Question

Using the digit 2 and 7 and addition or subtraction operations only, the number 2010 is written. The maximum number of 7 that can be used, so that the total numbers used is a minimum is
A. 284
B. 286
C. 288
D. 290

Answer 1

Hint: We try to reach as close as possible to the number 2010 with 7. Then we choose to delete one seven to compensate the rest with twos. We add up the numbers to find the final solution.

Complete step-by-step solution:
It’s given that only using the digit 2 and 7 and addition or subtraction operations only, we create the number 2010.
We have to maximize the use of number 7 keeping in mind that use of total numbers of 2 and 7 is minimum.
We have to add 7 that much number of times such that the addition doesn’t cross 2010.
It means we have to find the nearest number to 2010 (less than 2010) by 7.
We now use the long division of 2010 by 7.
$7\overset{287}{\overline{\left){\begin{align}
  & 2010 \\
 & \underline{14} \\
 & 61 \\
 & \underline{56} \\
 & 50 \\
 & \underline{49} \\
 & 1 \\
\end{align}}\right.}}$
We have 287 as the quotient and 1 as remainder.
This means the number that is divisible by 7 is $2010-1=2009$.
To reach 2009 we need to add 7 in total 287 number of times. The rest we have to complete by 2 but that’s not possible.
Now we can reduce one 7, then the number becomes $2009-7=2002$ and to reach 2010, we need to add four twos. So, the total number increases $4-1=3$.
We can also add one more 7, then the number becomes $2009+7=2016$ and to reach 2010, we need to subtract three twos. So, the total number increases $3+1=4$.
To minimize the use of total numbers of 2 and 7, we apply the first step.
This way we take $287-1=286$ number of sevens and four twos.
Total number is $286+4=290$. The correct option is D.

Note: The main condition of maximizing the use of number 7 keeping in mind the use of total numbers of 2 and 7 as minimum gives the scope of long division of 2010 with 7. The more we use two, the greater the number becomes which we have to minimize.