Reversed Checks
Five friends went to the bank to cash their weekly paychecks. Unfortunately, the
bank teller was quite confused that day and swapped the dollars and cents on
everyone's check. The friends all stopped at the corner store and made a
purchase before noticing the mistakes.
- Andy bought a 5 cent piece of gum and then found he had exactly twice as much
as his original check.
- Bill bought a 5 cent piece of gum and then found he had exactly half as much
as his original check.
- Carol bought an 11 cent piece of taffy and then found she had exactly twice as much
as her original check.
- Doris bought an 11 cent piece of taffy and then found she had exactly half as much
as her original check.
- Emily bought a 75 cent comic book and then found she had an even number of dollars
(no coins.)
After sharing their discoveries, the five friends realized that as a group, they had
neither lost nor gained any money from the teller's errors. What was the original amount on each check?
Source: Based on prior puzzle experience, reminded by
Dan's Problem of the
Week
#212.
Solutions were received from Al Zimmermann, Jeremy Galvagni, Dan Chirica,
Denis Borris, Radu Ionescu, Joseph DeVincentis, David Bachtel, Philippe
Fondanaiche, John Hewson, Kirk Bresniker, and Kitsuki Ikeda. Joseph's
solution follows. Several solvers were concerned that the actual number
of Emily's dollars ended up being an "odd" number. I apologize; I was
using the term "even" to mean non-fractional, but it confused matters.
If you change Emily's purchase to 74 cents, the confusion goes away (her
check becomes $74.90.)
Let A,B,C,D,E be the dollars from the checks and a,b,c,d,e be the cents for Andy, Bill, etc. respectively. The amounts they should have received were 100A+a, 100B+b etc. cents, but they actually got 100a+A,
100b+B, etc. a,b,c,d,e are limited to the range 0 to 99.
Each of the purchases and money left over gives us an equation on that person's check.
100a+A-5 = 2*(100A+a)
100b+B-5 = (100B+b)/2
100c+C-11 = 2*(100C+c)
100d+D-11 = (100D+d)/2
100e+E-75 = 100k or E=75 with no further constraint on e.
Solve the first four equations:
98a = 199A + 5
a = (199A + 5)/98 = 2A + (3A + 5)/98
3A + 5 must be a multiple of 98, so 3A = 93 mod 98, and A =31, so a = 63.
199b - 10 = 98B
B = (199b - 10)/98 = 2b + (3b-10)/98
3b-10 must be a multiple of 98 so 3b = 10 mod 98, b = 36, B=73.
98c = 199C + 11
c = 2C + (3C + 11)/98
3C = 87 mod 98 so C = 29, c = 59.
199d - 22 = 98D
D = 2d + (3d - 22)/98
3d = 22 mod 98, so d = 40, D=81.
In each case, the next larger solution to the mod 98 equation makes a,b,c,d too large,
so these are the only solutions. So the first four checks were for 31.63, 73.36, 29.59,
and 81.40. These checks totalled 215.98 but they received 200.14 collectively.
If the five got the correct amount in total from the teller, Emily got 15.84 too much.
Since she actually received e dollars and 75 cents, she should have gotten e-16
dollars and 91 cents, so e is 91 and her check was for 75.91.
Mail to Ken