Let pqr be a three digit number. Then, pqr + qrp + rpq will not be always divisible by: 9 37 3 p + q + r
Question
Let pqr be a three digit number. Then, pqr + qrp + rpq will not be always divisible by:
- 9
- 37
- 3
- p + q + r
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Answer
The correct option is A (9).
Given a three-digit number $$ pqr $$, we can express the permutations as follows:
$$ \begin{aligned} & pqr = 100p + 10q + r \ & qrp = 100q + 10r + p \ & rpq = 100r + 10p + q \end{aligned} $$
To find the sum of these permutations, we add them together:
$$ \begin{aligned} pqr + qrp + rpq &= (100p + 10q + r) + (100q + 10r + p) + (100r + 10p + q) \ &= 100p + 10q + r + 100q + 10r + p + 100r + 10p + q \ &= 111p + 111q + 111r \ &= 111(p + q + r) \end{aligned} $$
Since we have:
$$ 111(p + q + r) = 3 \times 37 \times (p + q + r) $$
This implies that $$ pqr + qrp + rpq $$ is always divisible by 3, 37, and (p + q + r).
However, this sum is not guaranteed to be divisible by 9 unless $$ (p + q + r) $$ is divisible by 9. Thus, the correct answer is 9.
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