1997 AJHSME Problems/Problem 23

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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 each 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. Smaller $y$ will force all the numbers to the smallest values, and $(w,x,y) = (1,2,3)$ will give a sum of squares that is too small.

$z=4$ 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 (ProblemsAnswer KeyResources)
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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