Suppose p,q and r are positive integers such that 6^p+2^p+q×3^r+2^q=332
Answers
Step-by-step explanation:
Given,
\longrightarrow 6^p+2^{p+q}\cdot3^r+2^q=332⟶6
p
+2
p+q
⋅3
r
+2
q
=332
or,
\longrightarrow (2\cdot3)^p+2^{p+q}\cdot3^r+2^q=332⟶(2⋅3)
p
+2
p+q
⋅3
r
+2
q
=332
\longrightarrow 2^p\cdot3^p+2^p\cdot2^q\cdot3^r+2^q=332⟶2
p
⋅3
p
+2
p
⋅2
q
⋅3
r
+2
q
=332
Divide both sides by 2^q.2
q
.
\longrightarrow\dfrac{2^p\cdot3^p+2^p\cdot2^q\cdot3^r+2^q}{2^p}=\dfrac{332}{2^p}⟶
2
p
2
p
⋅3
p
+2
p
⋅2
q
⋅3
r
+2
q
=
2
p
332
\longrightarrow3^p+2^q\cdot3^r+2^{q-p}=\dfrac{332}{2^p}\quad\quad\dots(1)⟶3
p
+2
q
⋅3
r
+2
q−p
=
2
p
332
…(1)
Case 1:- Assume p=q,p=q, then q-p=0.q−p=0.
So (1) becomes,
\longrightarrow3^p+2^q\cdot3^r+1=\dfrac{332}{2^p}⟶3
p
+2
q
⋅3
r
+1=
2
p
332
In the LHS, the terms 3^p3
p
and 11 are odd and 2^q\cdot3^r2
q
⋅3
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=1p=q=1 then RHS will be \dfrac{332}{2}=166
2
332
=166 which is even. So,
\longrightarrow3+2\cdot3^r+1=166⟶3+2⋅3
r
+1=166
\longrightarrow2\cdot3^r=162⟶2⋅3
r
=162
\longrightarrow3^r=81⟶3
r
=81
\longrightarrow r=4⟶r=4
So \underline{(p,\ q,\ r)=(1,\ 1,\ 4)}
(p, q, r)=(1, 1, 4)
is possible.
Take p=q=2p=q=2 then RHS will be \dfrac{332}{4}=83
4
332
=83 which is odd. So this is not possible.
For higher values of p,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,p<q, then q-p > 0.q−p>0.
In the LHS of (1), 3^p3
p
is odd, and 2^q\cdot3^r2
q
⋅3
r
and 2^{q-p}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}
2
p
332
is odd, then 2^p2
p
must be the highest power of 2 that can exactly divide 332, which is none other than 2^2=4,2
2
=4, as we said earlier, and it implies p=2.p=2.
Then (1) becomes,
\longrightarrow3^2+2^q\cdot3^r+2^{q-2}=\dfrac{332}{2^2}⟶3
2
+2
q
⋅3
r
+2
q−2
=
2
2
332
\longrightarrow9+2^q\cdot3^r+2^{q-2}=83⟶9+2
q
⋅3
r
+2
q−2
=83
\longrightarrow2^q\cdot3^r+2^{q-2}=74⟶2
q
⋅3
r
+2
q−2
=74
Divide both sides by 2^{q-2}.2
q−2
.
\longrightarrow\dfrac{2^q\cdot3^r+2^{q-2}}{2^{q-2}}=\dfrac{74}{2^{q-2}}⟶
2
q−2
2
q
⋅3
r
+2
q−2
=
2
q−2
74
\longrightarrow4\cdot3^r+1=\dfrac{74}{2^{q-2}}⟶4⋅3
r
+1=
2
q−2
74
Here the LHS is odd (even + odd = odd), so would the RHS be. Then 2^{q-2}2
q−2
must be the highest power of 2 that can exactly divide 74, which is none other than 2^1=2,2
1
=2, and it implies q=3.q=3. Also 2 < 3.
Then,
\longrightarrow4\cdot3^r+1=\dfrac{74}{2^1}⟶4⋅3
r
+1=
2
1
74
\longrightarrow4\cdot3^r+1=37⟶4⋅3
r
+1=37
\longrightarrow4\cdot3^r=36⟶4⋅3
r
=36
\longrightarrow3^r=9⟶3
r
=9
\longrightarrow r=2⟶r=2
So \underline{(p,\ q,\ r)=(2,\ 3,\ 2)}
(p, q, r)=(2, 3, 2)
is possible.
Case 3:- Let p > q,p>q, then q-p < 0.q−p<0.
In this case, in the LHS of (1), the terms 3^p3
p
and 2^q\cdot3^r2
q
⋅3
r
are integers but 2^{q-p}2
q−p
is a decimal number. So the RHS should have the same decimal part as that of 2^{q-p}.2
q−p
.
Then in RHS \dfrac{332}{2^p},
2
p
332
, 2^p2
p
must exceed the highest power of 2 that can exactly divide 332, i.e., 2^p > 2^2.2
p
>2
2
.
Take p=3.p=3. Then (1) becomes,
\longrightarrow3^3+2^q\cdot3^r+2^{q-3}=\dfrac{332}{2^3}⟶3
3
+2
q
⋅3
r
+2
q−3
=
2
3
332
\longrightarrow27+2^q\cdot3^r+2^{q-3}=41.5⟶27+2
q
⋅3
r
+2
q−3
=41.5
\longrightarrow2^q\cdot3^r+2^{q-3}=14.5⟶2
q
⋅3
r
+2
q−3
=14.5
Here 2^{q-3}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=2.2
q−3
=2
−1
=0.5⟹q=2.
So,
\longrightarrow2^2\cdot3^r+0.5=14.5⟶2
2
⋅3
r
+0.5=14.5
\longrightarrow4\cdot 3^r=14⟶4⋅3
r
=14
\longrightarrow3^r=3.5⟶3
r
=3.5
There is no value for rr to satisfy this equation. So this is not possible.
For higher values of p,\ 2^{q-p}p, 2
q−p
and \dfrac{332}{2^p}
2
p
332
would not have same decimal part, since \left[\dfrac{332}{2^3}\right]=\left[41.5\right]=41[
2
3
332
]=[41.5]=41 is odd.
Finally,
\longrightarrow\underline{\underline{p\in\{1,\ 2\}}}⟶
p∈{1, 2}
Hence the sum of possible values of pp is 3.
Answer:
Step-by-step explanation:
is the answer