Father of the Brides
Math brain teasers require computations to solve.
The wedding reception was winding down as Bill Johnson, the father of the bride, held forth to anyone who was left to hear his report on the fiscal damages.
"That's the last one married off  thank goodness! That's been one wedding a year for the last four years I've had to pay for. Alice got married three years ago, then Barbara, Cheryl, and now Denise. They've all gotten married on the same day, June 30th, for some reason, and each girl had to have her three sisters as the bridesmaids, but could they share a color scheme? No  each one had to have her own colors. God forbid they should be able to reuse the bridesmaids' dresses, let alone share a wedding dress. Alice tried to persuade Denise to reuse her color scheme, but Denise said she had lived with her older sister's handmedowns long enough.
Funny thing, though  each bride managed to get married when her age was just the same as the sum of the digits of the ages of her three sisters. You don't see that every day, I bet. But why couldn't they share something that costs money instead of this trivia?"
Bill didn't mention how old his daughters are, but he's given us enough information to work out all four ages. How old are they?
Hint1. Any nonnegative integer is congruent to the sum of its digits, mod 9. (They have the same remainder when divided by 9.)
2. Each daughter was between 17 and 33 at the time of her wedding.
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Answer
Alice is 27, Barbara is 28, Cheryl is 20, and Denise is 21, 21 = 2+7+2+8+2+0.
A year ago, they were 26, 27, 19, and 20, 19 = 2+6+2+7+2+0.
Two years ago, they were 25, 26, 18, and 19, 26 = 2+5+1+8+1+9.
Three years ago, they were 24, 25, 17, and 18, 24 = 2+5+1+7+1+8.
You can solve this one by trial and error (particularly with the hint), but you can also use the fact that any nonnegative integer is congruent to the sum of its digits, mod 9. If a, b, c, and d are the present ages of the four women, that gives us four linear congruences:
d = a + b + c (mod 9)
c1 = (a1)+(b1)+(d1) (mod 9)
b2 = (a2)+(c2)+(d2) (mod 9)
a3 = (b3)+(c3)+(d3) (mod 9).
Solving for the four variables, we get: a=0, b=1, c=2, d=3 (all mod 9). It remains to find four integers congruent to these numbers that will satisfy the digit addition property. If you use the age restriction in the hint, you can omit some of the following analysis, but I present it here in full to show that the solution is truly unique even without the hint.
If we go back 3 years to Alice's wedding, we have: a3=6, b3=7, c3=8, and d3=9 (mod 9). Now all numbers less than 79 that are congruent to 7 also have a digit sum of 7, and likewise numbers congruent to 8 less than 89 have a digit sum of 8, and numbers congruent to 9 (except 0) less than 99 have a digit sum of 9. We can rule out such high ages, because each woman must have a turn being the sum of her sister's ages' digits, and the maximum sum of digits for three ages less than 199 is 54 (for 99, 99, and 99). Likewise, we can rule out zero as a valid age for Denise three years ago, because she needs to have an age now that can be the sum of the digits of three other valid ages. So Alice three years ago must have been 7+8+9=24, making her 27 now.
Now go back 2 years to Barbara's wedding. Alice we now know was 25. Cheryl's age was congruent to 9 (mod 9), so she must have had a digit sum of 9. Denise's age was congruent to 1 (mod 9), so she must have had a digit sum of 10 unless she was 10 years old herself (we have previously ruled out 1 as a valid age). But if Denise was 10, then Barbara would have been 17 two years ago, and this year Alice and Barbara together would be contributing 2+7+1+9 = 19 years to Denise's age, contradicting the assumption that she was 10 two years ago. So Denise had a digit sum of 10 two years ago, and Barbara was therefore 2+5+9+10=26 at the time of her wedding, hence 28 now.
Now consider Cheryl's wedding last year. Alice was 26 and Barbara was 27. Denise's age was congruent to 2 (mod 9), so she must have had a digit sum of either 2 or 11. If it was 11, then Cheryl would have been 11+2+6+2+7=28 then, making her 29 now. But then Denise would have to be 2+7+2+8+2+9=30 now to Alice's 27, contradicting the information that Alice was her older sister. So Denise's digit sum must have been 2 last year, making Cheryl 19 then, and Denise 21 now.
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