QBM002
M151. What is the total number of different divisors including 1 and the number that can divide the number 3600?
M152. A set X consists of 100 natural nos, each of which is a perfect cube. The maximum no of elements of X that one can always find such that each of them leaves the same remainder when divided by 17 is
M153. Let x and y be positive whole numbers, and let p be any odd prime.
It is well known that x3 + y3 is never equal to an odd prime.
But given that n is a positive integer which contains an odd factor greater than one, prove that xn + yn = p has no solutions.
M154. The largest number amongst the following that will perfectly divide 101100 - 1 is
(1) 100 (2) 10,000 (3) 100100 (4) 100,000
M155. (x, y) is a pair of non-negative integers such that (x+ y - 5)^2 = 9xy.
What are the possible values of (x,y)
QM156.
QM157. How many four digit numbers exist which can be formed by using the digits 2, 3, 5 and 7 once only such that they are divisible by 25?
M158. A positive integer, n, is divided by d and the quotient and remainder are q and r respectively. In addition d, q, and r are consecutive positive integer terms in a geometric sequence, but not necessarily in that order.
M159. Given that x and y are integer, how many different solutions does the equation |x| + 2|y| = 100 have?
M160. A number consisting entirely of the digit one is called a repunit; for example, 11111.
Find the smallest repunit that is divisible by 63.
END
Q 151
total no.of factors = 45 + 2 = 47
Abi
HUMRAJ!!!!!!!
@Abhishek
Q.151
3600 =24 x 32 x 52
total factors = 5x3x3 = 45.
Q 157
32...4 digit numbers can be formed
Abi
Thanks
Hey..... yea..sorry... i thought 45 does not include 1 and the number 3600..jus referred a book n confirmed..
45 is the answer..
Thanks
Abi
Thanks
Hey..... yea..sorry... i thought 45 does not include 1 and the number 3600..jus referred a book n confirmed..
45 is the answer..
Thanks
Abi
M 160
Ans: 111111
clue... the number should be divisible by 7 and 3
Abi
Q157
Here unit place should be = 5 i.e 1 way
Ten place can be 2 or 7 i.e 2 way
Balance two number cane be arranged thus 1x2x2x1 =4 Ans
Q159
Here |x| + 2|y| = 100 or |y| = (100 -|x|)/2.
Thus Here |y)| .=0 thue the values for
x=0 y=-100 and +100 = 2sol
x=2 and x=-2 y=-46and+46 thus 4 ways
... x=100 or x=-100 then y = 0 2 ways
Solution = 2+4+4+...+2 where 2n-2 =100 or n =51 term is 100 thus 4+..4+(50times) = 4x50=200 Ans.
HUMRAJ!!!!!
M158. A positive integer, n, is divided by d and the quotient and remainder are q and r respectively. In addition d, q, and r are consecutive positive integer terms in a geometric sequence, but not necessarily in that order.
n = q.d + r
Then r< d and (d - r) is ethier 1 or 2.
i cant understand........., how it solve?????????
HUMRAJ!!!!
M155. (x, y) is a pair of non-negative integers such that (x+ y - 5)2 = 9xy.
What are the possible values of (x,y)?
Sol.
only two possible pairs of (x, y) are (0, 5) and (5, 0)
Some solution
M158. A positive integer, n, is divided by d and the quotient and remainder are q and r respectively. In addition d, q, and r are consecutive positive integer terms in a geometric sequence, but not necessarily in that order.
Ans. There is no solution like this.
M160. A number consisting entirely of the digit one is called a repunit; for example, 11111.
Find the smallest repunit that is divisible by 63.
The factors of 63 =3^2x7
Smallest unit divisibele by 9 is 1... n times. and four 7 it gives 1-2-3 ,-1,2,3,1,-2,-3 = 1-2-3=-4 not divisible by 7
The number divisible by 7 are 6 or 12 or 18 digit which is multiple of 9 also thus smallest repit is 111.. total 18 times.
159
Number of solutions
159
Number of solutions wud be infinite...Best way is to make a graph for the equation...
Q 152
answer seems to be 100 itself. if u take 17^3.. it will give u a reminder 0 when divided by 17. again the (17^3)^3 will agian leave 0 as reminder when div by 17.. so in that fasion there will be 100 no.s/.. Am i correct?
Q 158..
There can be a solution. thake n= 34, d=8, =then q= 4 and r=2...... 2 , 4, 8 are in GP !!!.... Am i correct?
cudn't understand the
cudn't understand the solution to 152.plz elaborate
Q 159
the answer will be 198 as two pairs (0,50) and (0 ,-50) will be common.
Q 155
i dont know how to solve this question.see there are many pairs (0,5) (5,0) (5,45) (45,5) (1,1) and even more i cany say.
(x+y-5)(x+y-5)=3x.3y
so
x+y-5=3x
x+y-5=3y and solve this
x+y-5=9x
x+y-5=y and solve this
x+y-5=x
x+y-5=9y and solve this
answer is..45 no need to add 2 as 1 and 3600 are already include
answer is..45 no need to add 2 as 1 and 3600 are already include
Ques 152 no dude...
correct answer is five only...
from 1^3 to 17^3 each willl give a different reminder fronm 18 onwards the repeation of reminder starts.
rem of 18^3 is equal ti 1^3 and so on..
finally we get 5 as answer check it..
M160:no dude check for it
1111..9times..
i think i'm correct..
here it goes..
111..ntimes./(9*7)
sum should be divisible by 9 then numerator willl be divisible by 9.
111...9times
now for 7 divide rest of numearator..122345679
it is divisible...
Q 160
d smallest such nmbr will be 18 1's...m i crrect???