I'm trying to solve the following problem:
for a^b when 0 < a,b < 100 I want to find the maximum sum of digits of the generated numbers. For example 9^3=729, sum=7+2+9=18.
I have the following code:
clear all clc digits(1000) tot=0; for a=99:-1:1 for b=99:-1:1 Num=vpa(sym(a^b)); dig=double(1+floor(log10(Num)+eps)); Num=Num/10^dig; sumA=0; for i=1:dig+1 sumA = sumA+(floor(Num*(10^(i-1)))-10*floor(Num*10^(i-2))); end temp=double(sumA); if temp == 972 disp(a) disp(b) end end end
Where I found that 88^99 yields the max sum of digits 978, but it's the wrong answer. (correct answer is 972). Does anyone find the error?
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Does sym(a^b) reply a symbolic variable? I'd assume, that at first a^b is calculated as double with the corresponding rounding errors. What about sym('a^b')? I do not have the required toolbox, such that this is a guessing only.
Well, I'd use my vpi tool. The problem is, the number is far too large for all those digits to be represented using a standard floating point form like a double.
(Actually, I'd use the replacement for it, that is essentially written, subject to heavy use for testing.)
vpi(88).^99 ans = 31899548991064687385194313314353745484864573065650712770111884048604 753593728365505650462765416702028265157186333205198215936166634716861519 60018780508843851702573924250277584030257178740785152
As a test, compare the symbolic toolbox:
sym(88)^99 ans = 3189954899106468738519431331435374548486457306565071277011188404860475359372836550565046276541670202826515718633320519821593616663471686151960018780508843851702573924250277584030257178740785152
Yep, they agree down to the last digit. The sum of those digits is 847.
sum(digits(vpi(88).^99)) ans = 847
Or, do it using the symbolic TB.
sum(char(sym(88)^99)-'0') ans = 847
Either works, and yields 847. See that I was careful to do the exponentiation on either a symbolic starting point or a vpi start. Otherwise, MATLAB forms the result 88^99 as a double, THEN tries to convert it to a high precision result. Of course that must fail.
So how do we find the maximum sum? This seems to work, using vpi.
[A,B] = ndgrid(1:99); A = vpi(A); B = vpi(B); C = A.^B; CC = mat2cell(C,ones(1,99),ones(1,99));
sumfun = @(N) sum(digits(N)); dsum = cellfun(sumfun,CC); [S,loc] = max2(dsum)
S = 972
loc = 99 95
So we learn that 99^95 is
vpi(99).^95 ans = 38489607889348486119277958028245967896084511560873660346586279535301 481260085342580322673837686274870946109685542866926973747267258531956576 79460590239636893953692985541958490801973870359499
With a sum of digits:
sum(digits(vpij(99).^95)) ans = 972
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