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What is the number of ways of forming 6 digit number whose sum of digits is even?
Wiki User
∙ 11y ago
I look at it this way:
-- There are 90,000 numbers with five digits. (100,000 minus 10,000 with less than 5 digits)
-- If the sum of the first 5 digits is odd, it can be made even by using 1, 3, 5, 7, or 9
for the 6th digit ... 5 choices.
-- If the sum of the first 5 digits is even, it remains even by using 2, 4, 6, 8, or 0
for the 6th digit ... 5 choices.
-- So, for any one of the 90,000 five-digit numbers, there are 5 choices for the 6th digit
that result in an even sum of all six.
-- So, there must be 450,000 suitable 6-digit numbers.
Wiki User
∙ 11y ago
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992
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