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 System : General : 109 : ABCDE*4 = EDCBA

ABCDE * 4 = EDCBA. Solve for A,B,C,D, and E where each is a unique integer from 0 to 9.

Solution

It is obvious that A can be no more than 2. If A were 3 then 3BCDE * 4 would be at least 120,000 which is more than five digits. Also A must be an even number because EDCBA is an even number since it is the product of at least one even number (4). We can eliminate A=0 because E would have to be 5 (5*4=0) but BCDE*4 could not hope to reach 50,000. So A must be 2.

Next consider E. E*4 must end in the digit 2. The only numbers that works for are 3 and 8. However with A=2 EDCBA must be at least 80,000. So 8 is the only number that satisifies both conditions.

Next consider B. We already know that 2BCD8*4 is at least 80000 and less than 90000. B can not be more than 2 because then 2BCD8 * 4 would be more than 80000. 2 is already taken so B must be 0 or 1. Lets consider the case that B=0. Then D8 * 4 must end in the digit 02. However there is no D that satisfies this condition. So B must be 1.

Next consider D. D8*4 must end in the digits 12. The only possiblity is D=7 (78*4=312).

Now solve for C:

21C78 * 4 = 87C12.
84312+400C = 87012 + 100C
2700 = 300C
C=2700/300=9.

So ABCDE=21978.

Basics : Number System Remainders 1

When dividend is of the form a ⁿ + b ⁿ or a ⁿ - b ⁿ :

8. What is the remainder when (5555) ²²²² + (2222) ⁵⁵⁵⁵ is divided by 7?

       The remainders when 5555 and 2222 are divided by 7 are 4 and 3 respectively.

       Hence, the problem reduces to finding the remainder when (4) ²²²² + (3) ⁵⁵⁵⁵ is divided by 7.

       Now (4) ²²²² + (3) ⁵⁵⁵⁵ = (4 ² ) ¹¹¹¹ + (3 ⁵ ) ¹¹¹¹ = (16) ¹¹¹¹ + (243) ¹¹¹¹ .

        Now (16) ¹¹¹¹ + (243) ¹¹¹¹ is divisible by 16 + 243 or
        it is divisible by 259, which is a multiple of 7.

       Hence the remainder when (5555) ²²²² + (2222) ⁵⁵⁵⁵ is divided by 7 is zero.

9. 20 ²⁰⁰⁴ + 16 ²⁰⁰⁴ - 3 ²⁰⁰⁴ - 1 is divisible by:    (a) 317 (b) 323 (c) 253 (d) 91

         20 ²⁰⁰⁴ + 16 ²⁰⁰⁴ - 3 ²⁰⁰⁴ - 1 = (20 ²⁰⁰⁴ - 3 ²⁰⁰⁴ ) + (16 ²⁰⁰⁴ - 1 ²⁰⁰⁴ )

         20 ²⁰⁰⁴ - 3 ²⁰⁰⁴ is divisible by 17 (Theorem 3) and
        16 ²⁰⁰⁴ - 1 ²⁰⁰⁴ is divisible by 17 (Theorem 2).
        Hence the complete expression is divisible by 17.

        20 ²⁰⁰⁴ + 16 ²⁰⁰⁴ - 3 ²⁰⁰⁴ - 1 = (20 ²⁰⁰⁴ - 1 ²⁰⁰⁴ ) + (16 ²⁰⁰⁴ - 3 ²⁰⁰⁴ ).

        Now 20 ²⁰⁰⁴ - 1 ²⁰⁰⁴ is divisible by 19 (Theorem 3) and
       16 ²⁰⁰⁴ - 3 ²⁰⁰⁴ is divisible by 19 (Theorem 2).

       Hence the complete expression is also divisible by 19. is divisible by 17 × 19 = 323.

 Fermat's Theorem

12. What is the remainder when n ⁷ - n is divided by 42?

Since 7 is prime, n ⁷ - n is divisible by 7.
 n ⁷ - n = n(n ⁶ - 1) = n (n + 1)(n - 1)(n ⁴ + n ² + 1)

Now (n - 1)(n)(n + 1) is divisible by 3! = 6

Hence n ⁷ - n is divisible by 6 x 7 = 42. Hence the remainder is 0.

 Wilson's Theorem

13. Find the remainder when 16! Is divided by 17.

              16! = (16! + 1) -1 = (16! + 1) + 16 - 17

              Every term except 16 is divisible by 17 in the above expression.
              Hence the remainder = the remainder obtained when 16 is divided by 17 = 16

Numbers : Remainder 105

find the remainder  3^5^7^9 divided by 41.

Soln :
3^4 = 81 = 41(2) - 1. So 3^8 will leave remainder 1 when divided by 41.
That means now we need to find the remainder of power of 3 i.e. 5^7^9 with 8.
Now we know that 5² = 25 = 8(3) + 1 i.e. 57^9 is of the form 52k + 1 and leave remainder 51 = 5 when divided by 8 and is of the form 8k + 5.So finally we need to find the remainder when 38k + 5 is divided by 41 or the remainder when 35 is divided by 41 which is -3 or 38.

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).

Numbers: Remainder 106

[17^36+19^36]/111 what is the reminder?
Yes the remainder will be 2.

111 = 37 * 3

Φ(111)  = LCM[Φ(37) ,Φ(3)]= 36

1736 mod 111 = 1
1936 mod 111 = 1

Hence 1 + 1 = 2

Numbers System : Remainder N/D, N/12D , N/6D

A no. N when divided by divisor D leaves remainder 23. when same no. N is divided by 12D remainder is 104. what will be the remainder when the no is divided by 6D ?

Soln : If I start from second statement that N = a*12D + 104, so this is same N which when divided by D gives a  remainder of 23. i.e. 104 = b*D + 23 or D is a divisor of 104-23 = 81 and also D is greater than 23. So D = 27 or 81.Thus in any case when same N is divided by 6D, remainder will be given by 104 as a*12D is certainly divisible by 6D. So the required remainder is = 104.

D has to be greater than 23 as remainder on division is 23.
So, 6D > 104
Hence, remainder will be 104

Else,
N = Dx + 23
N = 12Dy + 104
Equating them we will get.
D(x - 12y) = 81
So, D has to be a factor of 81 and greater than 104. So, possible values are 27 and 81
So, 6D is 162 or 324 (both greater than 104)
So, remainder will be 104

Numbers : 9 digit number

Find the number of 9 digits which has the following properties: 

1. The number comprising the leftmost two digits is divisible by 2,that comprising the leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. the number formed from the first n digits is divisible by n, 2<=n<=9.
2. Each digit in the number is different i.e. no digits are repeated.
3. The digit 0 does not occur in the number i.e. it is comprised only of the digits 1-9 in some orde





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answer is 381654729 

Numbers : Divisibility Osculation

finding the divisibility of a number by 7,13,17,19.the method is known as osculator method.first we have to know the concept of osculator.the osculator of 7 is 2 ( 7* 3 = 21 = 20+1 here we are adding 1 so we need to consider the osculator as negative osculator )

 similarly for 13 ,13 * 3 = 39 = 40-1 so the osculator is 40 and is one more osculator and the value is 4 and for 17 ,17 * 3 = 51 = 50+1 so again the osculator is negative and it is 5 and for 19 it is one more osculator  19 = 20-1 ,so the osculator is 2 .

if you didn't understand the concept of osculator just remember these

for 7 osculator = 2 , sign = ' - '

for 13 osculator = 4 , sign = ' + '

for 17 osculator = 5 , sign = ' - '

for 19 osculator = 2 , sign = ' + '

lets proceed with an example

55277838 is considered and to check it's divisibility by 7.

5527783 8  : 5527783 - 8 *2 =5527767 ( here we have taken the units digit and multiplied it by osculator 2 and subtracted it from the remaining part as the sign is negative as suggested above we have subtracted )

the next step is the following is repeated for the resulting value in the above step

552776 7   : 552776 - 7 *2 = 552762

55276 2 : 55276 - 2 * 2 = 55272

5527 2 : 5527 - 2 *2 = 5523

552 3 : 552 - 3 *2 = 546

54 6 : 54 - 6*2 = 42 as 42 is divisible by 7 so the number is divisible by 7.

Number System Euler's Theorum

Geometry: Quadrilateral 1011

Q)A circle is inscribed in the quadilateral ABCD. Given that AB = 27cm, BC=38m, DC= 25cm and AD is perpendicular to DC. Find the maximum limit of radius and the area of the circle
a)10cm,226cm^2 b) 14cm,616cm^2 c)14cm,216cm^2 d)28cm,616cm^2
e) none of these

Soln : Since OP=OS (radii) and PD=PS(tangents) and angle D= S=P=90 , hence OPDS is a square , i determined AD by foll. method
AD=AP+PD
= 27-TB+PD (since AP= AT as both are tangents and AT= 27-TB)
= 27-(38-SC)+PD
= 27-(38-SC)+(25-SC) (since PD=SD and SD= 25-SC)
AD = 14cm
and as from fig. AD= AP+PD , PD has to be smaller than 14cm under any condition and PD= OS radius of the circle , therefore the radius of the circle will definitely be less than 14cm but Arun Sharma is saying answer as option(d) radius = 28cm which is not possible .