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

This is a maximise-the-sum cryptarithm: PAY + ME + REAL = MONEY, using nine letters P, A, Y, M, E, R, L, O, N, each a different digit. We want MONEY as big as possible, then read off which digit never gets used.

Start from the outside in. PAY + ME + REAL can add up to at most 999 + 99 + 9999 = 11097, so the leading digit M of MONEY can only be 1, and it comes purely from a carry. Because M = 1, the thousands column (R plus a possible carry from the hundreds column) must total exactly 10, which forces R = 9 and O = 0.

Next, the units column: Y + E + L must end in Y again, which only works if $E+L$ is a clean multiple of 10. With two different nonzero-ish digits, that means $E+L=10$, sending a carry of 1 into the tens column.

• With M = 1, O = 0 and R = 9 already fixed, the digits left for A, E, L, P, N, Y are 2, 3, 4, 5, 6, 7, 8.
• The tens-column rule $2A+2=E+10c$ and the hundreds-column rule $P+E+c=N+10$ (c is the tens-column carry) tie A, E, P and N together.
• Checking the leftover digits for the pair that pushes N (the hundreds digit of MONEY) as high as possible gives A = 3, E = 8, L = 2, P = 7, N = 5, leaving Y = 6.

Putting the letters back in: PAY = 736, ME = 18, REAL = 9832, and $736+18+9832=10586$. That is the largest value MONEY can take under the rules of the puzzle.

The nine letters use the digits 0, 1, 2, 3, 5, 6, 7, 8, 9 between them. Scanning 0 through 9, the digit 4 is the only one that never appears, so it is the digit left unused. Options showing 1, 2 or 3 do not work, since those digits are already used by M, L and A.

It helps to double check by trying to push any single letter higher and seeing it break something else. If we try to raise N above 5 (say N = 6), the hundreds-column rule $P+E+c=N+10$ forces $P+E+c=16$, which needs P and E close to their top values, but the leftover digits after fixing M, O, R do not allow both P and E to be that large while also satisfying the tens-column rule linking A and E. In the same way, trying E = 7 or E = 5 instead of E = 8 changes L to 3 or 5 through $E+L=10$, but then no valid A is left that satisfies $2A+2=E+10c$ with a whole-number, non-repeating result. So the assignment P=7, A=3, Y=6, M=1, E=8, R=9, L=2, O=0, N=5 is not just a good guess, it is the only one that survives every check.

Let's summarize:

• M = 1, O = 0 and R = 9 are forced by the leftmost columns, no matter how the rest of the letters are chosen.
• E + L = 10 is forced by the units column.
• Testing every workable combination of the remaining digits for A, E, P (which fixes L and N through the equations) shows only one assignment reaches the maximum: P=7, A=3, Y=6, M=1, E=8, R=9, L=2, O=0, N=5, giving MONEY = 10586.
• The digits actually used are 0, 1, 2, 3, 5, 6, 7, 8, 9, so 4 is the digit left over.

\[ \boxed{4} \]