Difference between revisions of "1995 AHSME Problems/Problem 11"
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*For condition (iii), the restriction is put on <math>b</math> and <math>c</math>. The possible ordered pairs of <math>b</math> and <math>c</math> are <math>(3,4)</math>, <math>(3,5)</math>, <math>(3,6)</math>, <math>(4,5), (4,6),</math> and <math>(5,6),</math> and there are <math>6</math> of them. Alternatively, we are picking from the four digits 3, 4, 5, 6, and for every combination of two, there is exactly one way to arrange them in increasing order, so we have <math>\binom{4}{2} = 6</math> choices for <math>b</math> and <math>c</math> when we consider them together. | *For condition (iii), the restriction is put on <math>b</math> and <math>c</math>. The possible ordered pairs of <math>b</math> and <math>c</math> are <math>(3,4)</math>, <math>(3,5)</math>, <math>(3,6)</math>, <math>(4,5), (4,6),</math> and <math>(5,6),</math> and there are <math>6</math> of them. Alternatively, we are picking from the four digits 3, 4, 5, 6, and for every combination of two, there is exactly one way to arrange them in increasing order, so we have <math>\binom{4}{2} = 6</math> choices for <math>b</math> and <math>c</math> when we consider them together. | ||
| − | Multiplying the possibilities for each restriction, <math>2 | + | Multiplying the possibilities for each restriction, <math>2 \cdot 2 \cdot 6=24\Rightarrow \mathrm{(C)}</math>. |
==See also== | ==See also== | ||
Latest revision as of 10:37, 30 March 2023
Problem
How many base 10 four-digit numbers, 
, satisfy all three of the following conditions?
(i) 
(ii) 
is a multiple of 5; (iii) 
.
Solution
- For condition (i), the restriction is put on
; 
if 
, and 
if 
. Therefore, 
. - For condition (ii), the restriction is put on
; it must be a multiple of 
. Therefore, 
. - For condition (iii), the restriction is put on
and 
. The possible ordered pairs of and are 
, 
, 
, 
and 
and there are 
of them. Alternatively, we are picking from the four digits 3, 4, 5, 6, and for every combination of two, there is exactly one way to arrange them in increasing order, so we have 
choices for and when we consider them together.
; 
if 
, and 
if 
. Therefore, 
.
; it must be a multiple of 
. Therefore, 
.
and 
. The possible ordered pairs of 
, 
, 
, 
and 
and there are 
of them. Alternatively, we are picking from the four digits 3, 4, 5, 6, and for every combination of two, there is exactly one way to arrange them in increasing order, so we have 
choices for Multiplying the possibilities for each restriction, 
.
See also
| 1995 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 | ||
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