If A is the sum of the digits of 4444^4444 and B is the sum of digits of A, then find the sum of the digits of B.
Brainliest Answer will be marked....
Answers
Answer:
12 ≥ sum of the digits of B ≥ 1
Step-by-step explanation:
we can't exactly find the value of B, but we can find an estimate...
4444⁴⁴⁴⁴ < 10000⁴⁴⁴⁴
=> the number of digits in 4444⁴⁴⁴⁴ < the number of digits in 10000⁴⁴⁴⁴
=> the number of digits in 4444⁴⁴⁴⁴ < 4×4444 = 17776 (10000⁴⁴⁴⁴ has this many zeroes)
=> the number of digits in 4444⁴⁴⁴⁴ cannot exceed 17776
=> the sum of digits in 4444⁴⁴⁴⁴ cannot exceed 9×17776 = 159984 (since 9 is the biggest possible number)
=> the value of A cannot exceed 159984
number less than A which has the highest possible value of sum of digits = 99999
=> the highest possible sum = 5×9 = 45
=> the value of B cannot exceed 45
number less than 45 which has the highest possible sum of digits = 39
=> the sum of the digits in B cannot exceed 9+3 = 12
since the sum of digits cannot be negative or zero, the sum must be ≤12 and ≥1
we have been using so many 9s because 9 is the greatest of all single digits... and we should be happy that we got an estimate at least because the true value of 4444⁴⁴⁴⁴ is equal to...
4444⁴⁴⁴⁴ = 5103632503725508048204025019552439592478247548228451474782897598673947907686253161583300186402719787543417249678221281036093699881487102802328040951760428970423299725005981765145203390256498105347434214398243447701005507790297138587957809702679504701392721044806973761259973536289419464535239969236107882490228631837484556691021182679299509834007675026127520738056469675750825886610949248900802705474568038737403208649455941287069612... and I couldn't send more in 5000 words...
Please mark me as the brainliest
