Difference between revisions of "1997 AJHSME Problems/Problem 23"
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==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 | + | 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>. | |
| − | <math> | + | 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> |
| − | <math>z= | + | <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=5</math> will give <math>w^2 + x^2 + y^2 = 25</math>. <math>y=4</math> forces <math>x=3</math> and <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. |
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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>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=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. | ||
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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 == | ||
Revision as of 23:10, 31 July 2011
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
Solution
Five digit numbers will have a minimum of 
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 
, as 
.
Trying four digit numbers 
, we have 
with 
will not work, since the other digits must be at least 
, and the sum of the squares would be over 
.
will give 
. 
will work, giving the number 
. No other number with will work, as 
and 
would each have to be greater.
will give 
. 
forces 
and 
, which has a leading zero. Smaller will force all the numbers to the smallest values, and will give a sum of squares that is too small.
can only give the number 
, which does not satisfy the condition of the problem.
Thus, the number in question is , and the product of the digits is 
, giving 
as the answer.
See also
| 1997 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 | ||
