1976 AHSME Problems/Problem 15

Problem 15

If $r$ is the remainder when each of the numbers $1059,~1417$ , and $2312$ is divided by $d$ , where $d$ is an integer greater than $1$ , then $d-r$ equals

$\textbf{(A) }1\qquad \textbf{(B) }15\qquad \textbf{(C) }179\qquad \textbf{(D) }d-15\qquad \textbf{(E) }d-1$

Solution

We are given these congruences:

$1059 \equiv r \pmod{d} \qquad$ (i) $1417 \equiv r \pmod{d} \qquad$ (ii) $2312 \equiv r \pmod{d} \qquad$ (iii)

Let's make a new congruence by subtracting (i) from (ii), which results in $358 \equiv 0 \pmod{d}.$ Subtract (ii) from (iii) to get $895 \equiv 0 \pmod{d}.$

Now we know that $358$ and $895$ are both multiples of $d$ . Their prime factorizations are $358=2 \cdot 179$ and $895=5 \cdot 179$ , so their common factor is $179$ , which means $d=179$ .

Plug $d=179$ back into any of the original congruences to get $r=164$ . Then, $d-r=179-164= \boxed{\textbf{(B) }15}$ . ~jiang147369

1976 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1976 AHSME Problems/Problem 15

Problem 15

Solution

See Also