Difference between revisions of "1963 AHSME Problems/Problem 13"
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Assume <math>c,d \ge 0</math>, and WLOG, assume <math>a<0</math> and <math>a \le b</math>. This also takes into accout when <math>b</math> is negative. That means | Assume <math>c,d \ge 0</math>, and WLOG, assume <math>a<0</math> and <math>a \le b</math>. This also takes into accout when <math>b</math> is negative. That means | ||
− | <cmath>\frac{1}{2^-a} + 2^b = 3^c + 3^d</cmath> | + | <cmath>\frac{1}{2^{-a}} + 2^b = 3^c + 3^d</cmath> |
− | Multiply both sides by <math>2^-a</math> to get | + | Multiply both sides by <math>2^{-a}</math> to get |
− | <cmath>1 + 2^{-a+b} = 2^-a (3^c + 3^d)</cmath> | + | <cmath>1 + 2^{-a+b} = 2^{-a} (3^c + 3^d)</cmath> |
− | Note that both sides are integers. If <math>a \ne b</math>, then the right side is even while the left side is odd, so equality can not happen. If <math>a = b</math>, then <math>2^-a (3^c + 3^d) = 2</math>, and since <math>a<0</math>, <math>a = -1</math> and <math>3^c + 3^d = 1</math>. No nonnegative value of <math>c</math> and <math>d</math> works, so equality can not happen. Thus, <math>a</math> and <math>b</math> can not be negative when <math>c,d \ge 0</math>. | + | Note that both sides are integers. If <math>a \ne b</math>, then the right side is even while the left side is odd, so equality can not happen. If <math>a = b</math>, then <math>2^{-a} (3^c + 3^d) = 2</math>, and since <math>a<0</math>, <math>a = -1</math> and <math>3^c + 3^d = 1</math>. No nonnegative value of <math>c</math> and <math>d</math> works, so equality can not happen. Thus, <math>a</math> and <math>b</math> can not be negative when <math>c,d \ge 0</math>. |
Assume <math>a,b \ge 0</math>, and WLOG, assume <math>c < 0</math> and <math>c \le d</math>. This also takes into account when <math>d</math> is negative. That means | Assume <math>a,b \ge 0</math>, and WLOG, assume <math>c < 0</math> and <math>c \le d</math>. This also takes into account when <math>d</math> is negative. That means | ||
− | <cmath>2^a + 2^b = \frac{1}{3^-c} + 3^d</cmath> | + | <cmath>2^a + 2^b = \frac{1}{3^{-c}} + 3^d</cmath> |
− | Multiply both sides by <math>3^-c</math> to get | + | Multiply both sides by <math>3^{-c}</math> to get |
− | <cmath>3^-c (2^a + 2^b) = 1 + 3^{d-c}</cmath> | + | <cmath>3^{-c} (2^a + 2^b) = 1 + 3^{d-c}</cmath> |
That makes both sides integers. The left side is congruent to <math>0</math> modulo <math>3</math> while the right side is congruent to <math>1</math> or <math>2</math> modulo <math>3</math>, so equality can not happen. Thus, <math>c</math> and <math>d</math> can not be negative when <math>a,b \ge 0</math>. | That makes both sides integers. The left side is congruent to <math>0</math> modulo <math>3</math> while the right side is congruent to <math>1</math> or <math>2</math> modulo <math>3</math>, so equality can not happen. Thus, <math>c</math> and <math>d</math> can not be negative when <math>a,b \ge 0</math>. | ||
Assume <math>a,c < 0</math>, and WLOG, let <math>a \le b</math> and <math>c \le d</math>. This also takes into account when <math>b</math> or <math>d</math> is negative. That means | Assume <math>a,c < 0</math>, and WLOG, let <math>a \le b</math> and <math>c \le d</math>. This also takes into account when <math>b</math> or <math>d</math> is negative. That means | ||
− | <cmath>\frac{1}{2^-a} + 2^b = \frac{1}{3^-c} + 3^d</cmath> | + | <cmath>\frac{1}{2^{-a}} + 2^b = \frac{1}{3^{-c}} + 3^d</cmath> |
− | Multiply both sides by <math>2^-a \cdot 3^-c</math> to get | + | Multiply both sides by <math>2^{-a} \cdot 3^{-c}</math> to get |
− | <cmath>3^-c (1 + 2^{b-a}) = 2^-a (1 + 3^{d-c})</cmath> | + | <cmath>3^{-c} (1 + 2^{b-a}) = 2^{-a} (1 + 3^{d-c})</cmath> |
That makes both sides integers. The left side is congruent to <math>0</math> modulo <math>3</math> while the right side is congruent to <math>1</math> or <math>2</math> modulo <math>3</math>, so equality can not happen. Thus, <math>a</math> and <math>c</math> can not be negative. | That makes both sides integers. The left side is congruent to <math>0</math> modulo <math>3</math> while the right side is congruent to <math>1</math> or <math>2</math> modulo <math>3</math>, so equality can not happen. Thus, <math>a</math> and <math>c</math> can not be negative. | ||
Revision as of 17:18, 15 July 2018
Problem
If , the number of integers
which can possibly be negative, is, at most:
Solution
Assume , and WLOG, assume
and
. This also takes into accout when
is negative. That means
Multiply both sides by
to get
Note that both sides are integers. If
, then the right side is even while the left side is odd, so equality can not happen. If
, then
, and since
,
and
. No nonnegative value of
and
works, so equality can not happen. Thus,
and
can not be negative when
.
Assume , and WLOG, assume
and
. This also takes into account when
is negative. That means
Multiply both sides by
to get
That makes both sides integers. The left side is congruent to
modulo
while the right side is congruent to
or
modulo
, so equality can not happen. Thus,
and
can not be negative when
.
Assume , and WLOG, let
and
. This also takes into account when
or
is negative. That means
Multiply both sides by
to get
That makes both sides integers. The left side is congruent to
modulo
while the right side is congruent to
or
modulo
, so equality can not happen. Thus,
and
can not be negative.
Putting all the information together, none of can be negative, so the answer is
.
See Also
1963 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
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