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Bunuel
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Bunuel
What is the smallest possible sum of nonnegative integers a, b, and c such that 36a + 6b + c = 173?

A. 5
B. 7
C. 13
D. 16
E. 18

OFFICIAL SOLUTION



In order to minimize the total value of a, b, and c, we should make a as large as possible -- since we'll simply reach 173 much faster moving in chunks of 36 than we would by moving in chunks of 6 or 1.

Since 173÷36=4 remainder 29, we will choose a=4 and move on to consideration of b and c.

In the same manner, we will now maximize b, since moving in chunks of 6 is much more efficient than moving in chunks of 1.
Since 29÷6=4 remainder 5, we will choose b=4. At this point we will be left to choose c=5 to cover the remaining value.

Thus the minimum possible value of a+b+c is 4+4+5=13, and answer C is correct.
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My approach:
Write a list of all of the multiples of:

36 - 36, 72, 108, 144
6 - 6, 12, 18, 24, 30, 36, 42, 48
1 - 1, 2, 3, 4, 5

Look at the numbers for a while.

You would observe that 144 + 24 + 5 = 173

Therefore a is 4, b is 4 and c is 5

Add up all numbers together 4 + 4 + 5 = 13

Which is answer C
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My approach :
6(6a+b)+c = 173 => the nearest multiple of 6 for the number 173 is 168
therefore the expression can be expressed as 6[6{4}+4]+5 => a=4, b=4 and c=5. these are the least values of a,b and c that satisfy the equation.
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