Question

Suppose p, q, r are positive integer such that 6^p+2^p+q.3^r+2^q=332 Then sum of possible
values of p is​

Answer 1

Given :- p, q, r are positive integer such that 6^p+2^(p+q) •3^r+2^q = 332 .

To Find :- sum of possible values of p is ?

Answer :-

→ 6^p + 2^(p+q) •3^r + 2^q = 332 .

→ 2^(p+q) •3^r + 2^q = 332 - 6^p

possible values of p :-

• p = 1 => 6¹ = 6 => Possible
• p = 2 => 6² = 36 => Possible .
• p = 3 => 6³ = 216 => Possible .
• p = 4 => 6⁴ = 1296 => Not possible , since p, q, r are positive integer => 336 - 1296 = - ve .

taking p = 1,

→ 2^(1 + q) * 3^r + 2^q = 332 - 6

→ 2*2^q * 3^r + 2^q = 326

→ 2^q(2*3^r + 1) = 326

using hit and trial ,

→ 2¹(2*3⁴ + 1) = 326

→ 2(162 + 1) = 326

→ 2 * 163 = 326

→ 326 = 326 .

so, when p = 1 => q = 1, r = 4 . { All values are + ve integers.}

taking p = 2,

→ 2^(2+q) •3^r + 2^q = 332 - 6²

→ 4*2^q * 3^r + 2^q = 332 - 36

→ 2^q(4*3^r + 1) = 296

using hit and trial,

→ 2³(4*3² + 1) = 296

→ 4*9 + 1 = 296/8

→ 37 = 37 .

so, when p = 2 => q = 3, r = 2 . { All values are + ve integers.}

taking p = 3,

→ 2^(3+q) •3^r + 2^q = 332 - 6³

→ 8*2^q * 3^r + 2^q = 332 - 216

→ 2^q(8*3^r + 1) = 116

put, q = 1,

→ 8*3^r + 1 = 116/2

→ 8*3^r = 58 - 1

→ 3^r = 57/8

→ r ≠ positive integer .

put q = 2,

→ 8*3^r + 1 = 116/4

→ 8*3^r = 29 - 1

→ 3^r = 28/8

→ r ≠ positive integer .

at p = 3 , r will not be a positive integer .

therefore,

→ sum of possible values of p = (1 + 2) = 3 (Ans.)

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Answer 2

Answer:

Given :- p, q, r are positive integer such that 6^p+2^(p+q) •3^r+2^q = 332 .

To Find :- sum of possible values of p is ?

Answer :-

→ 6^p + 2^(p+q) •3^r + 2^q = 332 .

→ 2^(p+q) •3^r + 2^q = 332 - 6^p

possible values of p :-

p = 1 => 6¹ = 6 => Possible

p = 2 => 6² = 36 => Possible .

p = 3 => 6³ = 216 => Possible .

p = 4 => 6⁴ = 1296 => Not possible , since p, q, r are positive integer => 336 - 1296 = - ve .

taking p = 1,

→ 2^(1 + q) * 3^r + 2^q = 332 - 6

→ 2*2^q * 3^r + 2^q = 326

→ 2^q(2*3^r + 1) = 326

using hit and trial ,

→ 2¹(2*3⁴ + 1) = 326

→ 2(162 + 1) = 326

→ 2 * 163 = 326

→ 326 = 326 .

so, when p = 1 => q = 1, r = 4 . { All values are + ve integers.}

taking p = 2,

→ 2^(2+q) •3^r + 2^q = 332 - 6²

→ 4*2^q * 3^r + 2^q = 332 - 36

→ 2^q(4*3^r + 1) = 296

using hit and trial,

→ 2³(4*3² + 1) = 296

→ 4*9 + 1 = 296/8

→ 37 = 37 .

so, when p = 2 => q = 3, r = 2 . { All values are + ve integers.}

taking p = 3,

→ 2^(3+q) •3^r + 2^q = 332 - 6³

→ 8*2^q * 3^r + 2^q = 332 - 216

→ 2^q(8*3^r + 1) = 116

put, q = 1,

→ 8*3^r + 1 = 116/2

→ 8*3^r = 58 - 1

→ 3^r = 57/8

→ r ≠ positive integer .

put q = 2,

→ 8*3^r + 1 = 116/4

→ 8*3^r = 29 - 1

→ 3^r = 28/8

→ r ≠ positive integer .

at p = 3 , r will not be a positive integer .

therefore,

→ sum of possible values of p = (1 + 2) = 3 (Ans.)