5^3x + 5^2y + 5z + p = 264, where x, y, z and p are all positive integ

Since we're given that all of x, y, z and p are +ve integers, we must assign them a minimum value of 1 each. Doing so we get the LHS quantity as 125+25+5+1=156

Since 264-156=108 which is less than 125, we know that x cannot increase in count by 1, thereby making the maximum and the only possible value of x=1.

Now we have 25y+5z+p=139

The maximum value that 5z+p can take is 5*5+5=30 which means that 25y must take a minimum value of 139-30=109 or that y must take a minimum value of 109/25=4.36 but since y is an integer and can take a maximum value of 5, we get y=5

Now x=1, y=5, and 5z+p=14

Since p is a +ve quantity, we know that p must at least be 1 which means that z can be a maximum of 13/5=2.6 and since we also know that p can at max be 5, z must be a minimum of 9/5=1.8

This leaves us with the only feasible integral value of z as 2, and subsequently value of p=4

Overall, we get
x = 1
y = 5
z = 2
p = 4

Giving us x+y+z+w = 12, hence C.

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