yup same getting 3 as the ans!! :)book says 9.
Got 3 as the answer or is it right??
its 3 not 9.
hi can some one solve these from number systems(making annoying mistakes)!
1.find the 28383rd term od the series 123456789101112......
2.ifu form a subset of integers chosen from between 1 to 3000,such that no 2 integers add up to a multiple of 9, what can be the max no f elements in the subset?
Anyone for the2nd one?
hereIgo SaysAnyone for the2nd one?
if you form a subset of integers chosen from between 1 to 3000,such that no 2 integers add up to a multiple of 9, what can be the max no of elements in the subset?
we can start with 1,2,3,4,9, i.e. x-8,x-7,x-6,x-5 and 9.so,we get 1+333*4+3=1336 terms
if we start from 2,3,4,8,9,11,12,13,17, and so on.the number of terms will be lesser than the first case.
now,
3,4,7,8,9 i.e. x-6,x-5,x-2,x-1 again,there will be fewer terms.
so,i think,it should be 1336.
what does OA say??
the book gives the ans as 1333
can u explain ur method..din getthe 333 andy added another 3 to it
hereIgo Sayscan u explain ur method..din getthe 333 andy added another 3 to it
the if you form a subset of integers chosen from between 1 to 3000,such that no 2 integers add up to a multiple of 9, what can be the max no of elements in the subset?
2,3,4,8,9,11,12,13,17,...2990,2991,2992,2996,2999
see,it follows a pattern
(x-7),(x-6),(x-5),(x-1) here,x is 9 or multiple of 9.
from 9*0 to 9*332=>333*4=1332
9 comes at the start so,1 more
last number is 2999
1334 terms total.dunno where 1 term goes 😐
Please help me solve this question
There is a nine digit number, each digit a unique digit between 1-9 (including 1 and 9).
1. This number is divisible by 9
2. If I remove units digit my number is divisible by 8
3. If I remove 10's digit my number is divisible by 7 and it goes on till we are left with one digit which is divisible by 1.
What is the number?
In my Edition of this book this question is on page 7.17 (Logical reasoning section), question number 46 - 51.
Please help me in understanding the logic to solve this question.
Please help me solve this question
There is a nine digit number, each digit a unique digit between 1-9 (including 1 and 9).
1. This number is divisible by 9
2. If I remove units digit my number is divisible by 8
3. If I remove 10's digit my number is divisible by 7 and it goes on till we are left with one digit which is divisible by 1.
What is the number?
In my Edition of this book this question is on page 7.17 (Logical reasoning section), question number 46 - 51.
Please help me in understanding the logic to solve this question.
Hey i thought on this problem
Frst thing if you got options on this then no need to solve , ok
Else this problem is getting bigger
My approach is this one
__ __ __ __ __ __ __ __ __
9 places are there
5th place is fixed as 5 will come there ,(problem says when u remove 6 th place digit it should be divisible frm 5)
apart from that 2nd 4th 6th 8th place will be filled by even numbers which are 2, 4,6,8
now if we take 4th and 6th place ,as we know 5+ 4th digit+ 6th digit sum should be divisible by 3(frst 3 digit sum will be divisible by 3 and if we want the 6 digits figure should be divisible then these 3 digit sum also should get divide by 3 )
the interesting thing is only these 654 and 258 are only pair who will satisfy this thing
Now we need to check 10-12 combinations which was getting weary
so if neone else can help out in ne other thought line
......
the if you form a subset of integers chosen from between 1 to 3000,such that no 2 integers add up to a multiple of 9, what can be the max no of elements in the subset?
2,3,4,8,9,11,12,13,17,...2990,2991,2992,2996,2999
see,it follows a pattern
(x-7),(x-6),(x-5),(x-1) here,x is 9 or multiple of 9.
from 9*0 to 9*332=>333*4=1332
9 comes at the start so,1 more
last number is 2999
1334 terms total.dunno where 1 term goes :|
finally clear 😃 thnx...yup seems like 1334
Hey i thought on this problem
__ __ __ __ __ __ __ __ __
9 places are there
5th place is fixed as 5 will come there ,(problem says when u remove 6 th place digit it should be divisible frm 5)
Dude how can u b sure dat it wil b 5 in the 5 th place it can b 0 too , which is also divisible by 5.
Isn't it so?Wat do u say..........
Good initiative.
Arun Sharma is a book with very good questions but very bad explanations.
This thread will help us explore the solutions in a better way.
Alos, LOD3 for profit and loss sucks, the problems are simply long with nothing much interesting.
So, it can be skipped ( i wasted one whole day on it grrrrr )
hi all,
yes as a matter of fact, the questions of arun sharma quant sometimes make a standstill in solving the questions. while solving the LOD1 & LOD2 questions even i get stuck at times. just wanted to know, are'nt any solution books available for arun sharma quant????
coz each time while solving a question if i counter a doubt it is not possible to come acroos the internet to post questions and wait for answer....
hi all,
yes as a matter of fact, the questions of arun sharma quant sometimes make a standstill in solving the questions. while solving the LOD1 & LOD2 questions even i get stuck at times. just wanted to know, are'nt any solution books available for arun sharma quant????
coz each time while solving a question if i counter a doubt it is not possible to come acroos the internet to post questions and wait for answer....
Dude i dont know, solution for arun sharma's book available in the market or not
But if u will post question here then definitely u will get solution ASAP
:
NUMBER SYSTEM
95) A number is such that when divided by 4,5,6,7 it leaves the remainder 2,3,4,5 resp. Which is the largest no. below 4000 satisfying this condition.
89) Find the least number by which 30492 must be mltiplied or divided to make it a perfect square
Can some one also explain the concept underlying Q 95 type of questions, I don't seem to be getting it 
!!
Thnks
Dude how can u b sure dat it wil b 5 in the 5 th place it can b 0 too , which is also divisible by 5.
Isn't it so?Wat do u say..........
It can't be zero, because it is asked the digits from 1-9.
NUMBER SYSTEM
95) A number is such that when divided by 4,5,6,7 it leaves the remainder 2,3,4,5 resp. Which is the largest no. below 4000 satisfying this condition.
Can some one also explain the concept underlying Q 95 type of questions, I don't seem to be getting it!!
Thnks
The numbers are having negative remainder equal ,i.e,2
Hence it is of the form LCM of (4,5,6,7)-(-ive remainder)
=420x-2
largest number below 4000 will be 3780-2
=3778
Hope it helps
[89) Find the least number by which 30492 must be mltiplied or divided to make it a perfect square
Prime factors are 2*2*3*3*11*11*7
Hence needs to be multiplied by 7 to get a perfect square
Abhay is rite in solving the question. However at the end you have to add 2 instead of substracting 2.
Answer is 3780 + 2
Hi,
Your second question can be solved as follows
1. Break the number into factors. It will come out to be 2^2 * 3^2 * 11 ^ 2 * 7
2. Now 7 is the number, either you multiply by 7 or divide by 7 both will make the number a oerfect square.
NUMBER SYSTEM
95) A number is such that when divided by 4,5,6,7 it leaves the remainder 2,3,4,5 resp. Which is the largest no. below 4000 satisfying this condition.
89) Find the least number by which 30492 must be mltiplied or divided to make it a perfect square
Can some one also explain the concept underlying Q 95 type of questions, I don't seem to be getting it
Thnks