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Standard XII

Mathematics

Composite Function

Let m,n,p ∈ℕ ...

Question

Let $m, n, p \in N$ with $1 \leq m \leq 100;$ $1 \leq n \leq 50;$ $1 \leq p \leq 25.$ The number of possible ordered triplets $(m, n, p)$ such that $2^{m} + 2^{n} + 2^{p}$ is divisible by $3$ is $5 k .$ Then $k$ is

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Solution

$2^{m} + 2^{n} + 2^{p}$
$= (3 - 1)^{m} + (3 - 1)^{n} + (3 - 1)^{p}$
$= 3 u + (- 1)^{m} + (- 1)^{n} + (- 1)^{p},$ where $u \in N$
So, $2^{m} + 2^{n} + 2^{p}$ is divisible by $3$ if $m, n, p$ all are odd or all are even.
Number of possible ordered triplets
$= 50 \times 25 \times 13 + 50 \times 25 \times 12$
$= 50 \times 25 (12 + 13)$
$= 50 \times 625 = 5 \times 6250 = 5 \times k$
$∴ k = 6250$

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Let m,n,p∈N with 1≤m≤100; 1≤n≤50; 1≤p≤25. The number of possible ordered triplets (m,n,p) such that 2m+2n+2p is divisible by 3 is 5k. Then k is

2m+2n+2p =(3−1)m+(3−1)n+(3−1)p =3u+(−1)m+(−1)n+(−1)p, where u∈N So, 2m+2n+2p is divisible by 3 if m,n,p all are odd or all are even. Number of possible ordered triplets =50×25×13+50×25×12 =50×25(12+13) =50×625=5×6250=5×k ∴k=6250

Let $m, n, p \in N$ with $1 \leq m \leq 100;$ $1 \leq n \leq 50;$ $1 \leq p \leq 25.$ The number of possible ordered triplets $(m, n, p)$ such that $2^{m} + 2^{n} + 2^{p}$ is divisible by $3$ is $5 k .$ Then $k$ is