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,

$\longrightarrow 6^p+2^{p+q}\cdot3^r+2^q=332$

or,

$\longrightarrow (2\cdot3)^p+2^{p+q}\cdot3^r+2^q=332$

$\longrightarrow 2^p\cdot3^p+2^p\cdot2^q\cdot3^r+2^q=332$

Divide both sides by $2^q.$

$\longrightarrow\dfrac{2^p\cdot3^p+2^p\cdot2^q\cdot3^r+2^q}{2^p}=\dfrac{332}{2^p}$

$\longrightarrow3^p+2^q\cdot3^r+2^{q-p}=\dfrac{332}{2^p}\quad\quad\dots(1)$

Case 1:- Assume $p=q,$ then $q-p=0.$

So (1) becomes,

$\longrightarrow3^p+2^q\cdot3^r+1=\dfrac{332}{2^p}$

In the LHS, the terms $3^p$ and $1$ are odd and $2^q\cdot3^r$ is even. Then the whole LHS would be odd + even + odd = even. So would the RHS be. Also each term is an integer.

Take $p=q=1$ then RHS will be  $\dfrac{332}{2}=166$ which is even. So,

$\longrightarrow3+2\cdot3^r+1=166$

$\longrightarrow2\cdot3^r=162$

$\longrightarrow3^r=81$

$\longrightarrow r=4$

So $\underline{(p,\ q,\ r)=(1,\ 1,\ 4)}$ is possible.

Take $p=q=2$ then RHS will be $\dfrac{332}{4}=83$ which is odd. So this is not possible.

For higher values of $p,$ the RHS will not be an integer. Because the highest power of 2 that can exactly divide 332 is 2² = 4, and that is what we saw above.

Case 2:- Let $p<q,$ then $q-p>0.$

In the LHS of (1), $3^p$ is odd, and $2^q\cdot3^r$ and $2^{q-p}$ are even. Then the whole LHS would be odd + even + even = odd. So would the RHS be.

If the RHS $\dfrac{332}{2^p}$ is odd, then $2^p$ must be the highest power of 2 that can exactly divide 332, which is none other than $2^2=4,$ as we said earlier, and it implies $p=2.$

Then (1) becomes,

$\longrightarrow3^2+2^q\cdot3^r+2^{q-2}=\dfrac{332}{2^2}$

$\longrightarrow9+2^q\cdot3^r+2^{q-2}=83$

$\longrightarrow2^q\cdot3^r+2^{q-2}=74$

Divide both sides by $2^{q-2}.$

$\longrightarrow\dfrac{2^q\cdot3^r+2^{q-2}}{2^{q-2}}=\dfrac{74}{2^{q-2}}$

$\longrightarrow4\cdot3^r+1=\dfrac{74}{2^{q-2}}$

Here the LHS is odd (even + odd = odd), so would the RHS be. Then $2^{q-2}$ must be the highest power of 2 that can exactly divide 74, which is none other than $2^1=2,$ and it implies $q=3.$ Also 2 < 3.

Then,

$\longrightarrow4\cdot3^r+1=\dfrac{74}{2^1}$

$\longrightarrow4\cdot3^r+1=37$

$\longrightarrow4\cdot3^r=36$

$\longrightarrow3^r=9$

$\longrightarrow r=2$

So $\underline{(p,\ q,\ r)=(2,\ 3,\ 2)}$ is possible.

Case 3:- Let $p>q,$ then $q-p<0.$

In this case, in the LHS of (1), the terms $3^p$ and $2^q\cdot3^r$ are integers but $2^{q-p}$ is a decimal number. So the RHS should have the same decimal part as that of $2^{q-p}.$

Then in RHS $\dfrac{332}{2^p},$ $2^p$ must exceed the highest power of 2 that can exactly divide 332, i.e., $2^p>2^2.$

Take $p=3.$ Then (1) becomes,

$\longrightarrow3^3+2^q\cdot3^r+2^{q-3}=\dfrac{332}{2^3}$

$\longrightarrow27+2^q\cdot3^r+2^{q-3}=41.5$

$\longrightarrow2^q\cdot3^r+2^{q-3}=14.5$

Here $2^{q-3}$ must have same decimal part of 14.5, i.e., 0.5, thus $2^{q-3}=2^{-1}=0.5\,\implies\,q=2.$

So,

$\longrightarrow2^2\cdot3^r+0.5=14.5$

$\longrightarrow4\cdot 3^r=14$

$\longrightarrow3^r=3.5$

There is no value for $r$ to satisfy this equation. So this is not possible.

For higher values of $p,\ 2^{q-p}$ and $\dfrac{332}{2^p}$ would not have same decimal part, since $\left[\dfrac{332}{2^3}\right]=\left[41.5\right]=41$ is odd.

Finally,

$\longrightarrow\underline{\underline{p\in\{1,\ 2\}}}$

Hence the sum of possible values of $p$ is​​ 3.