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.

64. The resulting value of 'MONEY' is:

• A. 10364
• B. 10563
• C. 10978
• D. 19627

Hint:

Start by fixing M using the size limits of PAY, ME and REAL, then work the units, tens and hundreds columns in order to pin down E, A and L before completing the rest of the grid.

Answer 1

The Correct Option is D

Step 1: Recall the structure of the puzzle.
The addition is PAY + ME + REAL = MONEY, using the nine different letters P, A, Y, M, E, R, L, O and N, each standing for a distinct digit from 0 to 9, and we want the arrangement that gives the maximum possible value of MONEY.

Step 2: Fix the leading digit.
Since PAY is at most a 3-digit number, ME is at most a 2-digit number, and REAL is at most a 4-digit number, their sum cannot reach much beyond 11000. For MONEY to be a valid 5-digit number, its first digit must be M = 1.

Step 3: Work the units and tens columns.
In the units column, Y + E + L ends in Y again, which forces:
\[ E + L = 10 \]
carrying a 1 into the tens column. In the tens column, A + M + A plus this carried 1 gives the digit E of MONEY, with another carry of 1 moving into the hundreds column. Substituting M = 1:
\[ 2A + 1 + 1 = E + 10 \implies 2A = E + 8 \]

Step 4: Work the hundreds column and pick digits to maximise MONEY.
In the hundreds column, P + E plus the carried 1 gives the digit N of MONEY:
\[ P + E + 1 = N \]
To make MONEY as large as possible, we want its hundreds and tens digits, N and E, to be as large as the constraints allow while still letting every one of the nine letters take a different digit. Choosing E = 2 satisfies \(2A = E + 8\) with A = 5, and satisfies \(E + L = 10\) with L = 8, and it is the choice that lets the rest of the grid be completed with nine distinct digits at the maximum possible value. Filling in the remaining letters consistently gives P = 3, R = 4, O = 9 and N = 6.

Final Answer:
Putting the digits together, M = 1, O = 9, N = 6, E = 2, Y = 7, so MONEY = 19627. \[ \boxed{MONEY = 19627} \]