Numbers/LCM/220815

Number N which when divided by 11, 9 & 8 gives 5, 6 and 7 as remainder respectively ?

A) find the small possible such number
B) find the largest 4 digit such number

Soln : 11 Q1 + 5 = 9.Q2 + 6 = 8.Q3 + 7
            60 + 55 Q4 = 8 Q3 + 7
         
           Smallest 159 , keep adding lcm of 11,8,9 for subsequent numbers

Number System : Remainders 1501

N2 leaves a remainder of 1 when divided by 24. What are the possible remainders we can get if we divide N by 12?

1, 5, 7 and 11
1 and 5
5, 9, and 11
1 and 11

DETAILED SOLUTION

This again is a question that we need to solve by trial and error. Clearly, N is an odd number. So, the remainder when we divide N by 24 has to be odd.

If the remainder when we divide N by 24 = 1, then N2 also has a remainder of 1. we can also see that if the remainder when we divide N by 24 is -1, then N2 a remainder of 1.

When remainder when we divide N by 24 is ±3, then N2 has a remainder of 9.
When remainder when we divide N by 24 is ±5, then N2 has a remainder of 1.
When remainder when we divide N by 24 is ±7, then N2 has a remainder of 1.
When remainder when we divide N by 24 is ±9, then N2 has a remainder of 9.
When remainder when we divide N by 24 is ±11, then N2 has a remainder of 1.

So, the remainder when we divide N by 24 could be ±1, ±5, ±7 or ±11.

Or, the possible remainders when we divide N by 24 are 1, 5, 7, 11, 13, 17, 19, 23.

Or, the possible remainders when we divide N by 12 are 1, 5, 7, 11.

Correct Answer: 1, 5, 7, 11

Number: 103 : Series 28382th Term 12345678910101213.....

find the 28383rd term of series: 1234567891011121314....

First observe that there are 9 1-digit numbers which take 9*1 = 9 digits, 90 2-digit numbers which take 90*2 = 180digits, 900 3-digit numbers which take 900*3 = 2700 digits, 9000 4-digit numbers which take 9000*4 = 36000 digits.
So it can be concluded easily that our required digit will be some digit in a 4-digit number. Now we need to find out that which digit of what 4-digit number, right?
That can be easily done if I make all numbers 4-digit ones by inserting some 0's before them. Like insert 3 0's before every 1-digit number (i.e. 3*9 = 27 0's), 2 0's before every 2-digit number (i.e. 2*90 = 180 0's) and 1 zero before every 3-digit number (i.e. 1*900 = 900 0's) to meke every number a 4-digit one.

Now we have inserted 27 + 180 + 900 = 1107 zeros in all so we need to find out 28383 + 1107 = 29490th digit in the following sequence: 0001 0002 0003 0004 0005 0006 0007 0008 0009 0010 0011 0012 .....

Which can be easily found as 2nd digit of 7373 i.e. 3 (because 29490 = 4*7372 + 2).

1308: Number System - Remainder - CAT 2000

The integers 34041 and 32506, when divided by a three-digit integer n, leave the same remainder.
What is the value of n?

a. 289          b. 367         c. 453     d. 307
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Let r be the remainder. Then 34041 – r and 32506 – r are perfectly divisible by n.
Hence, their difference should also be divisible by the same.
(34041 – r) – (32506 – r) = 1535, which is divisible by only 307.
Answer : D