Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution)
 
(3 intermediate revisions by 3 users not shown)
Line 12: Line 12:
 
==Solution==
 
==Solution==
  
−
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.
+
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.
  
 
No digit will be greater than <math>7</math>, as <math>8^2 = 64</math>.
 
No digit will be greater than <math>7</math>, as <math>8^2 = 64</math>.
Line 20: Line 20:
 
<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>.
 
<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>.
  
−
<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 each have to be greater.  
+
<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.  Smaller <math>y</math> will force all the numbers to the smallest values, and <math>(w,x,y) = (1,2,3)</math> will give a sum of squares that is too small.
+
<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.
Line 29: Line 29:
  
 
== See also ==
 
== See also ==
−
{{AJHSME box|year=1997|num-b=21|num-a=23}}
+
{{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 15:15, 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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1997_AJHSME_Problems/Problem_23&oldid=192840"