1.
is it 381654729 ?
13785
@Logrhythm can u explain this a bit
13786
@nirzone said:
@Logrhythm can u explain this a bit
which part do you need an explanation in? kindly go through my post again, i guess i have explained it properly but if you still need an explanation in any part....pls let me know..
13787
@Logrhythm for starters on the remainders obtained by dividing a perfect cube by 13 and also on your deduction of 29..sorry but i am not able to comprehend this one
13788
@nirzone said:
@Logrhythm for starters on the remainders obtained by dividing a perfect cube by 13 and also on your deduction of 29..sorry but i am not able to comprehend this one
a number can be of the form (13k + n)^3 where k is a +ve integer and n is a natural number..
let's take k = 1
(13 + n)^3 is what we are left with
now put n from 0 to 12 and observe the remainders, you would get 5 different remainders...
we took 0 to 12 only coz after that the cycle would repeat if we take n = 13 the number becomes (13 + 13)^3 = (13*2)^3 so this is again 13k form..

and now 5 diff remainders are there and 143 such numbers in the set
till 140 we would have 140/5 = 28 numbers...
and in the rest 3 we will have 1 more..
so 29 in total..

i hope this is clear now
13789
@bs0409 approach please
13790
@bs0409 said:
Ki baba....!!! Kemon cholche tomar??So, u applied to any univ or waiting for CAT'13?I am assuming your company joining letter will come soon enough.Find the total number of values of n for which n^2 + 24n + 21 is a perfect square.Where n is natural number.
2 values..
13791
@veertamizhan said:
"If a student weighing 70 kgs joins a group of n students, the average of the group increases by 1 kgs. If the new student weighed 55 kgs, the average of the group would have declined by 2 kgs. Find n"3, 4, 5, 6, 7
4
13792

"The average of n consecutive natural numbers is 17.5. What is the largest value n can assume?"


options

infinite, 34, 35, 36, 18

edit: deleted foolish attempt to find the solution
13793
@bs0409 said:
Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number should satisfy the following requirements:a. The number should be divisible by 9.b. If the most right digit is removed, the remaining number should be divisible by 8.c. If then again the most right digit is removed, the remaining number should be divisible by 7.d. etc. until the last remaining number of one digit which should be divisible by 1.
381654729...:D atlast!!!! after 10minutes!
13794
@veertamizhan said:
"The average of n consecutive natural numbers is 17.5. What is the largest value n can assume?"options infinite, 34, 35, 36, 18Here is what I didlet a be the first of the natural number.then next will be a+1 and so on..so the series is a+(a+1)+(a+2)+(a+3)+......a+(n-1) a+(n-1) is the nth term so the equations is a+(a+1)+(a+2)+...a+(n-1)___________________________ = 17.5 n clubbing n's and digits together na+[1+2+3+4......(n-1)]___________________________ = 17.5 n calculating the sum of the digitssum= n-1/2*[1+(n-1)]n(n-1)/2 = sum of digits na+ n(n-1)/2___________________________ = 17.5 n n cancels outleaving me with a+(n-1)/2 = 17.52a+(n-1)=35putting the highest value in the question = 362a+35=352a=0a=0 (possible) infinite does not seem rightso my answer is 36, which is right, but I am pretty sure this isn't the right way to go about this question.Help me folks. :-/
is it 34...!..pls check the OA
13795
@sails oh shit. Yes.
13796
@veertamizhan said:
@sails oh shit. Yes.
😁
13797
@sails how did I do? what is the wrong with my approach? how did you solve this?
13798
@veertamizhan said:
@sails how did I do? what is the wrong with my approach? how did you solve this?
sappa matter macha..:D
See first n numbers summation is n(n+1)/2
average is sum/number of numbers...
so average is n(n+1)/2n = 17.5(given)
so n+1=35
n=34...:)
13799
@veertamizhan said:
@sails how did I do? what is the wrong with my approach? how did you solve this?
i didnt see ur approach sry..:P
13800
@veertamizhan said:
"The average of n consecutive natural numbers is 17.5. What is the largest value n can assume?"options infinite, 34, 35, 36, 18

Let's look at it differently. Average 17.5 could come from 17, 18, or 16, 17, 18, 19 or 15 - 20 or 14-21 and so on. At the lower end we can go only till 1 (16 steps down from our initial starting point of 17) so at the upper end we can go max 18 + 16 = 34.

Alternatively, the average of an AP is (1st + last term)/2. Since average is fixed, 17.5, and first term can be minimum 1, last term can be max (17.5 x 2) - 1 = 34.

regards
scrabbler
13801
@veertamizhan 34
13802
@sails damn. I take into consideration the fact that these are natural number which begin with 1. I am a god damn idiot.

thanks anna (i am not tamil)