Question

Directions for questions 63 and 64: Substitute different digits (0 to 9) for different letters in the addition below, so that the addition is correct and it gives the maximum possible value of MONEY.

		P	A	Y
			M	E
	R	E	A	L
M	O	N	E	Y

So the addition reads PAY + ME + REAL = MONEY, using nine different letters: P, A, Y, M, E, R, L, O, N.

63. There are nine letters and ten digits. The digit that remains unutilized is:

• A. 4
• B. 3
• C. 2
• D. 1

Hint:

Work out M, O and R first from the ten-thousands and thousands columns (they are forced), then push the rest as high as possible digit by digit.

Answer 1

The Correct Option is A

Step 1: Set up the addition and place values.
The puzzle stacks three numbers to give MONEY: PAY (a 3-digit number) plus ME (a 2-digit number) plus REAL (a 4-digit number) equals MONEY (a 5-digit number).
So PAY + ME + REAL = MONEY, and the nine letters P, A, Y, M, E, R, L, O, N must all stand for different digits from 0 to 9.

Step 2: Pin down M, O and R using the leftmost column.
Line up the numbers by place value. The leftmost (ten-thousands) column of MONEY holds only the letter M, and this digit can only come from a carry out of the thousands column. A carry out of a column addition can be at most 1, so M = 1.
The thousands column has R (from REAL) on top and O (the result) below, plus a possible carry-in from the hundreds column. Since M = 1 forces a carry of 1 out of this column, we need R + carry-in = 10, and the only way to get 10 is R = 9 with a carry-in of 1. This also fixes O = 0.

Step 3: Use the units column to link E and L.
The units column has Y (from PAY) plus E (from ME) plus L (from REAL), and it must produce Y again in the units place of MONEY, with some carry into the tens column.
\[ Y + E + L = Y + 10c_1 \]
The Y on both sides cancels, so \(E + L = 10c_1\). Since E and L are digits, the only workable choice is \(c_1 = 1\), which gives \(E + L = 10\).

Step 4: Push the hundreds and tens digits as high as possible.
The tens column gives \(2A + M + c_1 = E + 10c_2\), which is \(2A + 2 = E + 10c_2\). The hundreds column gives \(P + E + c_2 = N + 10c_3\), and since \(c_3\) must be 1 (that is what forced R = 9 in Step 2), this is \(P + E + c_2 = N + 10\).
With M = 1, O = 0, R = 9 already used, the digits left for A, E, L, P, N, Y are 2, 3, 4, 5, 6, 7, 8. To make MONEY biggest, we want the hundreds digit N as large as possible first, then E, then Y. Trying the remaining digits in the two equations above shows the best fit is A = 3, E = 8 (so L = 10 - 8 = 2, with the tens carry \(c_2 = 0\)), and then P = 7 gives \(N = 7+8+0-10=5\), leaving Y = 6 as the only digit left over.

Step 5: Check the full addition.
With P=7, A=3, Y=6, M=1, E=8, R=9, L=2, O=0, N=5:
\[ PAY = 736,\qquad ME = 18,\qquad REAL = 9832 \]
\[ 736 + 18 + 9832 = 10586 = MONEY \]
This checks out, and testing the other combinations of remaining digits (in the same two equations) never gives a bigger hundreds or tens digit than this one, so 10586 is the largest value MONEY can take.

Step 6: Find the unused digit.
The nine letters between them use the digits 0, 1, 2, 3, 5, 6, 7, 8, 9. Checking 0 through 9 one by one, only the digit 4 never shows up among the letters. Digits 1, 2 and 3 are not the answer because they are already used, by M, L and A respectively.

Final Answer:
The digit left unused is 4.
\[ \boxed{4} \]