Question

The digits p, q and r of a three-digit natural number pqr satisfies the condition 49p + 7q + r = 286. Find the sum of the digits of pqr.

• A. 16
• B. 18
• C. 10
• D. 9

Answer 1

The correct option is A

16

Since 49p+7q+r = 286
Or $7^{2}p+7^{1}q+7^{0}r=286$
In other words, this question is the conversion of 286 in base 10 to base 7.
pqr in the equivalent of 286 in base $7=(pqr{)}_7=(556{)}_7$

Thus, the sum of the digits = 16.