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2025 AMC 10B Problem 23

2025 AMC 10B Problem 23AMC 10 Step by Step

Name: Date: September 13, 2026

Diophantine Equations · difficulty 4 of 5 · about 6 min · position P21-25

A rectangular grid of squares has $141$ rows and $91$ columns. Each square has room for two numbers. Horace and Vera each fill in the grid by putting the numbers from $1$ through $141 \times 91 = 12{,}831$ into the squares. Horace fills the grid horizontally: he puts $1$ through $91$ in order from left to right into row $1$ , puts $92$ through $182$ into row $2$ in order from left to right, and continues similarly through row $141$ . Vera fills the grid vertically: she puts $1$ through $141$ in order from top to bottom into column $1$ , then $142$ through $282$ into column $2$ in order from top to bottom, and continues similarly through column $91$ . How many squares get two copies of the same number?

A)
$7$
B)
$10$
C)
$11$
D)
$12$
E)
$19$

Answer

Key insight

Both fillings are linear in the row and column indices, so setting 91(r-1)+c equal to 141(c-1)+r reduces to 9r = 14c - 5 and hence c = 1 mod 9.

Solution

Index the square in row $r$ and column $c$ , with $1 \le r \le 141$ and $1 \le c \le 91$ .

Horace works in rows of $91$ , so he finishes $r-1$ complete rows and then counts $c$ squares:
$H = 91(r-1) + c .$
Vera works in columns of $141$ , so she finishes $c-1$ complete columns and then counts $r$ squares:
$V = 141(c-1) + r .$

Set $H = V$ :
$91r - 91 + c = 141c - 141 + r \implies 90r - 140c = -50 \implies 9r - 14c = -5,$
after dividing by $10$ . So
$9r = 14c - 5 .$

Reduce modulo $9$ : $14c \equiv 5$ , i.e. $5c \equiv 5 \pmod 9$ , and $5$ is invertible mod $9$ , so $c \equiv 1 \pmod 9$ . Write
$c = 9k + 1 \implies 9r = 14(9k+1) - 5 = 126k + 9 \implies r = 14k + 1 .$

Now impose the size of the grid:
$1 \le 9k+1 \le 91 \implies 0 \le k \le 10, \qquad 1 \le 14k+1 \le 141 \implies 0 \le k \le 10 .$

Both give the same range, so $k = 0, 1, \dots, 10$ : that is $11$ squares. They run from $(r,c) = (1,1)$ , which holds $1$ twice, to $(141, 91)$ , which holds $12{,}831$ twice.

The answer is $\boxed{\textbf{(C)}\ 11}$ .

Why this works

Each filling is an affine function of the pair $(r,c)$ , so "the two numbers agree" is one linear equation in two unknowns — a linear Diophantine equation, not a search. Dividing out the common factor $10$ before reducing modulo $9$ keeps every number small, and the solutions then automatically form an arithmetic progression in both $r$ and $c$ , so the count is just the number of terms that stay inside the grid. In general, for an $m \times n$ grid the equation is $n(r-1) + c = m(c-1) + r$ , i.e. $(n-1)r = (m-1)c - (m-n)$ ; the two corner squares always agree, which is a useful check that the progression has been set up correctly.

The trap

Listing c = 1, 10, 19, ..., 91 and then counting the gaps rather than the terms, which gives the 10 in (B).

Common mistakes

Listing $c = 1, 10, 19, \dots, 91$ and then counting the gaps rather than the terms, which gives the 10 in (B).
Swapping the roles of $91$ and $141$ , for instance writing Horace's entry as $141(r-1)+c$ ; the multiplier is the length of the line being filled.
Solving the congruence and stopping, without checking that $r = 14k+1$ also stays within $141$ rows.

Techniques

Set up the equation/formula and compute; no special trick needed · Substitute to simplify (u = x+1/x, shifting, scaling)