M16, M17

M16 - 45

M17 - 38 (countr the numbers between 10 and 99 which satisfy the mentioned ppty)

Abi

M 18

Not sure.. but the answer shud be 599

Solution:

Ppty: X mod Y + Z mod Y = (X + Z) mod Y = (X mod Y + Z mod Y) mod Y

(431 + 379 ) mod y  = 810 mod y = 211

Hence divisor = 810-211 = 599

Abi

M 19

Ans :  21 numbers

Abi

M 20

Solution:

LCM(2,3,4,5,6,7) + 1 = 840 + 1 = 841

Abi

sorry...made a mistake

Slight change to the ppty but the answe still holds good

Ppty: X mod Y + Z mod Y = (X mod Y + Z mod Y) mod Y

Abi

M 21

Dont know the shortcut method..

But, the answer is 13

Abi

M 23

Dont know any proper method, but when u have choices this questionc an easily be done by substitution method.

Ans is a = 2 or 3, b = 2 or 3, c = 4

Abi

M 25

102 key strokes

Abi

M 27

C ... it cannot be even

Abi

M 29

Excellent question ......... The number to be multiplied is 9...

Solution:

No of 3's in 100 = 33

No of 3^2 's in 100 = 11

No of 3^3 's in 100 = 3

No of 3^4 in 100 = 1

Now... 3^5 is greater than 100 and hence we can stop with 3^ 4.

This means that 100! already has 48 3's in it.

When 100! / 3^ 50, the 48 3's in 100! gets cancelled out with 48 3's in 3 ^ 50. Hence 9 is left over in the denominator.

Hence for 100! to be exactly divisible by 9, it as to be multiplied to 100!.

Abi

M17 answer is 45

The nos. are 10,20,21,30,31,32,40,41,42,43,

Like this 1+2+3+4+5+6+7+8+9=45.

Answer should be 421

Calculate the lcm carefully.

Answer is incorrect

Since 16+21=37 which is a prime no. hence the expression will be only divided by 37 .

Answer is 192

See for 1to 9 he has to hit 9 strokes

From 10 to 99 he has to hit (90*2)=180 strokes

For 100 he has to hit 3 strokes

Therefore total he has to hit 192 strokes.

Some of solution

M16. The product of a two-digit number by a number consisting of the same digits written in the reverse order is equal to 2430. Find the lower number? 45
Ans The number is (10x+y )(10y+x) = 2430 => No. is 45x54

M17. For each of the numbers: 41, 83, 32, the first digit is greater in value than the second digit.How many 2-digit numbers have this property?
Ans. Two digit no in which the first digit is greater than 2 nd digit is for Unit place 0 = 9
        For 1 in unit place = 8 and for 2 =7 and 3 = 6 and 4 = 5 and 5 = 4 and 6 = 3 and 7 = 2 and 8 =1
        Thus total is 1+2+..+9 = 10x9/2 =45 Ans.

M18. Two numbers when divided by a certain divisor leave remainders of 431 and 379 respectively. When the sum of these two numbers is divided by the same divisor, the remainder is 211. What is the divisor?
Ans. N1 = 431 mod X and N2 = 379 mod X and N1+N2 = 211 mod X
       Here the 431+379 =  810 and 810  = 211 modX or X+211 = 810 minimum mumber i.e. 599 Ans

M19.How many 3-digit numbers exist for which the sum of the digits is six?
Ans. Let the digits be x,y,z and x+y+z = 6 and x varies from 0 to 6 is given as C(6+3-1,3-1) = C(8,2)= 8x7/2=28 ways
       Which includes
       It includes the first digit as 0 also thus z+y =6 the number of ways are C(8-1,1)=C(7,1) = C=7 ways
       i.e.006,060,015,051,033,042,024
       Thus total digits = C(8,2)-C(7,1) = 28 -7 = 21 numbers

M20. Find the smallest number, greater than 1, which has a remainder of 1 when divided by any of 2, 3, 4, 5, 6, or 7
Ans.  LCM of (2,3,4,5,6,7)+1 = 421 Ans

M21. 16^5 + 21^5 is divisible by 33 , 13 , 27 or 31?
Ans.  (16 + 21)(16^4-16^3.21+...21^4) = 37xP which is divisible by 13 thus answer is 13.

M22. How many pairs of natural numbers are there the difference of whose squares is 45?
Ans.  Here x^2-y^2 =45 = (x+y)(x-y) Now factors of 45 = 3^2x5 thus the
        Pairs are (x+y)(x-y) = 9x5 i.e. x = 7 and y =2 Pair is (7,2)
                      (x+y)(x-y)=  15x3 i.e (9,6)
        Only 2 pairs are possible.

M24. A five digit number 3A25B is divisible by 19 and 7. Find A and B?
Ans. The seed of 19 is 2 thus 3+2A+8+40+16B => 51+2A+16B is divisible by 19
        and   3 -2A -6 -5+2B is divisible by 7 which gives 8+2(B-A) is divisible by 7 or 2(B-A)+1=7 or B-A=-3
       Consider A = B+3 then the numbers A,B pairs are (3,0),(4,1),(5,2),(6,3),(7,4),(8,5),(9,6)
        For (3,0)  13+2(A+8B) =19 is divisible by 19
        For (4,1)  13+2(4+8)   =37 which is not divisible by 19
        For (5,2)  13+2(5+16) =55 which is not divisible by 19
        For (6,3)  13+2(6+24) =73 not divisible by 19
        For (7,4)  13+2(7+32) =91 and is not divisible
        For (8,5)  13+2(8+40) = 109 and not divisible
        For (9,6)  13+2(9+48)= 127 and is not divisible thus only solution is A=3 and B=0
      And number is 33250 Ans.

M25. How many keystrokes are needed to type numbers from 1 to 100?
Ans  Number of one digit no are 1,2,..9 = 9
        Number of two digit nos. 10,11,...,99 terms =10+(n-1)1=99 terms = 90 terms
        One digit is 3 digits so toal keys = 9+ 90x2+3=192 keystrokes

M26. How many 3-digit numbers have two digits the same?
Ans. Let the digits are 0,1,2,...,9 The all 3 digit no. possible = 9x10x10 = 900
        The numbers in which the 2 digit are same are 10 ways selection i.e 00,11,22,33,..,99
        Then third digit is out of balance digit 8x10 = 80 ways and each can be arranged in 3!/2!=3 ways i.e. 240 ways and in case of 00 the only 9 no is are there
        Total numbers are 240 +9 =249 Ans.

M27. 14^n + 11 can never be a
a. Prime b. odd c.even
Ans. 14^n is always even thus even + odd = odd thus it can never be even.

M28. The remainder obtained when 43^101 + 23^101 is divided by 66?
Ans.   Now (43+23)P thus remainder is zero.

M29. What is the least number that should be multiplied to 100! to make it perfectly divisible by 3^50
Ans. The exponent of 3 in 100 is [100/3]+[100/9]+[100/27]+[100/81] = 33+11+3+1= 48 Thus the number to be multiplied is 9

M30. How many times will the digit '0' appear between 1 and 10,000?
Ans.  The factors of 10000=5^3x2^3
         Number of 1 digit no = 0
         Number of 2 digit no. = 90 in which the 0 is in 9 times
         Number of 3 digit no. =100 to 999 i.e. 900 terms in which the 0 in unit place 90 and in ten place = 90= 90x2=180 times
         Number of 4 digit no = 1000 to 9999 i.e. 9000 = 900[1+1+1] = 2700
         Number of zeros in 10000 =4
         So total zeros = 9+180+2700+4 =2893 Ans.

s shahid basha I think

s shahid basha I think answer for M17 is 45. 1 2 3 4 5 6 7 8 9 when added up gives 45. 10 20 30 40 50 60 70 80 90 21 31 41 51 61 71 81 91 32 42 52 62 72 82 92 . . . . . . anything wrong in the above steps pls correct it. thank u.

M 21

s shahid basha

M21

   how come 37 x p is divisible by 13 ?  pls explain

thank u

    M22   what about the

M22

  what about the pairs 23 and 22  whose difference of squares is 45 ?

 You forgot the pair (1,45) during factorization.

thank u

Question M21

@ shabash_23

The solution is wrong and so is the question . .

I mean there is a typo among the given options

It should be 37 instead of 27   . .

Hoping it ‘ll be rectified soon.

s shahid

s shahid basha

 M23

         Even if  a is 2 or 3 and  b is 2 or 3  how can  c  be 4 ?

s shahid basha   M24   

s shahid basha

M24

   3A25B

        B x 1 + 5 x 3 + 2 x 2 + 6 x A +3 x 4 = B + 31 +6A   which should be divisible by 7

 by trial and error method we get A=1 and B=5

therefore the number is 31255 which is also divsible by 19.

thank u

M17

Answer to M17 should be

45 and not 38 since (9*10)/2

i don't think its divisible

i don't think its divisible by 13 instead by 37

here's the solution for M22..

Let the two numbers me a and b.

it is given that a²-b²=45
=> (a-b)*(a+b)=45

Now we know that addition and subtraction of two nos should be equal to 45.

Here is breakdown of all the natural numbers whose multiplication results in 45:
->5*9
->15*3
->45*1

For 5*9,
a-b=5 and a+b=9
a comes out to be 7 and b as 2
so we have our first pair as (7,2)

Similarly for 15*3 and 45*1 we will have (9,6) and (23,22) as the pairs.
So in all three natural pairs are possible.
Therefore,answer is 3.