QBM005
M31. If Sn represents the sum of the first n odd numbers, then 4Sn?
a. S2n b. S4n c. 2S2n d. ( S2n)2
M32. Given that p is prime, when is 8p+1 square?
M33. A tortoise and a hare race against each other. A hare runs at a constant speed of 36 km per hour for exactly ten seconds and waits for the tortoise to catch up. If the tortoise takes two hours to move 1 km, how long will it take to catch up?
M34. 22225555 + 555552222 is divided by 7, the remainder is
M35. When 242 is divided by a certain divisor the remainder obtained is 8. When 698 is divided by the same divisor the remainder obtained is 9. However, when the sum of the two numbers 242 and 698 is divided by the divisor, the remainder obtained is 4. What is the value of the divisor?
M36. Consider the infinite sequence:, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, ... What is the 1000th term?
M37. 32n-1 is divisible by 2n+3 for n= ?
M38. a!b! = a! + b! + 2c
M39. Find the value of the digit c in the following calculation ab - ba = c4
M.40 (323232)/9 will leave a remainder
M41. Find the sum of all even three digit numbers that are divisible by 8 and 10?
M42. A bag contains n discs, made up of red and blue colours. Two discs are removed from the bag.If the probability of selecting two discs of the same colour is 1/2, what can you say about the number of discs in the bag?
M43. Find the remainder when that the number 1989 x 1990 x 19923 gives when divided by 3.
M44. Find the greatest number of five digits, which is exactly divisible by 7, 10, 15, 21 and 28.
Q45. Four digits are selected from the set {1,2,3,4,5} to form a 4-digit number. Find the sum of all possible permutations
END
M34 :
2222^5555 will leave
M34 :
2222^5555 will leave the same remainder as 3^5555 when divided by 7.
5555^2222 (i am assumeing the 55555 is a typo or else the question doesn't look neat, anyways the solution process wont change) will leave the same remainder as 4^2222 when divided by 7.
the least power of 3 that leaves a remainder 1 when divided by 7 = 6
the least power of 4 that leaves a remainder 1 when divided by 7 = 3
remainder of 5555 when divided by 6 = 5
remainder of 2222 when divided by 3 = 2
hence the remainer will be the same as the reaminder of 3^5 + 4^2 when divided by 7 = 0 (checked by calculator)
if the question was assumed o be correct, then :
2222^5555 will leave the same remainder as 3^5555 when divided by 7.
55555^2222 will leave the same remainder as 3^2222 when divided by 7.
the least power of 3 that leaves a remainder 1 when divided by 7 = 6
remainder of 5555 when divided by 6 = 5
remainder of 2222 when divided by 6 = 2
neat, anyways the solution process wont change) will leave the same remainder as 4^2222 when divided by 7.
the least power of 3 that leaves a remainder 1 when divided by 7 = 6
the least power of 4 that leaves a remainder 1 when divided by 7 = 3
remainder of 5555 when divided by 6 = 5
remainder of 2222 when divided by 6 = 2
hence the remainer will be the same as the reaminder of 3^5 + 3^2 when divided by 7 = 0 (checked by calculator)
HEY! even the answer does't change !
M36: the total terms =
M36: the total terms = 1+2+3+4+....n = n(n+1)/2
now, by inspection: 44*45/2<1000<45*46/2
Hence the 1000th term = 45.
M43: since 1989 is divisible
M43: since 1989 is divisible by 3 , the entire number shall leave a remainder = 0 when divided by 3
M35: a= 8 mod(m)
b= 9
M35: a= 8 mod(m)
b= 9 mod (m)
thus a+b = 17 mod(m)
but a+b = 4 mod(m) given in question
thus 0 = 13 mod(m)
since 13 is prime , thus m =13
M32 : when p =3
proof:
8p+1
M32 : when p =3
proof:
8p+1 is an odd number . so it can be the sqaure of only a odd number.
thus (2m+1)^2 = 4m(m+1) + 1 = 8p+1
thus m(m+1) = 2p
for a prime number p this is only possible for p =3
M41:
M41:
since the numbers are divisible by 10 all end with 0.
from the divisibility rules of 8, the sum of the 3 digits is divisible by 8 , in this case we can find the sum of 2 digit noos divisible by 8 and simply multiply by10.
thus sum = 10*8*(12*13/2 - 1) = 6160
M43: the question can be
M44: the question can be reduced to : find the no. divisible by :
3*4*5*7 = 420.
M34: Another Approach
In Rashorshi's method, I am not able to comprehend the logic:
the least power of 3 that leaves a remainder 1 when divided by 7 = 6
the least power of 4 that leaves a remainder 1 when divided by 7 = 3
remainder of 5555 when divided by 6 = 5
remainder of 2222 when divided by 3 = 2
Will you plz illustrate the logic behind?
From some other source, I have searched the following solution:
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.
Ans:M31
It will be option a. S2n .
One should not go for formulas and brain tickling in such problems. Just take n=1,2 etc. and match it.
For n=2 Sn = 4 .... So 4Sn = 16
S2n...i.e. S4 = 1+3+5+7 = 16.
Sushil Singh
Software Engineer
eForce Global Inc.
KOlkata,India
M31
hello,
hw do u get S2n=1+3+5+7
plz xplain
Sonal Singh
Answers to some of the questions
M31. If Sn represents the sum of the first n odd numbers, then 4Sn?
a. S2n b. S4n c. 2S2n d. ( S2n)2
Ans. Sn = 1+3+..+1+(n-1)2 => Sn = nx(2+2n-2)/2=n^2 and 4n^2 = (2n)^2=S2n Ans is a)
M32. Given that p is prime, when is 8p+1 square?
Ans. Here 8p+1 = x^2 the 8p=(x-1)(x+1) then prime no factor which staisfies this condition=2^3xp
let x=5 then thus p =3
M33. A tortoise and a hare race against each other. A hare runs at a constant speed of 36 km per hour for exactly ten seconds and waits for the tortoise to catch up. If the tortoise takes two hours to move 1 km, how long will it take to catch up?
Ans. Distance of hare moved = 1sec min => 36x5/18 =10 m thus 10 sec => 100 mtr and tutoise has to cover 100 mtr in 2/10 hr =2x60/10=12 mts.
M34. 2222^5555 + 55555^2222 is divided by 7, the remainder is
Ans. 2222 when divided by 7 gives reminder = 3 and 3 added 5555 times gives 3x5555 = 16665 = 5 mod(7)
and 55555 = 3 mod(7) and 3x2222 gives 6666 = 2 mod(7) then the remainder in this case =0
M35. When 242 is divided by a certain divisor the remainder obtained is 8. When 698 is divided by the same divisor the remainder obtained is 9. However, when the sum of the two numbers 242 and 698 is divided by the divisor, the remainder obtained is 4. What is the value of the divisor?
Ans. 248 = 8 mod(X) and 698 = 9 mod(X) when sum divided the reaminder =4 then X =? here 8+9 =17=X+4 thus X=13 Ans
M.40 (32^32^32)/9 will leave a remainder
Ans. Now 32 = 5 mod 9 thus 32x5 = 160 = 7 mod 9 and 7x32 = 224 = 8 mod 9 thus the remainder is 9
M41. Find the sum of all even three digit numbers that are divisible by 8 and 10?
Ans. The even digit which are divisible 8 and 10 are LCM(8,10) = 40 thus the
40,80,120,160,200,...,960 ( 24 term) Their sum = 24x( 1000)/2 =12000 Ans
M42. A bag contains n discs, made up of red and blue colours. Two discs are removed from the bag.If the probability of selecting two discs of the same colour is 1/2, what can you say about the number of discs in the bag?
Ans. The possible outcome is {(R,R),(B,R),(R,B),(B,B)}
The total no of disc B and R are equal in number in the bag.
M43. Find the remainder when that the number 1989 x 1990 x 1992^3 gives when divided by 3.
Ans. The number 1989 = 0 mod 3 thus the remainder is 0.
M44. Find the greatest number of five digits, which is exactly divisible by 7, 10, 15, 21 and 28.
Ans. The LCM of (7,10,15,21,28) = 2100 and the 5 digit no greatest =98700 Ans
Q45. Four digits are selected from the set {1,2,3,4,5} to form a 4-digit number. Find the sum of all possible permutations
Ans. All possible 4 digit nos are 5x5x5x5 = 625 possible cases
The number in unit place is = 1 fo 125 times and 2 for 125 times and so on
Thus all = 1111x(125x(1+2+3+4+5)=1111x125x(15) =2083125 Ans.
@ 44
how can LCM (7,10,15,21,28) be equal to 2100 ??
LCM(7,10,15,21,28) = 420
and the largest five digit number than can be formed = 99980
Q 45
I think the answer is 1066560
count of numbers with units place = 1 would be 4 * 4 * 4 (allowing no repetition)
that is equal to 64. This is same for all the digits mentioned thus when you sum up the units place we get 64*(1+2+3+4+5) = 960.
count of numbers with tens place = 1 would again be 4*4*4 (allowing no repetition of digits)which would be the same for rest of digits menitoned in the set ... so suming it up again we 960.
following rules of addition ... we carry 96 from units place and add it to 960(total in tens place) = 960+96 = 1056
U|T|H
0 6
carry 105 to hundreds place and follow the same we get 1065. and doing it for rest of the digits we get
U|T|H|Thousands|10 thousandth|.....
0 6 5 6 6
final sum = 1066560