## Gem 17: The Bank Teller's Predicament Puzzle Solution

The Bank Teller's Predicament PuzzleIt was a strange lapse on the part of the bank teller. Evidently he

misread the check, for he handed out the amount of the dollars

in cents, and the amount of the cents in dollars. When the error

was pointed out to him he became flustered, made an absurd

arithmetical mistake, and handed out a dollar, a dime, and a cent

more. But the customer declared that he was still short of his due.

The teller pulled himself together, doubled the total amount he

had already given to the customer, and so precisely settled the

transaction to the latter’s satisfaction.

What was the amount of the check? Note: Here we must solve an equation where we require positive

integer solutions. This requirement places the problem in a large

and famous Class of problems known as Diophantine equations

named for third century Greek mathematician Diophantus. The

concept of zero was not invented yet in the third century.

The Bank Teller's Predicament Puzzle Solution

by Moshe ShweigerThe cents in a check take up two decimals after the decimal point.

The bank teller handed out the amount of cents in dollars, and we

can conclude that the check was in the form XY.ZW where X, Y, Z,

and W are positive integer digits. When the teller doubled the amount

he had already given, that implies that the amount after the teller

added only the amount $1.11 is 1/3 of the amount that was written on

the check. Let's write the check in the most general form of XY.ZW

and figure out what are X, Y, Z, and W. The first handout was ZW.XY

(mixing up the dollars for the cents), and then the teller added $1.11

as shown by the top two rows of Expression 1:Expression 1Z W . X Y

+ 1 . 1 1

--------------

T_{1}T_{2}.T_{3}T_{4 + }S_{1}S_{2 .}S_{3}S_{4 }--------------

X Y . Z W

This implies that Y cannot be the

digit 9, since if it were 9, then T_{4}

would be zero and S_{4}would be

zero as it is twice T_{4}and most

importantly W = T_{4}+S_{4}would be

zero which it cannot be.Since Y+1 = T _{4}and as was stated before, the amount T_{1}T_{2}.T_{3}T_{4}is 1/3

of the total amount, then 3T_{4}= W, 3(Y+1) = W or 3Y+3 = W and thus

Y = (W - 3)/3 equation 1We know that Y and W must be integers from 1 to 9. Also Z must be

less than 4, because if Z is 4 or greater, then T_{1}would also be 4 or

greater. But then X = 3T_{1}would be greater than 9, which it cannot be.

For Y to be a positive integer in equation 1, W must be 9 or 6. But 6

would not satisfy Expression 1 as shown below*. So W=9 and Y=2.* W cannot be 6 because if it were, then Y would

be 1 by Equation 1, and there would be no carryover

to the X column:

+

+

X

1

---

T_{3 }S_{3 }_{--- Z}Let's try for all the values of X from 9 down to

1. For X=9, Z=0; No Good. For X=8, Z=7; No

good, since Z must be less than 4.

For X=7, Z=4; No good, since Z must be

less than 4.

For X=6, Z=1 which we must check:

For X=5, Z=8; No good, since

Z must be less than 4.

For X=4, Z=5: No good, since

Z must be less than 4.16.61

1.11

---------

17.72

35.44

---------

53.16

No good.+ +

For X=3, Z=2, which

we must check:

For X=2, Z=9:

No good, since Z

must be

less than 4.

26.31

1.11

---------

27.42

54.84

---------

82.26

No good.+ +

For X=1, Z=6: No good, since Z must be

less than 4.

We conclude that W cannot be 6. So to satisfy

Equation 1, W=9 as it cannot be 6 or 3.Rewriting Expression 1 we get

Z 9 . X 2

1 . 1 1

-----------

T_{1}T_{2}.T_{3}3

S_{1}S_{2 .}S_{3}6

-------------

X 2 . Z 9X must be 8 or lower since if X=9

then Z would be 0 which cannot be.

So there is no carryover from

X + 1 therefore T_{2}= 0.Z 9 . X 2

1 . 1 1

-----------

T_{1}0 .T_{3}3

S_{1}S_{2 .}S_{3}6

-------------

X 2 . Z 9We know that X + 1 = T _{3}, so 3*T_{3}= n*10+Z, where n is 0, 1, or 2,

but not higher than 2. Let's deduce that n must be less than 3:

If X=8 (the highest it can be), then T_{3}= 9, and so 3*T_{3}= 27 and n=2.

Let's try n=2: Then 3T_{3}= 20+Z but T_{3}= X+1 so 3(X+1)=3X+3=20+Zso X = (17+Z)/3. For integer values for X and Z less than 4, Z

must be 1 and X=6. Let's check Z=1 and X=6:

We can try n=1: We get 3(X+1)=10+Z and X=(7+Z)/3.

For integer values for X and Z less than 4, Z must

be 2 and X=3. Let's check Z=2 and X=3:

Let's try n=0: We get 3(X+1)=Z and

X=(Z-3)/3. No good, since no value of Z

less than 4 gives a positive integer

value for X.29.32

1.11

---------

30.43

60.86

---------

91.29

No good.+ + So the amount on the check was 62.19.

(SOLVED)+ 19.62

1.11

---------

20.73

41.46

---------

62.19

Good.+

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