Gem 17: The Bank Teller's Predicament Puzzle Solution

	The Bank Teller's Predicament Puzzle
	It 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 Shweiger
	The 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 1	Z W . X Y + 1 . 1 1 -------------- T1T2 .T3T4 + S1S2 .S3S4 -------------- X Y . Z W		This implies that Y cannot be the digit 9, since if it were 9, then T4 would be zero and S4 would be zero as it is twice T4 and most importantly W = T4+S4 would be zero which it cannot be.
	Since Y+1 = T4 and as was stated before, the amount T1T2 .T3T4 is 1/3 of the total amount, then 3T4 = W, 3(Y+1) = W or 3Y+3 = W and thus Y = (W - 3)/3                                             equation 1 We 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 T1 would also be 4 or greater. But then X = 3T1 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 --- T3 S3 --- 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 ----------- T1T2 .T33 S1S2 .S36 ------------- X 2 . Z 9       X 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 T2 = 0.       Z 9 . X 2 1 . 1 1 ----------- T1 0 .T33 S1S2 .S36 ------------- X 2 . Z 9     We know that X + 1 = T3, so 3*T3 = 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 T3 = 9, and so 3*T3 = 27 and n=2. Let's try n=2: Then 3T3 = 20+Z but T3 = X+1 so 3(X+1)=3X+3=20+Z so 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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