M4maths Previous Puzzles

28 Nov 2014 Possible Combination

If x^2 - y^3 is a positive non zero integer for all possible integral values of x, y satisfying 1 =< x =< 9 and 1 =< y =< 9 then number of possible ordered combinations of (x, y) are.

OPtion

1) 8
2) 10
3) 12
4) 16
5) 18
6) 20
7) 21
8) 24
9) 25
10)27

Solution

If y = 1, then x can have values ranging 2 to 9 so 8 possibilities.
If x = 2, then x can have values ranging 3 to 9 so 7 possibilities.
If y = 3, then x can have values ranging 6 to 9 so 4 possibilities.
If y = 4, then x can have values only 9 so 1 is possibilities.

Total cases = 8+7+4+1 = 20

Option 6)

Correct Option: 6 Best Solution (28)

BHASKAR MANDAL   10 years AGO

possibilities are
1) y=1 , then x will be from 2 to 9 ---> 8
2)y=2 , then x will be from 3 to 9 -----> 7
3) y= 3, then x will be from 6 to 9------>4
4) y= 4, then x will be from 9 ------------> 1
so total = 8+7+4+1 = 20

Like? Yes (11) | No (2)  

kamal   10 years AGO

1,2,3,4,5,6,7,8,9(possible value of x,y)
1,4,,9,16,25,36,49,64,81 (x^2 possible)
1,8,27,64,125,216,343,512,729 (all possible y^3)
now take x^2=1 and y^3 from 1 to 729 there is no pair such possible (x,y) such that x^2-y^3=non zero positve number,,
solve further and u'll get 20 such pairs---(2,1) (3,1) (3,2) (4,1) (4,2) (5.1)----------------(9,4).

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Aditya Verma   10 years AGO

according to question...
x^2 - y^3 > 0
where...
1 =< x =< 9 and
1 =< y =< 9
so all possible combinations of (x,y) are...
(2,1),(3,1),(3,2),(4,1),(4,2),(5,1),(5,2),(6,1),(6,2),(6,3),(7,1),(7,2),(7,3),(8,1),(8,2),(8,3),(9,1),(9,2),(9,3),(9,4)
there are total 20 possible combinations of x and y
so answer is 20

Like? Yes (2) | No   

Jayanta Dutta   10 years AGO

First consider y = 1 so x will be (2,3,4,5,6,7,8,9)
Next y=2
So x will have to be(3,4,5,6,7,8,9)
Next y=3
X will hve to be (6,7,8,9)
Next y=4
X will hav to be (9).
So total combo =[8+7+4+1] optn 6

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RAKESH   10 years AGO

(2,1)
(3,1),(3,2)

------------------
(8,1),(8,2),(8,3)
(9,1),(9,2),(9,3),(9,4)

total = 1+2+2+2+3+3+3+4 = 20

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ANIL KUMAR   10 years AGO

by jst chkng all the possible combination of x^2 - y^3
got it as 20.

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Devendra Marghade   10 years AGO

For x^2 - y^3 to be positive,
possible ordered combinations of (x, y) are
(2,1), (3,1), (3,2), (4,1), (4,2), (5,1), (5,2), (6,1), (6,2), (6,3), (7,1), (7,2), (7,3), (8,1), (8,2), (8,3), (9,1), (9,2), (9,3), (9,4)

Total 20

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PAWAN KUMAR   10 years AGO

For
x=9, y=1,2,3,4
x=8, y=1,2,3
x=7, y=1,2,3
x=6, y=1,2,3
x=5, y=1,2
x=4, y=1,2
x=3, y=1,2
x=2, y=1

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Narasapu Hari prasad   10 years AGO

ans is option6 20:
the possible x,y combinations are
(2,1)(3,1)(4,1)(5,1)(6,1)(7,1)(8,1)(9,1)
(3,2)(4,2)(5,2)(6,2)(7,2)(8,2)(9,2)
(6,3)(7,3)(8,3)(9,3)
(9,4)

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Rajat Rokade   10 years AGO

check for all combinations of x and y
The answer is 20

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SUBHASH TULSYAN   10 years AGO

x^2-y^3>0
1

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Azhar Syed   10 years AGO

possible combinations if (x,y) are (2,1) (3,1) (3,2) (4,1) (4,2) (5,1) (5,2) (6,1) (6,2) (6,3) (7,1) (7,2) (7,3) (8,1) (8,2) (8,3) (9,1) (9,2) (9,3) (9,4) total of 20 combinations

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CHANDAN KUMAR   10 years AGO

x^2-y^3=+ve>0
xy=21,31,32,41,42,51,52,61,62,63,71,72,73,1,82,83,91,92,93,94
total=20

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sravani   10 years AGO

x can have values from 2 to 9
but to satisfy the condition y takes only two values 1 for x=2.
values 1,2 for x=3,4,5
values 1,2,3 for x=6,7,8.
values 1,2,3,4 for x=9.......
so total (x,y) pairs are 20.......

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Pramod Gaikwad   10 years AGO

put the number between 1 to 9,
number of solution =20.
so
option is 6)20

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prem   10 years AGO

(2,1)...........................(9,1) total 8 combination
(3,2).................(9,2) total 7 combination
(6,3).....................(9,3) total 4 combination
(9,4) 1 combination hence total 20 combination
hence 6 option

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Rishikesh Ranjan   10 years AGO

8+7+4+1

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poornima    10 years AGO

if k is the positive int, then assume X =6,Y=3
it satisfy x^2-Y^3=k
then ordered combination of (6,3) is 20
ans is 20

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khushboo goenka   10 years AGO

when y=1
8 combinations
y=2, 7 combn, y=3,4 cominatons
y=4, one comination total= 8+7+4+1=20

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MRITUNJAY   10 years AGO

X^2-y^3=non zero
by heat and trial
x=9&y=1,2,3,4
x=8&y=1,2,3
x=7&y=1,2,3
x=6&y=1,2,3
x=5&y=1,2
x=4&y=1,2
x=3&y=1,2
x=2&y=1
total 20.

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ARPAN SHAH   10 years AGO

for y only 4 possibilities 1,2,3,4 ...now for y=1 x can 8 values 2 to 9,for y=2 ,x=3 to 9,for y=3,x=6 to 9,for y=4,x=9.so total 20

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sanjeev kumar   10 years AGO

ans 20

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abeer naskar   10 years AGO

as the whole equation is positive then the only options for y are 1,2,3 and 4
for y=5, y^3 will be 125 but max value of x^2 will be 81
so for y=1 we get x from 2 to 9 that is 8 no. of x
for y=2 we get x from 3 to 9 that is 7 choice of x
for y=3 we get x from 6 to 9 that is 4
for y=4 we get x from 9 to 9 that is 1
so answer that is total choice is
8+7+4+1=20

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DEERGHA AGGARWAL   10 years AGO

solving through the equation

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Manju Prakash R   10 years AGO

Condition is that y^3 must be greater than x^2
There are 20 possible ordered pairs
y=9 -> 4 pairs
y=8 -> 3 pairs
y=7 ->3 pairs
y=6 -> 3 pairs
y=5 => 2 pairs
y=4=>2 pairs
y=3 =>2 pairs
y=2 =>1 pair

Ans 6) 20

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Sourabh Prasad   10 years AGO

x 2 - y 3 =
Satisfying values are

2 1
3 1
3 2
4 1
4 2
5 1
5 2
6 1
62
6 3
7 1
7 2
7 3
8 1
8 2
8 3
9 1
9 2
9 3
9 4

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Md.Sultan Khichi   10 years AGO

x^2-y^3= non zero +ve value.
By it and trial method
The combinations are
(2,1),(3,1)(3,2),(4,1),(4,2),(5,1),(5,2),(6,1),6,2),(6,3),(7,1),(7,2),(7,3),(8,1),(8,2),(8,3),(9,1),(9,2),(9,3)
Total=20.
Ans ob. 6

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Rahul Rathod   10 years AGO

check all combinations of all (x,y) pair
we get 20 solutions.

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