We need to find all possible integer solutions (values less than 100) of the equation 11x + 15y = -1

To answer this we need to find one solution first which can be found either using hit and try or using other methodical method

Let/s try some hit and try,

11x + 15y = -1
x = 3 and y = -2, 11x + 15y = 11*3 +15*(-2) = 3 ≠ -1
[b]x = 4 and y = -3, 11x + 15y = 11*4 +15*(-3) = -1[/b]

i.e. (x, y) = (4, -3) is the first solution

RULE: To find the next solution, value of x differs by co-efficient of y and value of y differs by co-efficient of x
i.e. x will differ by 15 and y will differ by 11
i.e .Next solution will be x = 4+15 = 19 and y = -3-11 = -14
@ (x, y) = (19, -14), 11x + 15y = 11*19 +15*(-14) = 209-210 = -1

i.e. solutions can be listed down as per rule as follows

(x, y) = (-131, 96)
(x, y) = (-116, 85)
(x, y) = (-101, 74)
(x, y) = (-86, 63)
(x, y) = (-71, 52)
(x, y) = (-56, 41)
(x, y) = (-41, 30)
(x, y) = (-26, 19)
(x, y) = (-11, 8)
(x, y) = (4, -3)
(x, y) = (19, -14)
(x, y) = (34, -25)
(x, y) = (49, -36)
(x, y) = (64, -47)
(x, y) = (79, -58)
(x, y) = (94, -69)

16 solutions

Answer: Option B
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11x= 0 mod 11
-15y = 1 mod 11
y= -3 mod 11

Maximum value of y= 11*9-3=96
value of x when y is equal to 96= -131 (minm possible value of x, as slope of the line is negative)

-131+n*15<100

n<15.xyz

total possible solutions= 15+1=16

nick1816 Can you explain your solution please.

I tried to make it as simple as possible (without using modular arithmetics ). If you still have any doubt, you can ask.

11x+15y=-1
11x+1= -15y

-3(11x+1)= -3*-15y
-33x-3 = 45y
-33x-3 = 44y+y
-33x-44y -3 =y

y= 11(-3x-4y) -3
y= 11k -3

11k-3<100

k< 9.xyz

max value of y is when k = 9

y= 11*9-3 = 96
at y = 96, x=-131


Now our equation is
y= (-11/15)x + (-1/15)

Slope is -11/15
for every increment of 15 units along x-axis, we have 11 units of decrement along y-axis.
As 15 and 11 are coprimes, if x=k is a integral solution of the equation, then next integral solution will be x=k+15



Also, we can infer that when y is max, x is minm.

so -131 is the minm value of x.

-131+n*15 <100
n<15.xyz

So there are 15 integral solutions between x=-131 and x=100. -131 is also a solution.

total solutions= 15+1=16.

Asked: How many integer solutions exists for the equation 11x + 15y = -1 such that both x and y are less than 100?

y = -(1+11x)/15

x = 4; y = -3; Solution
x = 19; y = -14; Solution
x = 34; y = - 25; Solution
x = 49; y = - 36; Solution
x = 64; y = - 47; Solution
x = 79; y = - 58; Solution
x = 94; y = - 69; Solution

x = -11; y = 8; Solution
x = - 26; y = 19; Solution
x = - 41; y = 30; Solution
x = - 56; y = 41; Solution
x = - 71; y = 52; Solution
x = - 86; y = 63; Solution
x = - 101; y = 74; Solution
x = - 116; y = 85; Solution
x = - -131; y = 96; Solution

Total number of solutions = 16

IMO B
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Asked: How many integer solutions exists for the equation 11x + 15y = -1 such that both x and y are less than 100?

(x,y) = (4,-3) satisfy the equation
x will change by 15 and y will change with opposite sign by 11 for each solution

(x,y) = {(-131,96),(-116,85),(-101,74),(-86,63),(-71,52),(-56,41),(-41,30),(-26,19),(-11,8),(4,-3),(19,-14),(34,-25),(49,-36),(64,-47),(79,-58),(94,-69)}

16 solutions are possible

IMO B
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First integer solution of the equation 11x + 15y = -1 using hit and trial method is
x = 4 and y = -3



i.e. Next solution will be

x = 4+15 = 19 and y = -3-11 = -14

x = 19+15 = 34 and y = -14-11 = -25

i.e. Positive values of x further will be {4, 19, 34, 49, 64, 79, 94} i.e. 7 solutions


Similarly, values of y will change by 11 and positive values of y will be (first positive value of y = -3+11 = 8}

so all positive values of y will be {8, 19, 30, 41, 52, 63, 74, 85, 96} = 9 SOlutions

Total Solutions = 7+9 = 16

Answer: Option B
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