Correction

Sum of nos + HCF = 77

Let HCF = h, nos hx, hy (x, y coprime)

nos. Are non coprime so h ≠ 1

Given that

h(1 + x + y) = 77

If h = 7

x + y = 10

(x, y) = (1, 9), (3, 7)

If h = 11

x + y = 6

(x, y) = (1, 5)

3 pairs are possible.

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(5,0,0,0,0) = 1

(4,1,0,0,0) = 2*4!/3! = 8

(3,2,0,0,0) = 2*4!/3! = 8

(3,1,1,0,0) =3*4!/2!*2! = 18

(2,2,1,0,0) = 3*4!/2!*2! = 18

(1,1,1,2,0) = 4*4!/3! = 16

(1,1,1,1,1) = 1

69

Solution

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Correction

Quadratic : mx² + nx + 1 = 0

For real roots D ≥ 0

n² ≥ 4m

case 1: n = 1, no values for m

Case 2: n = 2, 1 ≥ m, m = 1 (m,n) = (1, 1)

Case 3: n = 3, 9/4 ≥ m, m = 1,2

(m, n) = (1, 3), (2, 3)

Case 4: n = 4, 4 ≥ m, m = 1, 2, 3, 4

(m, n) = (1, 4), (2, 4), (3, 4), (4, 4)

7 values

If N = even means in form of 2n

Can be written as {2(n - 4) + 8} (sum of two composite numbers)

Where n ≥ 6

If N = odd means in form of 2n + 1

Can be written as 2(n - 4) + 9

Where n ≥ 5

By both cases n ≥ 6

2n ≥ 12

Means value of P = 11

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TRY

A+b===2, 11


From options
A if it is 2 (0, 2) (2, 0 ) possible

83502 83520 not divided by 7

If c (a-b=3 a+b=11) (7, 4

Possible numbers

(83574, 83547) not divided by 7


If b

(83583 not divided by 7

(83538 yes 7*11934


So b