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Difference between revisions of "1961 AHSME Problems/Problem 35"

(Solution to Problem 35)
 
m (→‎Solution)
 
Line 14: Line 14:
  
 
Since <math>120 < 695 < 720</math>, divide <math>695</math> by <math>120</math>.  The quotient is <math>5</math> and the remainder is <math>95</math>, so rewrite the number as
 
Since <math>120 < 695 < 720</math>, divide <math>695</math> by <math>120</math>.  The quotient is <math>5</math> and the remainder is <math>95</math>, so rewrite the number as
−
<cmath>695 = 5 \cdot 24 + 95</cmath>
+
<cmath>695 = 5 \cdot 120 + 95</cmath>
 
Similarly, dividing <math>95</math> by <math>24</math> results in a quotient of <math>3</math> and a remainder of <math>23</math>, so the number can be rewritten as
 
Similarly, dividing <math>95</math> by <math>24</math> results in a quotient of <math>3</math> and a remainder of <math>23</math>, so the number can be rewritten as
−
<cmath>695 = 5 \cdot 24 + 3 \cdot 24 + 23</cmath>
+
<cmath>695 = 5 \cdot 120 + 3 \cdot 24 + 23</cmath>
 
Repeat the steps to get
 
Repeat the steps to get
−
<cmath>695 = 5 \cdot 24 + 3 \cdot 24 + 3 \cdot 6 + 2 \cdot 2 + 1</cmath>
+
<cmath>695 = 5 \cdot 120 + 3 \cdot 24 + 3 \cdot 6 + 2 \cdot 2 + 1</cmath>
 
The answer is <math>\boxed{\textbf{(D)}}</math>.  One can also stop at the second step by noting <math>23 < 24</math>.
 
The answer is <math>\boxed{\textbf{(D)}}</math>.  One can also stop at the second step by noting <math>23 < 24</math>.
  

Latest revision as of 20:43, 14 September 2023

Problem

The number $695$ is to be written with a factorial base of numeration, that is, $695=a_1+a_2\times2!+a_3\times3!+ \ldots a_n \times n!$ where $a_1, a_2, a_3 ... a_n$ are integers such that $0 \le a_k \le k,$ and $n!$ means $n(n-1)(n-2)...2 \times 1$. Find $a_4$

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4$

Solution

This problem can be approached similarly to other base number problems.

Since $120 < 695 < 720$, divide $695$ by $120$. The quotient is $5$ and the remainder is $95$, so rewrite the number as \[695 = 5 \cdot 120 + 95\] Similarly, dividing $95$ by $24$ results in a quotient of $3$ and a remainder of $23$, so the number can be rewritten as \[695 = 5 \cdot 120 + 3 \cdot 24 + 23\] Repeat the steps to get \[695 = 5 \cdot 120 + 3 \cdot 24 + 3 \cdot 6 + 2 \cdot 2 + 1\] The answer is $\boxed{\textbf{(D)}}$. One can also stop at the second step by noting $23 < 24$.

See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 34 		Followed by
Problem 36
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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