SOLUTIONS
Exercise 1.1
Ans 1) Since 41 is prime and (5,41)=1 , by Fermat’s Theorem
, Hence 52039=5 40.50+39=(540)50.539=150.539(mod41)=539(mod41)
, , , , ,
Ans 2) We can write
,
= -1000(mod(n+100)
Ie. n+10 should divide -900. The largest such N is 900-10=890. As n+10 cnnot be greater than and the greatest devisor of
So the largest +ve integer n, such that n3+100 is divisible by n+10 is 890
Ans 3) 25=32, 255=(25)11
. This implies
Hence it is divisible by 11.
Ans 4)We shall group the terms as follows
(11997+19961997)+ (21997+19951997)+ (31997+19941997)+…..+ (9981997+9991997)
AS a 2n+1 + b 2n+1 is divisible by a+b, Value in Each bracket is divisible by 1997, Hence it is divisible by 1997
Ans5.(a) There are two possibilities-
(i) If digits can be repeated-
Largest number of six digits = 999999
Smallest number of six digits=100000 Difference=899999
(ii) If digits are not repeated-
Largest number of six digits = 987654
Smallest number of six digits=102345 Difference=885305
Ans (b) Let numbers be x. x+1, x+2,x+3,x+4, x+5,x+6,x+7,x+8
Average=81 =>9x+36/9=81 => x+4=81 =>x=77
Hence Largest number =85
Ans (c) 55% of x=240 => x=240*100/55
77%of x=(77*240*100)/55*100=336
Ans1(d) Flowers in basket become double after every minute.
Therefore, Basket was half full 1 minute before it becomes full.
i.e. in 9 minutes.
Ans 6). Let, Digit at Hundred’s place=x , Digit at Ten’s place=y ,Then Digit at one’s place=2y. Also sum of digits=7
=> x+y+2y=7 => x=7-3y , Number=100(7-3y)+10y+2y=700-288y
On reversing digits new Number=100(2y)+10y+(7-3y)=207y+7
Since on adding 297 digits are reversed .
Therefore, 700-288y+297=207y+7 , => 495y=990 ,y=2
Hence, Number=124
7(a). Let, The Number be n, when divided by 1995 leaves remainder 75.
=> n=1995xq+75
=> n=57x35q+57+18 , => n=57(35q+1) +18.
Hence, remainder will be 18 when the number is divided by 57.
Ans7(b). 3 5 9 7
7 5 3
------
1 0 7 9 1
1 7 9 8 5
2 5 1 7 9
------
2 7 0 8 5 4 1
------
{Solve and justify answers yourself, Explanation required in Examination}
Ans 8):-
314+313-12
=313 (3+1) -12 =3.4(312-1)
=3.4(36-1)(36+1)
=3.4.(32-1)(34+32+1)(32+1)(34-32+1) =3.4.8.91.10.73 =26.3.5.7.13.73
Largest prime factor of 314+313-12 =73
Ans) 9). a) Required number will be 1 more than greatest four digit multiple of 2,3,4,5,6,7 .
LCM of 2,3,4,5,6,7 =420
10000=420x23+340
Greatest four digit multiple of 2,3,4,5,6,7 =420x23=9660
Required number=9660+1=9661
Ans 9b) There are 9 numbers between 10 and 99 which remain prime when the order of their digits is reversed.
These are- 11,13,17,31,37,71,73, 79 and 97.
Ans 9c) 234,23456,2345678 are even.
2345 is divisible by 5 and 234567 is divisible by 3.
Hence, 23456789 is prime
Ans. 10) Let, Number=10x+y If decimal is placed- (x+y)/10=1/4(x+y)
y=5x Only possible value for x is 1 Therfore, Number=15
Ans 11) Let number be x
X=56a+48=72b+64=84c+72=96d+88
=56(a+1)-8=72(b+1)-8=84(c+1)-8=96(d+1)-8
Number must be 8 less than a multiple of 56, 72, 84, 96
L.C.M of 56,72,84 and 96 = 2016
Greatest number of 5 digits =99999
=2016X49+1215
Largest multiple of 5 digits = 99999-1215= 98784
Required number = 98784 -8= 98776
Ans 12) 3111-1714=1711((31/17)11-173 )
=1711((31/17)11-4913 )
But 1<31/17 <2
=>(31/17)11<211
=>(31/17)11<2048
=>(31/17)11<4913
=>(31/17)11 - 4913< 0
3111-1714 <0
1714 Greater Than 3111
Ans 13)
199+299+399 +499+599 =(199+499)+(299+399)+599 ,each is divisible by 5
{Since x n + y n is divisible by x + y when n is odd}
199+299+399 +499+599is exactly divisible by 5
Ans 14) Number of perfect cubes between 1 and 1000001, which are exactly divisible by 7 must be cubes of numbers between 1 and 100 that are exactly divisible by 7. Therefore, requied number of such cubes = 14
Ans 15) Number of numbers divisible neither by 5 nor by 7
= 50 -10 – 7 +1=34
Numbers having 5 or/and 7 as digit in above numbers are 17 , 27, 37 and 47
Hence, Required number of numbers = 34 - 4=30
Ans 16) Let Number be 10x+y
(10x+y)2 = 4.(10y+x)
Number is even and 10y+x is a perfect squre .
Possible values=25,49,64 and 81
Square root of 4(10y+x) may be 10,14,16 ,18
10 ,14 and 16 does not satisfy other conditions
Required number is 18
Ans 17) 22000 .52004
= 54.22000.52000 = 625. 102000
Therefore , Sum of digits =6+2+5=13
Ans 18) 25555 =(25)1111 =321111
33333 =(33)1111 =271111
62222 = (62)1111=361111
Since exponents are equal therefore bases will decide the order of numbers Hence, Ascending order: 33333, 25555, 62222
Ans 19) 99= (93)3 =7293
Cubes of all factors of 729 will divide 99
Factors are= 1, 33, 93, 273, 813, 2433 , 7293
= 1,33,36,39, 312, 315, 318
Ans 20) √143 =11.958 app
100(12-√143) =100(12- 11.958)
=100x.042
= 4.2
Nearest integer =4
Ans 21) Here (123456)2 +123456 +123457
= (123456)2 +123456 +123456 +1
= (123456)2 +2*123456 *1+12
= (123456+1)2
= 1234572
Required number is 123457
Ans 22) Following cases are possible:
- 1 used thrice and 2 once
- 1 used twice and 2 twice
- 1 used once and 2 thrice
Number of four digit numbers in i and iii case
= 4 each ( one different digit can be placed at any of the four places)
Number of four digit numbers in ii case
= (4*3*2*1)/2*2
= 6 ( 4 digits can be arranged in 24 ways(Nr) but 1 and 2 occur twice so actual number will be half for each)
Required number of numbers=14
Ans 23) LCM of 2,3,4,5,6,and 7=420
Largest four digit number which is a multiple of 420= 9999-339
=9660
Therefore required number = 9660-1 =9659
(Since required number is 1 less than the exact multiple)
Ans 24) Last two digits is remainder when number is divided by 100
(2003)2 =32 (mod 100 ) =9 ( mod 100 )
(2003)4 = 92 (mod 100 ) = -19 (mod 100 )
(2003)8 =(-19)2 (mod 100 ) =61 (mod 100 )
(2003)16 = 612 (mod 100 ) = 21 (mod 100 )
(2003)32 = 212 (mod 100 ) = 41 (mod 100 )
(2003)40 =(2003)32 .(2003)8 = 41.61 (mod 100 )
=1(mod100)
(2003)2000 =(200340)50 =150 (mod100) =1(mod100)
(2003)2003 =20032000.20032.20031(mod100) =1.9.3(mod100)
=27(mod100)
Last two digits of 20032003 =27
Ans 25) Perfect cubes divisible by 9 will be cubes of multiples of 3.
Since, 1<x3 <1000009
ie. 1<x <101
Also x is a multiple of 3
But, 101=3x33 + 2
Between1 and101 there are 33multiples of 3
Required number of perfect cubes =33
Ans 26)
No. of multiples of 3 = [300 /3 ]=100
No. of multiples of 5 =[300 /3 ] = 60
No. of multiples of 3 and 5 both =[300 /15 ] = 20
No. of multiples of 10 = [300 /10 ]= 30
N0 of multiples of 15 =[300 /15 ] = 20
No. of multiples of 10nd 15both =[300 /30 ] =10
Therefore, Required number of numbers
= (100 +60-20) - (30+20-10)
= 140-40 = 100
Here [ ] denotes greatest integer less than or equal to x .
Ans 27) 24 can be written as a product of three numerals as
1x 3 x 8 , 1x 6 x4 , 2 x4 x3 , 2 x6x 2. For three different numerals there are 6 arrangements of each possible product and for fourth product having 2 two’s number of arrangements will be 3
No of three digit numbers having product of their digits 24 is 21.
They are 138, 183, 318, 381, 813, 831, 164, 146, 461, 416, 614, 641, 243, 234, 342, 324, 432, 423, 262, 226, 622
Ans 28 a)
22005.52000 = 25.22000.52000 = 32. 102000
Number of digits =2 +2000
=2002
Ans 28 b) 22005=22000.25
25=6(mod 13)
210= (25)2 =62(mod13) =10(mod13)
220= (210)2 =102(mod13)= 9(mod13)
240= (220)2 =92(mod13) =3(mod13)
2200= (240)5 =35(mod13)= 9(mod13)
2400=(2200)2 =92(mod13)= 3(mod13)
22000=(2400)5 =35(mod13) =9(mod13)
22005=22000.25 =6.9 (mod13)= 2(mod13)
Remainder is 2 when 22005 is divided by 13.
Ans 29) 248-1=(26-1)(26+1)(212+1)(224+1)
212+1 and 224+1 are greater than 70 . Therefore, Numbers between 60 and 70 are 26-1 and 26+1
i.e. 63 and 65
Ans 30) Let, X= 7q1+6 = 7(q1+1)-1
X= 11q2+10 = 11(q2+1)-1
X= 13q3+12 = 13(q3+1)-1
Hence number is 1 less than common multiple of 7,11 and 13
LCM of 7,11 and 13=1001
Hence, X=1001q-1 =1001(q-1)+1000
Therefore when X is divided by 1001 will leave remainder 1000.
Ans: 31) Let, The Number = 3x + 1 = 5y + 3
= 7z + 5 = 9u + 7
= 3(x + 1) – 2 = 5(y+1) – 2=7(z+1) – 2=9(u +1)– 2
i.e. Number is 2 less than common multiple of 3,5,7 and 9.
L.C.M. of 3,5,7 and 9 = 315
Greatest no. of 4 digits = 9999 = 315×31+ 234.
Greatest number of 4 digits which is a multiple of 315 = 9999-234=9765
Therefore, required number = 9765-2= 9763
Ans:32)
No. of digits used in 1 digit number = 9×1 = 9
No.of digits used in 2 digit number = 90×2 = 180
No. of digits used in 3 digit number = 900×3 = 2700
No.digits used till three digit numbers = 9 + 180 + 2700 = 2889
Remaining digits used for 4 digit numbers = 3189 – 2889 = 300
Therefore, number of 4 digit numbers = 300/4 = 75
Number of pages = 1074
Ans: 33)
312 + 212 - 2.66
= (36)2 + (26)2 – 2.36.26
= (36 – 26)2 = {(33 – 23) (33 + 23)}2
= {(3–2)(32+3.2 +22).(3+2)(32–3.2+ 22) = {19.5.7}2
Therefore, Largest Prime Factor = 19
Ans 34)
S= 12-22+32-42+………-982+992
=(12-22 )+(32-42)+…+(972-982)+(992-1002) +1002
= ( -3-7-11…………….-199) +10000
{ n2-(n+1)2 =-(2n+1) }
= -50/2[ 2*3+49*4] +10000
= 4950
Ans 35) Prime factors of 15 are 3 ,5.
Therefore any multiple of 15 must be divisible by 3 and 5.
As the required no has to be divisible by 5, it should end in zero (the option 5 is not applicable here)
Also, the given no must be divisible by 3.
Therefore if you put one 8 or two eights or one 8 and zero before zero
i.e. 80 or 880or800or8080 are not divisible by 3.
Also, we want the smallest multiple of 15
Therefore the only possibility is 8880. The required no is 8880.
Ans 36) According to the given condition,
5814 = ax + r,
5430 = bx + r, 5958 = cx +r
This implies the difference of any of the above 3 numbers is divisible by x.
5814 – 5430 = 384, 5958 – 5430 = 528,
5958 – 5814 = 144.
The required number is H.C.F of 384, 528, 144.
The H.C.F is 48. The required number here is 48.
Ans 37) 999 x abc = def132
LHS = (1000 – 1) abc
= abc000-abc
10 – c =2 c = 8
9 – b = 3 b = 6
9 – a = 1 a = 8
c – 1 = f f = 7
d=a=8 e=b=6
Ans 38) Let,N=.d25d25…
Solving we get N = d25/999 , But d25/999 = n/810
Now, 925 = 37 X 25 , n/30 = 37 X 25/37
n = 25 X 30 , n = 750
Ans 39):
32002=(34)500 X 32
As 34= 81 = (Unit digit 1) X 9
= unit digit of 32002 is 9
Unit place digit of (32002-1) = 8
Unit place digit of (32002+1 )= 0
5 & 2 are the factors of their LCM
Factor of LCM must be 2X5 = 10
If 10 is factor of LCM, then it’s unit digit will be 0
Ans40): fn(x) = 1/ 1-f n-1(x)
f1(x) = f0(f0(x)) ie.f1(x) = (x-1)/x
f2(x)=x and f3(x) = 1/(1-x) ie. f3(x) = f0(x)
Similarly
fo(x) = f3(x) = f6(x) = ……f2007(x) = 1/(1-x)
f2008(x) = 1/(1- f2007(x)) , f2008(x) = 1/1-(1/(1-x))
= (x – 1)/x
f2009(x) = 1/1-((x-1)/x) = x
f2009 (2009) = 2009
NOTE:
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