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Difference between revisions of "1997 AJHSME Problems/Problem 23"

(Created page with "==Problem== There are positive integers that have these properties: * the sum of the squares of their digits is 50, and * each digit is larger than the one to its left. The pr...")
 
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==Solution==
 
==Solution==
  
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Five digit numbers will have a minimum of <math>1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55</math> as the sum of their squares if the five digits are distinct and non-zero.  If there is a zero, it will be forced to the leading point by rule #2.
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Five-digit numbers will have a minimum of <math>1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55</math> as the sum of their squares if the five digits are distinct and non-zero.  If there is a zero, it will be forced to the left by rule #2.
  
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Trying four digit numbers <math>WXYZ</math>, we have <math>w^2 + x^2 + y^2 + z^2 = 50</math> with <math>0 < w < x < y < z</math>
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No digit will be greater than <math>7</math>, as <math>8^2 = 64</math>.
  
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<math>z=7</math> will not work, since the other digits must be <math>1^2 + 2^2 + 3^2 = 14</math>.
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Trying four digit numbers <math>WXYZ</math>, we have <math>w^2 + x^2 + y^2 + z^2 = 50</math> with <math>0 < w < x < y < z < 8</math>
  
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<math>z=6</math> will give <math>w^2 + x^2 + y^2 = 14</math>.  <math>(w,x,y) = (1,2,3)</math> will work, giving the number <math>1236</math>.  If <math>y</math> were bigger, <math>y^2</math> would be <math>16</math>, and if it were smaller, the number would be smaller.
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<math>z=7</math> will not work, since the other digits must be at least <math>1^2 + 2^2 + 3^2 = 14</math>, and the sum of the squares would be over <math>50</math>.
  
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<math>z=5</math> will give <math>w^2 + x^2 + y^2 = 25</math>.  <math>y=4</math> forces <math>x=3</math> and <math>z=0</math>, and smaller <math>y</math> will not work.
+
<math>z=6</math> will give <math>w^2 + x^2 + y^2 = 14</math>.  <math>(w,x,y) = (1,2,3)</math> will work, giving the number <math>1236</math>.  No other number with <math>z=6</math> will work, as <math>w, x, </math> and <math>y</math> would have to be greater.
 +
 
 +
<math>z=5</math> will give <math>w^2 + x^2 + y^2 = 25</math>.  <math>y=4</math> forces <math>x=3</math> and <math>w=0</math>, which has a leading zero, and then we have 345 which is a 3-digit number.  
  
 
<math>z=4</math> can only give the number <math>1234</math>, which does not satisfy the condition of the problem.
 
<math>z=4</math> can only give the number <math>1234</math>, which does not satisfy the condition of the problem.
−
 
−
Five digit numbers will have a minimum of <math>1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55</math> as the sum of their squares if the five digits are distinct and non-zero.
 
  
 
Thus, the number in question is <math>1236</math>, and the product of the digits is <math>36</math>, giving <math>\boxed{C}</math> as the answer.
 
Thus, the number in question is <math>1236</math>, and the product of the digits is <math>36</math>, giving <math>\boxed{C}</math> as the answer.
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== See also ==
 
== See also ==
−
{{AJHSME box|year=1997|num-b=21|num-a=23}}
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{{AJHSME box|year=1997|num-b=22|num-a=24}}
 
* [[AJHSME]]
 
* [[AJHSME]]
 
* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
 +
{{MAA Notice}}

Latest revision as of 14:15, 6 May 2023

Problem

There are positive integers that have these properties:

  • the sum of the squares of their digits is 50, and
  • each digit is larger than the one to its left.

The product of the digits of the largest integer with both properties is

$\text{(A)}\ 7 \qquad \text{(B)}\ 25 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 60$

Solution

Five-digit numbers will have a minimum of $1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55$ as the sum of their squares if the five digits are distinct and non-zero. If there is a zero, it will be forced to the left by rule #2.

No digit will be greater than $7$, as $8^2 = 64$.

Trying four digit numbers $WXYZ$, we have $w^2 + x^2 + y^2 + z^2 = 50$ with $0 < w < x < y < z < 8$

$z=7$ will not work, since the other digits must be at least $1^2 + 2^2 + 3^2 = 14$, and the sum of the squares would be over $50$.

$z=6$ will give $w^2 + x^2 + y^2 = 14$. $(w,x,y) = (1,2,3)$ will work, giving the number $1236$. No other number with $z=6$ will work, as $w, x,$ and $y$ would have to be greater.

$z=5$ will give $w^2 + x^2 + y^2 = 25$. $y=4$ forces $x=3$ and $w=0$, which has a leading zero, and then we have 345 which is a 3-digit number.

$z=4$ can only give the number $1234$, which does not satisfy the condition of the problem.

Thus, the number in question is $1236$, and the product of the digits is $36$, giving $\boxed{C}$ as the answer.

See also

1997 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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
All AJHSME/AMC 8 Problems and Solutions

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