AIMO 2015 Q3

Letting $w,d,t$ represent the number of whatsits, doovers and thinghys, respectively, we have that

• (i) $3w + 7d + t = 329$ and

• (ii) $4w + 10d + t = 441.$

Now as we have three variables and only two (linearly independent) equations there will be multiple positive integer solutions for $(w,d,t).$ However, this does not rule out the possibility of finding a unique value for $w + d + t.$ To this end, we need to find values $a,b$ such that

$a*(3w + 7d + t) + b*(4w + 10d + t) = w + d + t \Longrightarrow (3a + 4b)w + (7a + 10b)d + (a + b)t = w + d + t.$

Equating like coefficients, we require that $3a + 4b = 1, 7a + 10b = 1$ and $a + b = 1.$

Then $3a + 4b - 3(a + b) = 1 - 3 = -2 \Longrightarrow b = -2, a = 3,$ which also satisfies $7a + 10b = 1.$

Thus $w + d + t = 3*329 - 2*441 = \boxed{105}.$

Possible positive integer solutions for $(w,d,t)$ include $(22,30,53), (52,20,33)$ and $(82,10,13).$ (There are in fact $34$ such solutions, ranging from $(1,37,67)$ to $(100,4,1)$ .)

$3w + 7d + t = 329$ ..........(1)

$4w + 10d + t = 441$ ..........(2)

(2) - (1): $w + 3d = 112$ ..........(3)

(1) - 2*(3): $w + d + t = 105$