Maths questions

                                      Maths questions

Q.1. How many ordered pairs of integers satisfy (x,y) the equation 1/x+1/y=1/12 

Lets try and solve it using good old school algebra.

1x+1y=112${\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}={\displaystyle \frac{1}{12}}$

Or, (x+y)xy=112${\displaystyle \frac{(x+y)}{xy}}={\displaystyle \frac{1}{12}}$

Or, 12x+12y−xy=0$12x+12y-xy=0$

Or, 12x−xy+12y−144+144=0$12x-xy+12y-144+144=0$

Or, x(12−y)−12(12−y)+144=0$x(12-y)-12(12-y)+144=0$

Or, (x−12)(12−y)=(−144)$(x-12)(12-y)=(-144)$

Or, (x−12)(y−12)=144$(x-12)(y-12)=144$… Eqn1

Now, 144=(2)4×(3)2$144=(2{)}^4\times (3{)}^2$

• So, it has (1+4)×(1+2)$(1+4)\times (1+2)$ = 15 factors.

Factors: 1, 2, 3, 4, 6,……., 36, 48, 72, 144

• Thus it can be expressed as product of 2 factors in (1+15)2${\displaystyle \frac{(1+15)}{2}}$ = 8 ways.

For e.g, 144=1×144=2×72=3×48=12×12

Q.2. for how many natural numbers a is the expression sqrt((a+64)/(a-64)) also a natural number

Q.3. How many ordered pairs of integers satisfy (x,y) the equation 1/x+1/y=1/13

Q.4. In how many ways can 45 be written as the sum of consecutive positive integers?

Let’s see … we know 1+2+3+4+5+6+7+8+9 = 45
So if lets see if we can chop some numbers from the left side that sum to the next number on the right:

We can chop off 1+2+3+4 (=10) and put 10 on the right.

Then we can chop off 5+6 and put 11 on the right of 10.

Unfortunately if we chop off 7 and 8 that’s more than 12, and if we additionally chop off 9 that’s 24 which does not equal 12+13 (25) .

So the answer is 3 ways (1..9, 5..10, 7..11)