Well there is. Just find any integer less than 100 that is a 1, 2, 3 or have their prime factorization be 2^a * 3^b.
You can do easy counting from there. No need to brute force.
Say the perimiter of DGV is P, then from the given, we know that FPM's Perimiter : DGV's Perimiter = 8:5
Substituing in, we have 72:P = 8:5. In the scale factor 8:5, we know that each unit is 9 for FPM, likewise, each unit for 5 would also be nine. In other words, solving it as an equation 72/p = 8/5 you obtain p = 45.
Perimiter of DGV is 45 units.
a + b + c + d + e + f + g + h + i + j = 10
a, b, c, d, e, f, g, h, i, j = 1, 2 or 3.
Try stars and bars with removal of some cases. Brute forcing it will take too long and isn't smart
prime numbers greater than 4 are 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
you can count from there
A similar question has been asked here: (It should help formulate the ideas to solve this problem)
https://web2.0calc.com/questions/help-counting_85243#r1
Values T > 21 work, and then Max value is 28.
So 22 - 28 work.
That is 7 values.
When graphed on desmos, it turns out any real value of x works.
Well why? Because all the values of x after applying zero product property are imaginary, meaning the parabola is above the x-axis, hence the y values are always positive. Positive x Positive = Positive, so any real value of x will obtain >=0
\((2^{3})^{1\over6}\cdot2^x=32^{1\over2}\)
\(2^{1\over2}\cdot2^x=32^{1\over2}\)
\(2^{{1\over2}+x}=2^{6\cdot{1\over2}} = 2^3\)
\({1\over2}+x=3\)
\(x={5\over2}\)
Children a, b, c, and 2t need 7 identical candies:
Case 1: t = 0
a + b + c = 7, nonnegative a and b and c: stars and bars = 9 choose 2 = 36
Case 2: t = 1
a + b + c = 5, nonnegative a and b and c: stars and bars = 7 choose 2 = 21
Case 3: t = 2
a + b + c = 3, nonnegative a and b and c: stars and bars = 5 choose 2 = 10
Case 4: t = 3
a + b + c = 1, nonnegative a and b and c: stars and bars = 3 choose 2 = 3
36 + 21 + 10 + 3 = 70 total ways.
Given BE is 5, and CD is 3, we can tell that BED is a 3 - 4 - 5 triangle, so BC = 1. Via a similar translation, AGC is congruent to BED. By the same logic, GFE is congruent to AHB. The ratio of GF to FE is BC to CintersectionBE. Which is 4:3. Given GE is 3, and GFE is a 3-4-5 triangle, FE is 1.8 units long, which is equal to BH which in mixed number, is 1 and 4/5.