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

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There are positive integers that have these properties:
 
There are positive integers that have these properties:
  
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* the sum of the squares of their digits is 50, and
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*The sum of the squares of their digits is equal to 50
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* each digit is larger than the one to its left.
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*Each digit is larger than the one on it's left
  
 
The product of the digits of the largest integer with both properties is
 
The product of the digits of the largest integer with both properties is

Latest revision as of 11:07, 24 December 2024

Problem

There are positive integers that have these properties:

  • The sum of the squares of their digits is equal to 50
  • Each digit is larger than the one on it's 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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