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# help

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Find the number of 10-digit numbers where the sum of the digits is divisible by 5.

Jan 16, 2020

#1
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The answer is 4,000,000. Jan 16, 2020
#2
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%%time
# Efficient Python 3 program to sum up the #
# multiples of digit n in a range of 2 - 10-digit integers #
# E.G. "What is the total sum of all 3-digit integers that are
#  mutiples of 7?". Answer =70,336 #

# find the Sum of having n digit
# and divisible by the number
def totalSumDivisibleByNum(digit, number):

# compute the first and last term
firstnum = pow(10, digit - 1)
lastnum = pow(10, digit)

# first number which is divisible
# by given number
firstnum = (firstnum - firstnum % number)+number

# last number which is divisible
# by given number
lastnum = (lastnum - lastnum % number )

# total divisible number
count =int ((lastnum - firstnum) / number+1  )
print("Total count =", f"{count:,d}")
# return the total sum
return int(((lastnum + firstnum) * count) / 2)

# Driver code
digit =10 ; num = 5

print("Total Sum =", f"{totalSumDivisibleByNum(digit, num):,d}")

Total count = 1,800,000,000
Total Sum = 9,900,000,002,700,000,000
Wall time: 0 ns

Jan 16, 2020
#3
+2

Find the number of 10-digit numbers where the sum of the digits is divisible by 5.

The first digit cannot be zero, so 9 ways, the next 8 digits can be anything, so there are 10 ways for each of them.
We can then only choose two digits as last digit to satisfy the condition.

sum = $$9\times 10^8 \times 2 = 1~ 800~ 000~ 000$$

$$\begin{array}{|c|c|} \hline \text{the sum of the digits 1 until 9} & \text{last digit to satisfy the condition }\\ \hline \ldots 0 & 0~ \text{ or } ~5 \\ \ldots 1 & 4~ \text{ or } ~9 \\ \ldots 2 & 3~ \text{ or } ~8 \\ \ldots 3 & 2~ \text{ or } ~7 \\ \ldots 4 & 1~ \text{ or } ~6 \\ \ldots 5 & 0~ \text{ or } ~5 \\ \ldots 6 & 4~ \text{ or } ~9 \\ \ldots 7 & 3~ \text{ or } ~8 \\ \ldots 8 & 2~ \text{ or } ~7 \\ \ldots 9 & 1~ \text{ or } ~6 \\ \hline \end{array}$$

Example: 1 000 000 08(1) or 1 000 000 08(6)

4 567 123 98(0) or 4 567 123 98(5) Jan 17, 2020