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1. x! ends with k zeros; (x+3)! ends with (k+3) zeros . 0<x<1000; how many values of x are possible?
Ans.
5 changes 1 zero25 2 zeros125 3 zerosand 625 4 zerosso multiples of 125 till 6254..and since its k+3 so 3 for each..4*3=12
2. How many pairs of positive integers x,y exist such that HCF(x,y) + LCM (x,y) = 91?
Ans.
3. Find remainder when 3(14!)-15(13!) divides 28(13!)+ 27!
(1) 13! (2)10! (3)8! (4)12!
Ans. (28(13!)+ 27!) / 3(14!)-15(13!)
= ( 28(13!)+ 27!) / 13! (27)
= 28(13!) / 13! (27) + 27! / 13! (27)
= (27+1)13! / 13! (27) + 27! / 13! (27)
= (27) 13! / 13! (27) + 13! / 13! (27) + 27! / 13! (27)
Clearly 1st and 3rd are divisible and therefore from 2nd term remainder= 13!
4. x! ends with k zeros; (x+3)! ends with (k+3) zeros . 0<x<1000; how many values of x are possible?
Ans. x! =k zeros ; (x+3)! = (x+3)(x+2)(x+1)x! = k+3 zeros ; so extra 3 zeros must be from (x+3)(x+2)(x+1)-----since it's a product of 3 consecutive number----------all 3 cannot be multiple of 5-----------at the max only one of them can be multiple of 5---------but we need 3 extra multiple of 5------------so it must be multiple of 5^3 but not of any higher multiple of 5. so 5^3 can be either (x+1)or (x+2)or(x+3),so for every multiple of 5 there are 3 choices. From 1 to 125 there are 8 multiples of 125 but we will not take 625 (as it is 5^4 & it will give 4 extra powers); so answer is 7*3 = 21
5. Euler’s Formula
E(z) =z*(1-1/p) (1-1/q) (1-1/r).....
z: the no.
p, q, r : different prime factors of z
it is usually used for finding the remainders
6. The concept is that every prime number can be expressed in the form 6k -/+ 1 but not every number which is of the form 6k+/-1 is prime.
7. A no. when divided 88 leaves remainder 3. What is the remainder when its divided by 11?
Ans. D=88n+3. Now D/11=(88n%11)+3%11=0+3=3
8. Let x be the arithmetic mean of all positive integers k<577 such that k^4 when divided by 577, leaves a remainder of 144. Find the greatest integer less than or equal to x.
Ans.
9. 32^232 + 17^232 is definitely divisible by....
(a.) 49 (b.) 15 (c.) 49 & 15 (d.) none of these.
Ans. a^n +b^n when n is even is divisible by (a+b) and (a-b). Hence, ©.
10. To find no. of odd factors of a number. leave power of 2 and multiply powers of all other prime nos. after adding one to them i.e. (8 + 1)*(4+1)*(4+1) = 225 = no. of odd factors of 1260^4
11. To find total no. of factors of a number. =>first of all...express it in terms of prime numbers
e.g 1260^4. Express it as (2x2x3x3x5x7)^4 = 2^8 * 3^8 * 5^4*7^4.
=>now add 1 to the powers of every prime no. n multiply them all...u get the total no. of factors of 1260^4.
i.e (8+1)*(8+1)*(4+1)*(4+1) = 2025 = total no. of factors of 1260^4.
12. To find no. of even factors of a number. Take the difference of total factors and odd factors.
i.e. 2025 - 225 = 1800. = no. of even factors of 1260^4.
13. To find in how many ways can a given no. be represented as product of two relatively prime factors.
In this case, the power of prime no. becomes irrelevant as all the powers shud lie with the same factor else the two factors won’t be relatively prime i.e. in the above case 1260 = (2x2x3x3x5x7)^4 = 2^8 * 3^8 * 5^4*7^4.
Now powers 8,8,4,4 have no importance...what’s important is how many prime nos. are there...they are four...viz 2,3,5,7. Hence 4.
Now we have to form nos. using these 4 prime nos. It’s a like a question asking you...if you have 4 sweets...in how many ways can u eat them? The answer is 4C0. When u eat none... + 4C1 when eat any one...+4C2 + ...4C4...when u eat all.
Similarly here, answer would be 4C0 + 4C1 + 4C2 ...4C4 = 2^4 = 16.
But the factors have to exist in pairs...hence 2^4/2 = 8 factors are possible. On a general note...a no. formed of n prime nos. has 2^(n-1) pair of co-prime factors.
14. Find the smallest no. that has exactly.....
a. 16 factors
b. 12 factors
c. 60 factors
Ans. a) 120 - 2x3x5x7 (Factors of 16 - 2x2x2x2 or 4x4) check for smallest value.
b) 60 - (12 - 2x2x3 or 4x3) so 2^2x3x5 =60
15. Smallest no. that leaves remainders 4 every time when successively divided by 7,5,10,13 respectively.
Ans. 7 5 10 13 4 4 4 4
Cross multiply and add like this
(((((4x10)+4)x5)+4)x7)+4)
16. How many values of a are possible if x^2 + ax + 2400 has...
a.)integral roots
b.)roots which are natural nos.
Ans. To solve ax2 + bx + c =0,
we break it as ax2 + mx + nx + c = 0, such that m*n = a*c.
Here, a*c = 1*2400 = 2400. so v need to find in how many ways can 2400 be expressed as product of 2 nos. every such pair of nos. will give a new value of the coefficient of x.
2400 = 2^5 x 5^2 x 3
no. of factors = 6x3x2 = 36.
but these factors have to exist in pairs...e.g when v use one factor 2 (to express 2400 as 2x1200) the other factor...1200 is automatically used...so total pairs possible are ...
36/2 = 18.
But relax...this is not it....again...equal no. of negative pairs exist...i.e. 2x1200 corresponds to -2 x -1200. although the product is same as the required product i.e. 2400...the sum is different...its 1202 n -1202...n v need to find different values of sum...hence the answer would be 18x2 = 36 again...
a) for integral roots...the answer would be 36
b) values would be 1/2 the total possible values as negative roots aint allowed...36/2 = 18 is the answer
17. 32^2321 + 17^2321 + 13^2321 + 18^2321 is definetly divisible by....
a. 40 b. 20 c. 80 d. all of these. e. none of these.
Ans.
18. A number is having exactly 72 factors. What can be the maximum and minimum number of prime factors of this number?
a.71 &1 b. 6 & 2 c. 5 & 2 d. None of These
Ans. min1, max 5
19. How many natural numbers are factors of 540 but not a factor of 720?
a.6 b.9 c.12 d. 15
Ans. 540=2^2 * 3^3 * 5
72=2^4*3^2*5
thus 3^3 is common to all factors of 540 dat aren factors of 720
so we have (2+1)(1+1)=6
alternative method: no. of factor of 540 - no. of factor of hcf(540,720)
20. A number is having 15 composite factors. What can be the maximum number of prime factors of this number?
a.2 b. 3 c.4 d. 5
Ans.
N = p1^a*p2^b*p3^c*p4^d*p5^e, where p1,p2,p3,p4,p5 are prime numbers. for minimum no. of factors a=b=c=d=e = 1, so min. no. of total factors = 2*2*2*2*2 =32 ( if 5 prime factors). so min no. of composite factors = 32-5-1 = 26. (1 is neither prime nor composite).so if 5 prime no. are involved then they must have a min of 26 composite factors. but here 15 composite factors are there so ans has to be less than 5.
total no. of factors = 1 + no. of prime factors + no.of composite factors
here if 4 prime no. are involved then N = p1^a*p2^b*p3^c*p4^d; (a+1)(b+1)(c+1)(d+1) = 1+4+15 =20 = 2*2*5; not possible
here if 3 prime no. are involved then N = p1^a*p2^b*p3^c; (a+1)(b+1)(c+1) = 1+3+15 =19; not possible
if 2prime no. are involved then N = p1^a*p2^b; (a+1)(b+1) = 1+2+15 =18= 2*9 =3*6; possible
so maximum = 2
21. How many A.Ps are possible such that first term is 1235 and the last term is 3535 and there are at least ...
a. 3 terms
b. 4 terms
c. 5 terms
d. 6 terms
Ans. 3535 - 1235 = 2300.
2300 = 23x2^2x5^2 . hence, no. of factors = 3x3x2 = 18.
how many APs...18.
how many with atleast 3 terms?
since 2300%1 (number of terms exist between 1&2300 i.e. 1,x,2300, just one) = 0, 18-1 = 17
how many with atleast 4 terms?
2300%2 = 0, hence 17-1 = 16
how many with atleast 5 terms?
2300%3 =/ 0, answer remains 16. where =/ means not equal to
how many with atleast 6 terms?
2300%4 = 0, 16-1 = 15
how many with atleast 7 terms?
2300%5 = 0, 15-1 = 14
how many with atleast 8 terms?
2300%6 =/ 0 , answer remains 14
how many with atleast 9 terms?
2300%7 =/ 0, answer remains 14
and so on...
22. Find HCF and LCM of:
a.2222...250 times and 8888...300 times
b. 333....120 times and 1111...400 times
c. 111...700 times and 9999...200 times.
d. HCF of 33333...200 times. and 777777.....300 times
e. 32^250 -1 & 16 ^ 100 - 1.
f. 81^100 -1 & 243 ^ 200 - 1.
Ans.
23. N= 2^8 × 3^10 ×5^8 × 7^2 .w many factors of N are multiple of 360 and but not a multiple of 540?
a.210 b. 62 c.144 d. 124
Ans. no. of factors which are multiple of 360 - no. of factors which are multiple of lcm(360,540)
24. How many natural numbers are there which give a remainder of 41 after dividing 1997?
1.2 2.4 3.6 4.None of these
Ans. 3
25. How many factors of 125000 are perfect cubes?
Ans. Cube root of the number is 50=5^2x2...>>3x2=6(factors)
26. Albert starts to make a list, in increasing order, of the positive integers that have a first digit of 1. He writes 1,10,11,12-------------but by the 1,000th digit he (finally) realizes that the list would contain an infinite number of elements. Find the three-digit number formed by the last three digits he wrote (the 998th, 999th, and 1000th digits, in that order).
Ans. 116
27. N leaves a remainder of 4 when divided by 33, what are the possible remainders when N is divided by 55?
Ans. N=33d1+4, put d1=0,1,2.. remainders with 55 --> 37, 15, 48, 26, 2
28. A student wrote all the natural no. from 2 to 10000 on a blackboard, one after the other. Another student came and erased all perfect cubes. If student come this way and erase all the higher powers, find the no. of students who erase at least one number.
a. 6 b. 7 c. 12 d. None of These
Ans. Shortcut: Determine what is highest power of 2 can be deleted. Count all primes less than it. Here 2,3,5,7,11,13.
29. 27^27^54 /7 Rem?
Ans. 27^(6k)%7=1.. 27^54%6=3^54%6=3. Hence 27^54 is of the form 6k+3. Hence remainder = 27^3 with 7 = 6
See, 27^27^54 /7 =6^27^54 or (-1)^27^54... so, when further divided by 7 we need to find a power of 6 which gives 1 as a rem when divided by 7. Here, 36(power 2) gives rem 1 with 7. So, divide 27^54 which can be writen in the form 2k+1. Hence, 6^(2k+1) where 6^2k is divisible by 7 while 6^1 is the remainder. Hence, 6-->OA.
30. 32^32^32 divided by 7. Rem =?
Ans. for such questions , first find power of 32 when divided by 7, gives a remainder as +1/-1..in dis case 32^3%7 gives rem +1....this means dat d remainders will repeat with a cycle of 3...so i need 2 find 32^32nd term.....for which find 32^32%3...i.e.1...therefore, 32^32^32 will give d same remainder as 32^1 whn divided by 7....so i feel ans shud be 4....
31. Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum of the two-digit number and the three-digit number?
Ans.
32. LCM of three numbers is equal to 1080; HCF of the three numbers is equal to 2. How many such
triplets are possible?
Ans. HCF = 2. Let the numbers be 2x, 2y and 2z. HCF of (x,y,z) =1
LCM = 2xyz. => xyz = 540
540 = 2^2 * 3^3 * 5
x = 2^2 , y = 3^3, z = 5
x = 2^2 * 3^3, y = 5, z = 1
x = 2^2 * 5, y = 3^3, z = 1
x = 3^3 * 5, y = 2^2, z = 1
x = 2^2 * 3^3 * 5, y=1, z=1
These are the basic combinations. The first 4 yield 6 ordered triplets each, the last one yields 3 ordered triplets.
So, there are 5 unordered triplets and 27 ordered triplets
33. What is the remainder when7^7^7^7^7^7-------infinity
is divided by 13?
a. 5 b. 6 c. 7 d. None of These
Ans. E(13) = 12
E(12) = 4
7^7^7^7.... mod 4 = -1^odd = -1 = 3
7^3 mod 12 = 7
7^7 mod 13 = 6
34. The integer n is the smallest positive multiple of 15 such that every digit of n is either 0 or 8. Compute n/15
Ans. 8880/15 =592
35. let N be the number of consecutive zero’s at the right end of the decimal representation of the product 1!2!3!4!.......99!100!.find the remainder when N is divided by 1000.
Ans. the answer is 5(1+2+3 + 4+6+7....+10+12 +13+..16+18+19...+22) +24
36. A reduction of 20% in the price of sugar enables a person to purchase 6kg more for Rs 240. What is the original price per kg of sugar?
10, 8, 6,5, None?
Ans.100x*y = 240
80x*(y+6) = 240
100xy = 80xy+480x
=> y = 24
so , X= 10 = answer
20 percent decrease will enable him to buy 25 percent more than actual
therefore 25% of x=6
that is x=24
actual price is 240/24=10
Price * consumption - constant. so when price goes down by 1/5, consumption shud increase by 1/4. 1/4 = 6, so 24kg is the normal consumption. orig price = rs 10
37. The increasing sequence 3,15,24,48,…..consists of those positive multiples of three that are one less than a perfect square. What is the remainder when the 2010th term of the sequence is divided by 1000?
Ans. All numbers one less than a perfect square may be written as
x^2 - 1, factorable into (x+1)(x-1). For a number like this
to be divisible by 3, either x+1 or x-1 must be divisible by 3.
This case occurs whenever x is not divisible by three (because
when it isn't divisible by 3, it is either 1 more than a factor
of 3 or one less than a factor of 3).
The first couple of values of x are 2,4,5,7,8,10,11, etc.
The odd number terms may be determined with the relation 3n-1
for the nth odd term. Likewise, the even number terms may be
determined with the relation 3n+1 for the nth even number.
2010 is the 2010/2 = 1005th even number. Thus, the value for
x is 3(1005) + 1 = 3016.
The value of x^2 - 1 (the 2010th term in the sequence) may be
written as (3000 +16)^2 - 1 ==>.
When this number is divided by 1000, Thus, the remainder is just 16^2 – 1
38. The number r can be expressed as a four-place decimal 0.abcd, where a, b, c, and d represent digits, any of which could be zero. It is desired to approximate r by a fraction whose numerator is 1 or 2 and whose denominator is an integer. The closest such fraction to r is 2/7. What is the number of possible values for r?
Ans.
39. What is the largest positive integer n for which n^3 + 100 is divisible by n + 10?
Ans. (n+10)^3 = n^3 + 30n^2 + 300n + 1000
So n^3 + 100 = (n+10)^3 - 30(n^2 + 10n + 30)
= (n+10)^3 - 30[(n+10)^2 - 10n - 70]
= (n+10)^3 - 30[(n+10)^2 - 10(n + 10) + 30]
= k*(n+10) - 900... hence 900 shud be divisible by (n+10) highest such n is 890
40. What is the remainder when 17! is divided by 23?
a. 15 b. 14 c. 3 d. None of These
Ans. wilson's theorem: (p-2)! % p = 1 where p is a prime no.
21!%23=1. hence 21*20*19*18*17! % 23=1 .. (-2)(-3)(-4)(-5)*17! % 23=1. 120*17! % 23 = 1 --> 17! * 5 % 23 = 1. RHS gives remainder 1, hence LHS also should. realize that 70%23 = 1, hence remainder = 70/5=14
41. Remainder....
when
1) 35^65^95/13
2) 35^65^95/1001
Ans. 9^65^95/13, 9^3 gives 1 as remainder with 13.. so 65^95 should be divided by 5 which could be written in the form of 3k+2.. Hence 9^(3k+2). 9^3k divided by 13 while 81 gives rem 3 with 13. OA=3 13k+3, divide by 11 and it should give 10 rem with 11.. so 2k+3=10 or 21 (21 is possible which gives k=9) hence.. 120+143n(rem of 11&13) divide it further by 7.. 1+3n=7 hence n=2. sp 286+120=406
42. Last three digits of 57^802
Ans. Last 3 digit means remainder with dividing with 1000
1000=2^3*5^3
so euler=400
so last 3 digit=57^2/1000=3249/1000=249
43. By a proper divisor of a natural number we mean a positive integral divisor other than 1 and the
number itself. A natural number greater than 1 will be called “nice” if it is equal to the product of its
distinct proper divisors. What is the sum of the first ten nice numbers?
Ans.
44. How many pairs of consecutive integers can be added without a carry in the sequence (1000,1001,1002.............2000) ?
Ans. The possible number has the following form
1abc, 1ab9, 1a99, 1999, where a, b, c -- {0, 1, 2, 3, 4}
5^3+5^2+5+1 = 156
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