Alfred_F

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1 years, 56 days

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These are questions asked by Alfred_F

We are looking for the smallest natural number n with the property that both the digit sum Q(n) of the number n and the digit sum Q(n + 1) of the successor of n are divisible by 5.

During a birthday party, the birthday child realizes: In 1968, I was the same age as the sum of the digits of my birth year. How old will I be now at the end of 2025?

(Please no AI solution)

In the decimal system, specify the smallest natural number k that begins with the digit 7 and has the following additional property:
If you delete the first digit 7 and write it at the end, the newly created number z = (1/3)*k.

I would like to experiment with error estimation in the symbolic solution of ordinary differential equations. I've written a simple example in the attached file. I would now like to plot both the left side of the ODE and the solution y(x) together in the same coordinate system. I can't do this, and I'm asking for help. How can the cumbersome numerical terms in the solution y(x) be converted to floating-point numbers?

As I said, this is a recreational experiment ;-) .DGL_test.mw

restart

ode := x^2 = sum(x*(diff(y(x), [`$`(x, i)]))/factorial(i), i = 0 .. 3)

x^2 = x*y(x)+x*(diff(y(x), x))+(1/2)*x*(diff(diff(y(x), x), x))+(1/6)*x*(diff(diff(diff(y(x), x), x), x))

(1)

``

``

ics := y(0) = 0, (D(y))(0) = 0, ((D@@2)(y))(0) = 0

y(0) = 0, (D(y))(0) = 0, ((D@@2)(y))(0) = 0

(2)

dsolve({ics, ode})

y(x) = x-1+((1/12)*(2-2^(1/2))*(1+2^(1/2))^(2/3)+(1/12)*2^(1/2)*(1+2^(1/2))^(1/3)+1/3)*exp(((1+2^(1/2))^(2/3)*2^(1/2)-(1+2^(1/2))^(2/3)-(1+2^(1/2))^(1/3)-1)*x)+((1/12)*(2^(1/2)-2)*(1+2^(1/2))^(2/3)-(1/12)*2^(1/2)*(1+2^(1/2))^(1/3)+2/3)*exp(-(1/2)*((1+2^(1/2))^(2/3)*2^(1/2)-(1+2^(1/2))^(2/3)-(1+2^(1/2))^(1/3)+2)*x)*cos((1/2)*3^(1/2)*(1+2^(1/2))^(1/3)*(2^(1/2)*(1+2^(1/2))^(1/3)-(1+2^(1/2))^(1/3)+1)*x)-3^(1/2)*(-1+(2^(1/2)-1)*(1+2^(1/2))^(1/3))*2^(1/2)*exp(-(1/2)*((1+2^(1/2))^(2/3)*2^(1/2)-(1+2^(1/2))^(2/3)-(1+2^(1/2))^(1/3)+2)*x)*sin((1/2)*3^(1/2)*(1+2^(1/2))^(1/3)*(2^(1/2)*(1+2^(1/2))^(1/3)-(1+2^(1/2))^(1/3)+1)*x)/((1+2^(1/2))^(2/3)*(12*2^(1/2)-12))

(3)

NULL``

Download DGL_test.mw

...but it's easily solvable with the help of a trick. My main concern is the path to the solution for the attached Diophantine equation. I was able to solve the problem both with pen and paper and then in Maple. There's certainly a more elegant way. I'm particularly interested in a special Maple command. However, I don't want to ask for it here yet, as it would give a hint of the trick and spoil the fun of solving the puzzle.

 

419*x^2+116*x*y-426*x*z+78*y^2-142*y*z+133*z^2-1604*x-682*y+1086*z+2306 = 0

NULL

Download Diophant.mw

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