(A)Consider the set of all possible positive quaternary (base 4) integers each having exactly twelve digits consisting of precisely three 1's, three 2's, three 3's and three 0's. The first digit cannot be 0. Determine the sum of all these numbers in the decimal notation.
(B)Consider the set of all possible positive quinary (base 5) positive integers each having exactly fifteen digits consisting of precisely three 1's, three 2's, three 3's, three 4's, and three 0's. The first digit cannot be 0. Determine the sum of all these numbers in the base ten (decimal) notation.
A: 2889036460800
B: 2456066894465184000
First practice on a smaller problem, verify with a program to make sure the logic is correct.
start with [0,0,1,1,2,2,3,3]; 8 digits
avg 1st digit is 2, avg other digits are all 10/7 (remove a 2 from set of integers)
Total number by computer progra is 1890
Sums of each digit individually, by program is:
[3780, 2700, 2700, 2700, 2700, 2700, 2700, 2700]
how to get 1890?
3 ways to have first digit. Then you have 7 more to allocate from groups of 2,2,2, and 1
3*7!/(2!2!2!1!)
The average 8 digit number is 2*4^7 + (16/11)*sum of powers of 4 from zero to 6
4^7 is 16384
sum 4^0 + ... + 4^6 = (4^7 - 1)/(4-1) = 5461
(2*16384 + (10/7)*5461) * 1890 = 76676220
print out from program ..... 76676220 it checks
(A) The average of the first digit is 2.
The average of any other digit is sum([0,0,0,1,1,1,2,2,3,3,3])/11
The average 12 digit number is 2*4^11 + (16/11)*sum of powers of 4 from zero to 10
2*4^11 + (16/11)* (4^11 - 1)/(4 - 1)
2*4194304 + (16/11)*1398101 is the average 12 digit number
How many of them are there?
The start with either 1,2, or 3, so there are 3 ways for the first digit
Then you have 11 more to allocate from groups of 3,3,3, and 2
3*11!/(3!3!3!2!)
3 * 39916800 / 432 = 277200 different 12 digit numbers.
So the total sum is
(2*4194304 + (16/11)*1398101 ) * 277200 = 2889036460800
(B) The average of the first digit is 2.5
The average of any other digit is
the sum of all digits (30) minus 2.5 is 27.5; there are 14 more digits.
So the average "non first" digit is 27.5/14
The average 12 digit number is 2.5*5^14 + (27.5/14)*sum of powers of 5 from zero to 13
2.5*5^14 + (27.5/14) * (5^14 - 1)/(5 - 1)
2.5 * 6103515625 + (27.5/14)*1525878906 is the average 15 digit number
How many of them are there?
The start with either 1,2,3, or 4, so there are 4 ways for the first digit
Then you have 14 more to allocate from groups of 3,3,3,3, and 2
4*14!/(3!3!3!3!2!)
4 * 87178291200 / 2592 = 134534400 different 15 digit numbers.
So the total sum is
( 2.5 * 6103515625 + (27.5/14)*1525878906 ) * 134534400 = 2456066894465184000
---- code below is for my "practice" version ----
from itertools import permutations
count = 0
allofthem = []
placesums = [0 for i in range(8)]
for perm in permutations([0,0,1,1,2,2,3,3]):
if perm[0] == 0:
continue
if perm in allofthem:
continue
allofthem.append(perm)
for i,v in enumerate(perm):
placesums[i] += v
count += 1
# print(perm)
print(count)
print(placesums)
runningsum = 0
for a in allofthem:
x = ''.join([str(d) for d in a])
runningsum += base2base(x, 4, 10)
print('runningsum is ', runningsum)
ccount = 0
for i in range(7):
ccount += 4**i
print(ccount)
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Posted by Larry
on 2023-07-01 12:29:52 |
A and B, analytic backed by program
