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How do you find the answer?Maureen likes to play a game in which she reduces a number to a single digit. She adds the digits of the number together. When the total is still greater than nine, she adds the digits of the total together and continues in this way until she ends up with a single digit number. If Maureen does this for each of the integers from one to 100, how many times will she end up with a final result equal to one?


I am not sure if this helps you find the answer, but this process is used in the divisibility rules for 3 and 9.

If you add the digits up to a single digit number that it 3, 6, or 9, the number is divisible by 3.

If you add up the digits to the number 9, then the number is divisible by 9.


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Garnet H. | Calculus, Physics, Stat, Linear Algebra, GRE, GMAT, SAT... TutoringCalculus, Physics, Stat, Linear Algebra,...
5.0 5.0 (904 lesson ratings) (904)
check the number of digits in the number you have 3 possibilities: 1-digit numbers 2-digit numbers 3-digit numbers the only 3-digit candidate you have is 100 and that results in a 1, so this is an easy case for a 1-digit number the only way for the digits to add to 1 is for that number to be = 1 now for a 2-digit number the biggest possible sum is 18 (in the case of 99) and the smallest is 1 (in the case of 10). only way for the digits to add up to 1 is for the digits to either add up to 1 or 10. the numbers that work are: 10, 19, 28, 37, 46, 55, 64, 73, 82, 91 adding 2 (for 1 and 100) to the total gives 12
Kevin S. |
5.0 5.0 (4 lesson ratings) (4)

Since she's dealing with (mostly) two digit numbers, then look at the numbers where the digits total 10 (1 + 0 = 1)

1, 10, 19, 28, 37...

If you look at the pattern, it's 1 + 9n where 0<=n<=11 (1 + 99 = 100; 1 + 0 + 0 = 1)

So it's 12 times .