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MATHS OLYMPIAD WORKSHOP
For
KVS TEACHERS
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Ans1. 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
Q1 Find the difference between the largest and the smallest numbers that can be formed with six digits.
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q2.The average of nine consecutive natural numbers is 81. Find the largest of these numbers.
Ans. 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
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q3.What will be 77% of a number whose 55% is 240? ANS. 55% of x=240 => x=240*100/55 77%of x=(77*240*100)/55*100=336
Q4.Flowers are dropped in a basket which become double after every minute. The basket became full in 10 minutes. After how many minutes the basket was half full?
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Ans. Flowers in basket become double after every minute. Therefore, Basket was half full 1 minute before it becomes full i.e. in 9 minutes.
Q5. A number consists of 3 digits whose sum is 7. The digit at the units place is twice the digit at the ten’s place. If 297 is added to the number, the digits of the number are reversed. Find the number.
6/27/2015
PREPARED BY:- M. GOVINDU, Principal,Kvs )
Ans. 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 5 => 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
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q6.When an integer ‘n’ is divided by 1995.The remainder is 75. What is the remainder when ‘n’ is divided by 57?
Ans.
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. 6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q7.Find the missing digits in the following multiplication. 3 5 9 7
* * * ------------------------------ * * * * * * * * * * * * * * *** -------------------------------- * * * * 5 4 1 ---------------------------------
Ans. 3 5 9 7 x 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 ---------------------------------
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
6/27/2015
PREPARED BY:- M. GOVINDU, Principal,Kvs )
Q8.Consider the following multiplication in
decimal notations (999)X(abc) = def132,
determine the digits a,b,c,d,e,f.
Ans. Given, 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
Q9.Find the largest prime factor of 314+313-12
Ans :- 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
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q10.Find the greatest number of four digits which when divided by 2,3,4,5,6,7 leaves a remainder 1 in each case. Ans)
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
6/27/2015
PREPARED BY:- M. GOVINDU, Principal,Kvs
Q11.
Find the greatest number of four digits
which when increased by 1 is exactly
divisible by 2,3 ,4,5,6 and 7 Ans.
LCM of 2,3,4,5,6,and 7=420
Largest four digit number which is a multiple
of 420= 9660
Therefore required number = 9660-1
=9659
(Since required number is 1 less than the exact multiple)
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
6/27/2015
PREPARED BY:- M. GOVINDU, Principal,Kvs )
Q12.Find the greatest number of 4 digits,
which when divided by 3,5,7 and 9 leaves
remainder 1,3,5 and 7 respectively.
Ans.
Greatest four digit number = 9999
LCM of 3,5,7 and 9 = 315
Highest four digit multiple of 315 = 9765
Required Number = 9765 – 2
= 9763
As 3 - 1 = 5 - 3 = 7 – 5 = 9 – 7 = 2
6/27/2015
PREPARED BY:- M. GOVINDU, Principal,Kvs
Aliter:
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
Q13. Find the greatest number of five digits which is divisible by 56, 72, 84 and 96 leaves remainders 48, 64, 76 and 88 respectively. Ans. 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
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q14. A number when divided by 7,11 and 13(the prime
factors of 1001) successively leave the remainders
6,10 and 12 respectively. Find the remainder if the
number is divided by 1001.
Ans.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.
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q15. A number ‘X’ leaves the same remainder
while dividing 5814, 5430, 5958. What is
the largest possible value of ‘X’.
Ans. 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.
Q16.How many prime numbers between 10
and 99 remain prime when the order of
their digits is reversed?
Ans.
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.
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q17. Exactly one of the numbers 234, 2345,
23456, 234567, 2345678, 23456789 is a
prime. Which one must it be?
Ans.
234,23456,2345678 are even.
2345 is divisible by 5
234567 is divisible by 3.
Hence, 23456789 is prime.
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q18. A two-digit number is such that if a decimal point is placed between its two digits, the resulting number is one- quarter of the sum of two digits. What is the original number?
Ans. 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
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q19. Which is greater: 31 11 or 17 14
Ans. 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
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Without actually calculating, find which
is greater: 3111 or 1714.
Ans.
3111 < 3211 => 3111<(25)11
=> 3111<255
AND 1614 < 1714 => (24)14 < 1714
=> 256 < 1714
Hence 3111<255<256<1714
=> 3111<1714
Q20. Show that 199+299+399 +499+599 is exactly divisible by 5
Ans. 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+599 is exactly divisible by 5
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q21. Find the number of perfect cubes between 1 and 1000001 which are exactly divisible by 7
Ans. 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, required number of such cubes = 14
6/27/2015
PREPARED BY:- M. GOVINDU, Principal,Kvs
Q22. How many numbers from 1 to 50 are divisible by neither 5 nor 7, and have neither 5 nor 7 as a digit.
Ans.
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
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q23.The square of a number of two digits is four times the number obtained by reversing its digits . Find the number. Ans. Let Number be 10x+y (10x+y)2 = 4.(10y+x) Number is even and 10y+x, IS A PERFECT SQUARE .
Possible values= 25,36,49,64 and 81 Square root of 4(10y+x) may be 10,12,14,16 ,18. 10 ,12,14 and 16 does not satisfy other conditions. Required number is 18.
6/27/2015
PREPARED BY:- M. GOVINDU, Principal,Kvs
Q24.Find the sum of the digites in
2 2000 . 52004
Ans.
22000 .52004
= 54.22000.52000
= 625. 102000
Therefore , Sum of digits =6+2+5=13
6/27/2015
PREPARED BY:- M. GOVINDU, Principal,Kvs
Q25.Find the last two (ten’s and unit’s) digit
of (2003)2003 Ans.
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
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q26. Find the remainder when 22005 is divided by
13.
Ans.
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.
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q27. Find the number of digits in the number
22005 .52000 when written in full.
Ans.
22005.52000
= 25.22000.52000
= 32. 102000
Number of digits =2 +2000
=2002
Q28. Arrange the following in ascending
order: 25555, 33333, 62222.
Ans.
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
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q29. Find all the positive perfect cubes that
divide 99.
Ans.
99= (93)3 =7293
Cubes of all factors of 729 will
divide 99
Factors are: 1,33,93,273,813,2433,
7293
Required numbers are: 1,33,36,39,
312, 315, 318 6/27/2015
PREPARED BY:- M. GOVINDU, Principal,Kvs
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q31.(123456)2 +123456 +123457 is the square of…….
Ans.
Here, (123456)2 +123456 +123457
= (123456)2 +123456 +123456 +1
= (123456)2 +2*123456 *1+12
= (123456+1)2
= 1234572
Required number is 123457
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q32.How many four digit numbers can be formed
using the digits 1, 2 only so that each of
these digits is used at least once ?
Ans.
Following cases are possible:
i. 1 used thrice and 2 once
ii. 1 used twice and 2 twice
iii. 1 used once and 2 thrice
Number of four digit numbers in i and iii cases
= 4 each
Number of four digit numbers in ii case
= 6
Required number of numbers=14
Q33. Find the number of perfect cubes
between 1 and 1000009 which are exactly
divisible by 9.
Ans. Since, 1<x3 <1000009 ie. 1<x <101 Perfect cubes divisible by 9 will be cubes of multiples of 3. i.e. x is a multiple of 3
But, 101=3x33 + 2
Between1 and101 there are 33multiples of 3
Required number of perfect cubes =33
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q34. Find the number of positive integers less than or
equal to 300 that are multiples of 3 or 5, but are
not multiples of 10 or 15.
Ans. No. of multiples of 3 = [300 /3 ]=100
No. of multiples of 5 =[300 /5 ] = 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 [ X ] denotes greatest integer less than or equal
to x .
6/27/2015
PREPARED BY:- M. GOVINDU, Principal,Kvs
Q35.The product of the digits of each of the three – digit
numbers 138, 262 and 432 is 24. Write down all
three digit numbers having 24 as the product of the
digits.
Ans.
24 can be written as a product of three numerals as –
1x 3 x8 ,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 6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
6/27/2015
PREPARED BY:- M. GOVINDU, Principal,Kvs
Q36. Find the numbers lying between 60 and 70,
each of which divides 248-1.
Ans.
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
6/27/2015
PREPARED BY:- M. GOVINDU, Principal,Kvs
Q37. A printer numbers the pages of a book starting
with 1. He uses 3189 digits in all. How many
pages does the book have?
Ans.
No. of digits used in one digit number = 9×1 = 9
No.of digits used in two 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
6/27/2015
PREPARED BY:- M. GOVINDU, Principal,Kvs
Q38. Find the largest prime factor of
312 +212 – 2.66.
Ans.
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
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q39. Find the value of
S= 12- 22 +32-42+…………-982+992
Ans.
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
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q40. Find the smallest multiple of 15 such that each digit
of the multiple is either‘0’or ‘8’.
Ans. 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.
Also, we want the smallest multiple of 15.
Therefore the only possibility is 8880.
The required no is 8880.
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q41. If n is a positive integer such that n/810 =
0.d25d25… where d is a single digit in
decimal base. Find ‘n’.
Ans. Let N=.d25d25… => N=d25/999 But d25/999=n/810 => n=d25X90/111 It is possible when d=9 Hence , n=750
6/27/2015
PREPARED BY:- M. GOVINDU, Principal,Kvs
Q42. Let x be the LCM of 32002-1 and 32002+1. Find
the last digit of x.
Ans.
32002=(34)500 X 32
As 34= 81 is having unit digit is 1
hence, unit digit of 32002 is 9
Unit place digit of (32002-1) = 8
Unit place digit of (32002+1 )= 0
The units digit of the product is 0
5 & 2 are the factors of their LCM
If 10 is factor of LCM, then it’s unit digit
will be 0
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
Q43. Let f0(X)=1/(1-X) and fn(x) = f0(fn-1(x)) Where
n = 1,2,3….Calculate f2009(2009)
THAN Q &
6/27/2015 PREPARED BY:- M. GOVINDU, Principal,Kvs
ALL THE BEST